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Partial Derivatives

2.9 Maximum and Minimum Values

35 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1Past exam · M200 2016D

  1. Some level curves of a function f(x,y)f(x,y) are plotted in the xyxy–plane below.

    Figure from prob_s2.9, line 11

    Figure from prob_s2.9, line 11

    For each of the four statements below, circle the letters of all points in the diagram where the situation applies. For example, if the statement were “These points are on the yy–axis”, you would circle both PP and UU, but none of the other letters. You may assume that a local maximum occurs at point TT.

    (i) ∇f is zeroP Q R S T U(ii) f has a saddle pointP Q R S T U(iii)  the partial derivative fy is positiveP Q R S T U(iv)  the directional derivative of f in the direction <0,−1> isP Q R S T U negative\begin{align*} \text{(i) }& \vnabla f\text{ is zero} & &\text{P Q R S T U} \\ \text{(ii) }& f\text{ has a saddle point} & &\text{P Q R S T U} \\ \text{(iii) }& \text{ the partial derivative }f_y\text{ is positive} & &\text{P Q R S T U} \\ \text{(iv) }& \text{ the directional derivative of }f\text{ in the direction }\llt 0,-1\rgt\text{ is} & &\text{P Q R S T U} \\[-0.07in] & \text{ negative} & & \end{align*}

  2. The diagram below shows three “yy traces” of a graph z=F(x,y)z=F(x,y) plotted on xzxz–axes. (Namely the intersections of the surface z=F(x,y)z=F(x,y) with the three planes (y=1.9y=1.9, y=2y=2, y=2.1y=2.1). For each statement below, circle the correct word.

    (i)  the first order partial derivative Fx(1,2) ispositive/negative/zero (circle one)(ii) F has a critical point at (2,2)true/false (circle one)(iii)  the second order partial derivative Fxy(1,2) ispositive/negative/zero (circle one)\begin{align*} \text{(i) }& \text{ the first order partial derivative }F_x(1,2)\text{ is} & &\text{positive/negative/zero (circle one)} \\ \text{(ii) }& F\text{ has a critical point at }(2,2) & &\text{true/false (circle one)} \\ \text{(iii) }& \text{ the second order partial derivative }F_{xy}(1,2)\text{ is} & &\text{positive/negative/zero (circle one)} \end{align*}

    Figure from prob_s2.9, line 11

    Figure from prob_s2.9, line 11

Answer

(a) (i) TT, UU

(a) (ii) UU

(a) (iii) SS

(a) (iv) SS

(b) (i) Fx(1,2)>0F_x(1,2)>0

(b) (ii) FF does not have a critical point at (2,2)(2,2).

(b) (iii) Fxy(1,2)<0F_{xy}(1,2)<0

Full solution

a) (i) ∇f\vnabla f is zero at critical points. The point TT is a local maximum and the point UU is a saddle point. The remaining points PP, RR, SS, are not critical points.

(a) (ii) Only UU is a saddle point.

(a) (iii) We have fy(x,y)>0f_y(x,y)>0 if ff increases as you move vertically upward through (x,y)(x,y). Looking at the diagram, we see

fy(P)<0fy(Q)<0fy(R)=0fy(S)>0fy(T)=0fy(U)=0\begin{align*} f_y(P) <0\qquad f_y(Q) <0\qquad f_y(R) =0\qquad f_y(S) >0\qquad f_y(T) =0\qquad f_y(U) =0 \end{align*}

So only SS works.

(a) (iv) The directional derivative of ff in the direction <0,−1>\llt 0,-1\rgt is ∇f⋅<0,−1>=−fy\vnabla f\cdot \llt 0,-1\rgt =-f_y. It is negative if and only if fy>0f_y>0. So, again, only SS works.

(b) (i) The function z=F(x,2)z=F(x,2) is increasing at x=1x=1, because the y=2.0y=2.0 graph in the diagram has positive slope at x=1x=1. So Fx(1,2)>0F_x(1,2)>0.

(b) (ii) The function z=F(x,2)z=F(x,2) is also increasing (though slowly) at x=2x=2, because the y=2.0y=2.0 graph in the diagram has positive slope at x=2x=2. So Fx(2,2)>0F_x(2,2)>0. So FF does not have a critical point at (2,2)(2,2).

(b) (iii) From the diagram the looks like Fx(1,1.9)>Fx(1,2.0)>Fx(1,2.1)F_x(1,1.9) > F_x(1,2.0) > F_x(1,2.1). That is, it looks like the slope of the y=1.9y=1.9 graph at x=1x=1 is larger than the slope of the y=2.0y=2.0 graph at x=1x=1, which in turn is larger than the slope of the y=2.1y=2.1 graph at x=1x=1. So it looks like Fx(1,y)F_x(1,y) decreases as yy increases through y=2y=2, and consequently Fxy(1,2)<0F_{xy}(1,2)<0.

Q2Stage 1

Find the high and low points of the surface z=x2+y2z=\sqrt{x^2+y^2} with (x,y)(x,y) varying over the square ∣x∣≤1|x|\le 1, ∣y∣≤1|y|\le 1. Discuss the values of zx, zyz_x,\ z_y there. Do not evaluate any derivatives in answering this question.

Hint

Interpret the height x2+y2\sqrt{x^2+y^2} geometrically.

Answer

The minimum height is zero at (0,0,0)(0,0,0). The derivatives zxz_x and zyz_y do not exist there. The maximum height is 2\sqrt{2} at (±1,±1,2)(\pm 1,\pm 1,\sqrt{2}). There zxz_x and zyz_y exist but are not zero — those points would not be the highest points if it were not for the restriction ∣x∣,∣y∣≤1|x|,|y|\le 1.

Full solution

The height x2+y2\sqrt{x^2+y^2} at (x,y)(x,y) is the distance from (x,y)(x,y) to (0,0)(0,0). So the minimum height is zero at (0,0,0)(0,0,0). The surface is a cone. The cone has a point at (0,0,0)(0,0,0) and the derivatives zxz_x and zyz_y do not exist there. The maximum height is achieved when (x,y)(x,y) is as far as possible from (0,0)(0,0). The highest points are at (±1,±1,2)(\pm 1,\pm 1,\sqrt{2}). There zxz_x and zyz_y exist but are not zero. These points would not be the highest points if it were not for the restriction ∣x∣,∣y∣≤1|x|,|y|\le 1.

Q3Stage 1

If t0t_0 is a local minimum or maximum of the smooth function f(t)f(t) of one variable (tt runs over all real numbers) then f′(t0)=0.f'(t_0)=0. Derive an analogous necessary condition for x0\vx_0 to be a local minimum or maximium of the smooth function g(x)g(\vx) restricted to points on the line x=a+td .\vx=\va+t\vd\ . The test should involve the gradient of g(x)g(\vx).

Hint

Define f(t)=g(a+td)f(t)=g(\va+t\vd).

Answer

∇g(x0)⋅d=0\vnabla g(\vx_0)\cdot\vd=0, i.e. ∇g(x0)⊥d\vnabla g(\vx_0)\perp\vd, and x0=a+t0d\vx_0=\va+t_0\vd for some t0t_0. The second condition is to ensure that x0x_0 lies on the line.

Full solution

Define f(t)=g(a+td)f(t)=g(\va+t\vd) and determine t0t_0 by x0=a+t0d\vx_0=\va+t_0\vd. Then f′(t)=∇g(a+td)⋅df'(t)=\vnabla g(\va+t\vd)\cdot\vd. To see this, write a=<a1,a2,a3>\va=\llt a_1,a_2,a_3\rgt and d=<d1,d2,d3>\vd=\llt d_1,d_2,d_3\rgt. Then

f(t)=g(a1+td1,a2+td2,a3+td3)\begin{equation*} f(t)=g(a_1+td_1,a_2+td_2,a_3+td_3) \end{equation*}

So, by the chain rule,

f′(t)=∂g∂x(a1+td1,a2+td2,a3+td3) d1+∂g∂y(a1+td1,a2+td2,a3+td3) d2+∂g∂z(a1+td1,a2+td2,a3+td3) d3=∇g(a+td)⋅d\begin{align*} f'(t) &= \pdiff{g}{x}(a_1+td_1,a_2+td_2,a_3+td_3)\,d_1 +\pdiff{g}{y}(a_1+td_1,a_2+td_2,a_3+td_3)\,d_2 \\&\hskip2in +\pdiff{g}{z}(a_1+td_1,a_2+td_2,a_3+td_3)\,d_3 \\ &=\vnabla g(\va+t\vd)\cdot\vd \end{align*}

Then x0\vx_0 is a local max or min of the restriction of gg to the specified line if and only if t0t_0 is a local max or min of f(t)f(t). If so, f′(t0)f'(t_0) necessarily vanishes. So if x0\vx_0 is a local max or min of the restriction of gg to the specified line, then ∇g(x0)⋅d=0\vnabla g(\vx_0)\cdot\vd=0, i.e. ∇g(x0)⊥d\vnabla g(\vx_0)\perp\vd, and x0=a+t0d\vx_0=\va+t_0\vd for some t0t_0. The second condition is to ensure that x0x_0 lies on the line.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q4Stage 2Past exam · M200 2005D

Let z=f(x,y)=(y2−x2)2z = f(x,y) = {(y^2 - x^2)}^2.

  1. Make a reasonably accurate sketch of the level curves in the xyxy–plane of z=f(x,y)z = f(x,y) for z=0z = 0, 11 and 1616. Be sure to show the units on the coordinate axes.

  2. Verify that (0,0)(0,0) is a critical point for z=f(x,y)z = f(x,y), and determine from part (a) or directly from the formula for f(x,y)f(x,y) whether (0,0)(0, 0) is a local minimum, a local maximum or a saddle point.

  3. Can you use the Second Derivative Test to determine whether the critical point (0,0)(0, 0) is a local minimum, a local maximum or a saddle point? Give reasons for your answer.

Hint

Write down the equations of specified level curves.

Answer

(a)

Figure from prob_s2.9, line 228

Figure from prob_s2.9, line 228

(b) (0,0)(0,0) is a local (and also absolute) minimum.

(c) No. See the solutions.

Full solution

(a)

  • The level curve z=0z=0 is y2−x2=0y^2-x^2=0, which is the pair of 45∘45^\circ lines y=±xy=\pm x.

  • When C>0C>0, the level curve
    z=C4z=C^4 is (y2−x2)2=C4{(y^2-x^2)}^2=C^4, which is the pair of hyperbolae y2−x2=C2y^2-x^2=C^2, y2−x2=−C2y^2-x^2=-C^2 or

    y=±x2+C2x=±y2+C2\begin{equation*} y=\pm\sqrt{x^2+C^2}\qquad x=\pm\sqrt{y^2+C^2} \end{equation*}

    The hyperbola y2−x2=C2y^2-x^2=C^2 crosses the yy–axis (i.e. the line x=0x=0) at (0,±C)(0,\pm C). The hyperbola y2−x2=−C2y^2-x^2=-C^2 crosses the xx–axis (i.e. the line y=0y=0) at (±C,0)(\pm C,0).

Here is a sketch showing the level curves z=0z=0, z=1z=1 (i.e. C=1C=1), and z=16z=16 (i.e. C=2C=2).

Figure from prob_s2.9, line 239

Figure from prob_s2.9, line 239

(b) As fx(x,y)=−4x(y2−x2)f_x(x,y)=-4x(y^2-x^2) and fy(x,y)=4y(y2−x2)f_y(x,y)=4y(y^2-x^2), we have fx(0,0)=fy(0,0)=0f_x(0,0) = f_y(0,0) = 0 so that (0,0)(0,0) is a critical point. Note that

  • f(0,0)=0f(0,0)=0,

  • f(x,y)≥0f(x,y)\ge 0 for all xx and yy.

So (0,0)(0,0) is a local (and also absolute) minimum.

(c) Note that

fxx(x,y)=−4y2+12x2fxx(x,y)=0fyy(x,y)=12y2−4x2fyy(x,y)=0fxy(x,y)=−8xyfxx(x,y)=0\begin{alignat*}{3} f_{xx}(x,y) &=-4y^2+12x^2\qquad& f_{xx}(x,y)&=0 \\ f_{yy}(x,y) &= 12y^2-4x^2\qquad& f_{yy}(x,y)&=0 \\ f_{xy}(x,y) &=-8xy\qquad& f_{xx}(x,y)&=0 \\ \end{alignat*}

As fxx(0,0)fyy(0,0)−fxy(0,0)2=0f_{xx}(0,0)f_{yy}(0,0)-f_{xy}(0,0)^2=0, the Second Derivative Test (Theorem 2.9.16 in the CLP-3 text) tells us absolutely nothing.

Q5Stage 2Past exam · M200 2006A

Use the Second Derivative Test to find all values of the constant cc for which the function z=x2+cxy+y2z = x^2 + cxy + y^2 has a saddle point at (0,0)(0,0).

Answer

∣c∣>2|c|>2

Full solution

Write f(x,y)=x2+cxy+y2f(x,y) = x^2 +cxy +y^2. Then

fx(x,y)=2x+cyfx(0,0)=0fy(x,y)=cx+2yfy(0,0)=0fxx(x,y)=2fxy(x,y)=cfyy(x,y)=2\begin{align*} f_x(x,y)&= 2x+cy & f_x(0,0) = 0 \\ f_y(x,y)&= cx+2y & f_y(0,0) = 0 \\ f_{xx}(x,y) &= 2 \\ f_{xy}(x,y) &= c \\ f_{yy}(x,y) &= 2 \end{align*}

As fx(0,0)=fy(0,0)=0f_x(0,0)=f_y(0,0)=0, we have that (0,0)(0,0) is always a critical point for ff. According to the Second Derivative Test, (0,0)(0,0) is also a saddle point for ff if

fxx(0,0)fyy(0,0)−fxy(0,0)2<0  ⟺  4−c2<0  ⟺  ∣c∣>2\begin{align*} f_{xx}(0,0) f_{yy}(0,0) - f_{xy}(0,0)^2 <0 \iff 4-c^2 <0 \iff |c|>2 \end{align*}

As a remark, the Second Derivative Test provides no information when the expression fxx(0,0)fyy(0,0)−fxy(0,0)2=0f_{xx}(0,0) f_{yy}(0,0) - f_{xy}(0,0)^2 =0, i.e. when c=±2c=\pm 2. But when c=±2c=\pm 2,

f(x,y)=x2±2xy+y2=(x±y)2\begin{equation*} f(x,y) = x^2 \pm 2xy + y^2 =(x\pm y)^2 \end{equation*}

and ff has a local minimum, not a saddle point, at (0,0)(0,0).

Q6Stage 2Past exam · M200 2006D

Find and classify all critical points of the function

f(x,y)=x3−y3−2xy+6.\begin{equation*} f(x, y) = x^3 - y^3 - 2xy + 6. \end{equation*}
Answer
criticalpoint\Atop{\text{critical}}{\text{point}}type
(0,0)(0,0)saddle point
(−23,23)\left(-\frac{2}{3},\frac{2}{3}\right)local max
Full solution

To find the critical points we will need the gradient of ff, and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

f=x3−y3−2xy+6fx=3x2−2yfxx=6xfxy=−2fy=−3y2−2xfyy=−6yfyx=−2\begin{alignat*}{3} f&=x^3 - y^3 - 2xy + 6 \\ f_x&=3x^2-2y & f_{xx}&=6x \qquad & f_{xy}&= -2\\ f_y&=-3y^2-2x \qquad & f_{yy}&=-6y\qquad & f_{yx}&= -2 \end{alignat*}

The critical points are the solutions of

fx=3x2−2y=0fy=−3y2−2x=0\begin{equation*} f_x=3x^2-2y=0 \qquad f_y=-3y^2-2x = 0 \end{equation*}

Substituting y=32x2y=\frac{3}{2}x^2, from the first equation, into the second equation gives

−3(32x2)2−2x=0  ⟺  −2x(3323x3+1)=0  ⟺  x=0, −23\begin{align*} -3\left(\frac{3}{2}x^2\right)^2-2x =0 &\iff -2x\left(\frac{3^3}{2^3}x^3+1\right)=0 \\ &\iff x=0,\ -\frac{2}{3} \end{align*}

So there are two critical points: (0,0)(0,0), (−23,23)\left(-\frac{2}{3},\frac{2}{3}\right).

The classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)0×0−(−2)2<00\times 0 -(-2)^2 < 0saddle point
(−23,23)\left(-\frac{2}{3},\frac{2}{3}\right)(−4)×(−4)−(−2)2>0(-4)\times (-4)-(-2)^2>0−4-4local max
Q7Stage 2Past exam · M200 2007A

Find all critical points for f(x,y)=x(x2+xy+y2−9)f(x,y) = x(x^2 + xy + y^2 - 9). Also find out which of these points give local maximum values for f(x,y)f(x,y), which give local minimum values, and which give saddle points.

Answer
criticalpoint\Atop{\text{critical}}{\text{point}}type
(0,3)(0,3)saddle point
(0,−3)(0,-3)saddle point
(−2,1)(-2,1)local max
(2,−1)(2,-1)local min
Full solution

To find the critical points we will need the gradient of ff, and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

f=x3+x2y+xy2−9xfx=3x2+2xy+y2−9fxx=6x+2yfxy=2x+2yfy=x2+2xyfyy=2xfyx=2x+2y\begin{alignat*}{3} f&=x^3 + x^2y + xy^2 - 9x \\ f_x&=3x^2+2xy+y^2-9 \qquad & f_{xx}&=6x+2y \qquad & f_{xy}&= 2x+2y\\ f_y&=x^2+2xy & f_{yy}&=2x\qquad & f_{yx}&= 2x+2y \end{alignat*}

(Of course, fxyf_{xy} and fyxf_{yx} have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

fx=3x2+2xy+y2−9=0fy=x(x+2y)=0\begin{alignat*}{3} f_x&=3x^2+2xy+y^2-9&&=0 \tag{E1} \\ f_y&=x(x+2y) &&= 0 \tag{E2} \end{alignat*}

Equation (E2) is satisfied if at least one of x=0x=0, x=−2yx=-2y.

  • If x=0x=0, equation (E1) reduces to y2−9=0y^2-9=0, which is satisfied if y=±3y=\pm 3.

  • If x=−2yx=-2y, equation (E1) reduces to

    0=3(−2y)2+2(−2y)y+y2−9=9y2−9\begin{align*} 0=3(-2y)^2+2(-2y)y+y^2-9=9y^2-9 \end{align*}

    which is satisfied if y=±1y=\pm 1.

So there are four critical points: (0,3)(0,3), (0,−3)(0,-3), (−2,1)(-2,1) and (2,−1)(2,-1). The classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,3)(0,3)(6)×(0)−(6)2<0(6)\times (0)-(6)^2< 0saddle point
(0,−3)(0,-3)(−6)×(0)−(−6)2<0(-6)\times (0)-(-6)^2<0saddle point
(−2,1)(-2,1)(−10)×(−4)−(−2)2>0(-10)\times (-4)-(-2)^2>0−10-10local max
(2,−1)(2,-1)(10)×(4)−(2)2>0(10)\times (4)-(2)^2>01010local min
Q8Stage 2Past exam · M200 2007A

Find the largest and smallest values of x2y2zx^2 y^2 z in the part of the plane 2x+y+z=52x + y + z = 5 where x≥0x \ge 0, y≥0y \ge 0 and z≥0z \ge 0. Also find all points where those extreme values occur.

Answer

The minimum value is 00 on

{ (x,y,z) ∣ x≥0, y≥0, z≥0, 2x+y+z=5, at least one of x,y,z zero }\begin{equation*} \Set{(x,y,z)}{x\ge 0,\ y\ge 0,\ z\ge 0,\ 2x+y+z=5,\ \text{at least one of }x,y,z\text{ zero}} \end{equation*}

The maximum value is 44 at (1,2,1)(1,2,1).

Full solution

The region of interest is

D={ (x,y,z) ∣ x≥0, y≥0, z≥0, 2x+y+z=5 }\begin{equation*} D=\Set{(x,y,z)}{x\ge 0,\ y\ge 0,\ z\ge 0,\ 2x+y+z=5} \end{equation*}

First observe that, on the boundary of this region, at least one of xx, yy and zz is zero. So f(x,y,z)=x2y2zf(x,y,z)=x^2 y^2 z is zero on the boundary. As ff takes values which are strictly bigger than zero at all points of DD that are not on the boundary, the minimum value of ff is 00 on

∂D={ (x,y,z) ∣ x≥0, y≥0, z≥0, 2x+y+z=5, at least one of x,y,z zero }\begin{equation*} \partial D = \Set{(x,y,z)}{x\ge 0,\ y\ge 0,\ z\ge 0,\ 2x+y+z=5,\ \text{at least one of }x,y,z\text{ zero}} \end{equation*}

The maximum value of ff will be taken at a critical point. On DD

f=x2y2(5−2x−y)=5x2y2−2x3y2−x2y3\begin{align*} f &= x^2 y^2 (5-2x-y) =5x^2 y^2 - 2x^3 y^2 -x^2 y^3 \end{align*}

So the critical points are the solutions of

0=fx(x,y)=10xy2−6x2y2−2xy30=fy(x,y)=10x2y−4x3y−3x2y2\begin{align*} 0&=f_x(x,y) = 10xy^2 -6x^2y^2 -2xy^3 \\ 0&=f_y(x,y) = 10x^2y -4x^3y -3x^2y^2 \end{align*}

or, dividing by the first equation by xy2xy^2 and the second equation by x2yx^2y, (recall that x,y≠0x,y\ne 0)

10−6x−2y=0or3x+y=510−4x−3y=0or4x+3y=10\begin{alignat*}{3} 10 -6x -2y &=0 &\qquad\text{or}\qquad 3x+y&=5 \\ 10 -4x -3y &=0 &\qquad\text{or}\qquad 4x+3y&=10 \end{alignat*}

Substituting y=5−3xy=5-3x, from the first equation, into the second equation gives

4x+3(5−3x)=10  ⟹  −5x+15=10  ⟹  x=1, y=5−3(1)=2\begin{equation*} 4x+3(5-3x)=10 \implies -5x +15 =10 \implies x=1,\ y=5-3(1)=2 \end{equation*}

So the maximum value of ff is (1)2(2)2(5−2−2)=4(1)^2(2)^2(5-2-2)=4 at (1,2,1)(1,2,1).

Q9Stage 2

Find and classify all the critical points of f(x,y)=x2+y2+x2y+4f(x,y)=x^2+y^2+x^2y+4.

Answer
criticalpoint\Atop{\text{critical}}{\text{point}}type
(0,0)(0,0)local min
(2,−1)(\sqrt{2},-1)saddle point
(−2,−1)(-\sqrt{2},-1)saddle point
Full solution

To find the critical points we will need the gradient of ff, and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

f=x2+y2+x2y+4fx=2x+2xyfxx=2+2yfxy=2xfy=2y+x2fyy=2\begin{alignat*}{5} f&=x^2+y^2+x^2y+4 \\ f_x&=2x+2xy\qquad & f_{xx}&=2+2y\qquad & f_{xy}&= 2x\\ f_y&=2y+x^2 & f_{yy}&=2 \end{alignat*}

The critical points are the solutions of

fx=0fy=0  ⟺  2x(1+y)=02y+x2=0  ⟺  x=0 or y=−12y+x2=0\begin{alignat*}{5} & & &f_x=0 & &f_y=0 \\ &\iff\quad & &2x(1+y)=0 & &2y+x^2=0 \\ &\iff & &x=0\text{ or }y=-1\qquad & &2y+x^2=0 \end{alignat*}

When x=0x=0, yy must be 00. When y=−1y=-1, x2x^2 must be 22. So, there are three critical points: (0,0)(0,0), (±2,−1)\big(\pm\sqrt{2},-1\big).

The classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)2×2−02>02\times 2-0^2>02>02>0local min
(2,−1)(\sqrt{2},-1)0×2−(22)2<00\times 2-(2\sqrt{2})^2<0saddle point
(−2,−1)(-\sqrt{2},-1)0×2−(−22)2<00\times 2-(-2\sqrt{2})^2<0saddle point
Q10Stage 2Past exam · M200 2008D

Find all saddle points, local minima and local maxima of the function

f(x,y)=x3+x2−2xy+y2−x.\begin{equation*} f(x,y) = x^3 + x^2 - 2xy + y^2 - x. \end{equation*}
Answer
criticalpoint\Atop{\text{critical}}{\text{point}}type
(13,13)\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)local min
−(13,13)-\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)saddle point
Full solution

To find the critical points we will need the gradient of ff, and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

f=x3+x2−2xy+y2−xfx=3x2+2x−2y−1fxx=6x+2fxy=−2fy=−2x+2yfyy=2fyx=−2\begin{alignat*}{3} f&=x^3 + x^2 - 2xy + y^2 - x \\ f_x&=3x^2+2x-2y-1 \qquad & f_{xx}&=6x+2 \qquad & f_{xy}&= -2\\ f_y&=-2x+2y & f_{yy}&=2\qquad & f_{yx}&= -2 \end{alignat*}

(Of course, fxyf_{xy} and fyxf_{yx} have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

fx=3x2+2x−2y−1=0fy=−2x+2y=0\begin{alignat*}{3} f_x&=3x^2+2x-2y-1&&=0 \tag{E1} \\ f_y&=-2x+2y &&= 0 \tag{E2} \end{alignat*}

Substituting y=xy=x, from (E2), into (E1) gives

3x2−1=0  ⟺  x=±13=0\begin{align*} 3x^2-1=0 \iff x=\pm\frac{1}{\sqrt{3}}=0 \end{align*}

So there are two critical points: ±(13,13)\pm\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big).

The classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(13,13)\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)(23+2)×(2)−(−2)2>0(2\sqrt{3}+2)\times (2)-(-2)^2> 023+2>02\sqrt{3}+2>0local min
−(13,13)-\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big)(−23+2)×(2)−(−2)2<0(-2\sqrt{3}+2)\times (2)-(-2)^2<0saddle point
Q11Stage 2Past exam · M200 2009D

For the surface

z=f(x,y)=x3+xy2−3x2−4y2+4\begin{equation*} z = f (x, y) = x^3 + xy^2 - 3x^2 - 4y^2 + 4 \end{equation*}

Find and classify [as local maxima, local minima, or saddle points] all critical points of f(x,y)f(x,y).

Answer
criticalpoint\Atop{\text{critical}}{\text{point}}type
(0,0)(0,0)local max
(2,0)(2,0)saddle point
Full solution

To find the critical points we will need the gradient of ff and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

f=x3+xy2−3x2−4y2+4fx=3x2+y2−6xfxx=6x−6fxy=2yfy=2xy−8yfyy=2x−8fyx=2y\begin{alignat*}{3} f&=x^3 + xy^2 - 3x^2 - 4y^2 + 4 \\ f_x&=3x^2+y^2-6x\qquad & f_{xx}&=6x-6 \qquad & f_{xy}&= 2y\\ f_y&=2xy -8y & f_{yy}&=2x-8\qquad & f_{yx}&= 2y \end{alignat*}

(Of course, fxyf_{xy} and fyxf_{yx} have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

fx=3x2+y2−6x=0fy=2(x−4)y=0\begin{equation*} f_x=3x^2+y^2-6x=0 \qquad f_y=2(x-4)y = 0 \end{equation*}

The second equation is satisfied if at least one of x=4x=4, y=0y=0 are satisfied.

  • If x=4x=4, the first equation reduces to y2=−24y^2=-24, which has no real solutions.

  • If y=0y=0, the first equation reduces to 3x(x−2)=03x(x-2)=0, which is satisfied if either x=0x=0 or x=2x=2.

So there are two critical points: (0,0)(0,0), (2,0)(2,0).

The classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)(−6)×(−8)−(0)2>0(-6)\times(-8)-(0)^2> 0−6-6local max
(2,0)(2,0)6×(−4)−(0)2<06\times(-4)-(0)^2<0saddle point
Q12Stage 2

Find the maximum and minimum values of f(x,y)=xy−x3y2f(x,y)=xy-x^3y^2 when (x,y)(x,y) runs over the square 0≤x≤10\le x\le 1, 0≤y≤10\le y\le 1.

Answer

min=0max=233≈0.385\text{min}=0\qquad \text{max}=\frac{2}{3\sqrt{3}}\approx0.385

Full solution

The specified function and its first order derivatives are

f(x,y)=xy−x3y2fx(x,y)=y−3x2y2fy(x,y)=x−2x3y\begin{align*} f(x,y)=xy-x^3y^2\qquad f_x(x,y)=y-3x^2y^2\qquad f_y(x,y)=x-2x^3y \end{align*}

  • First, we find the critical points.

    fx=0  ⟺  y(1−3x2y)=0  ⟺  y=0 or 3x2y=1fy=0  ⟺  x(1−2x2y)=0  ⟺  x=0 or 2x2y=1\begin{alignat*}{5} f_x&=0 & &\quad\iff\quad &y(1-3x^2y)&=0 & &\quad\iff\quad & &y=0 \text{ or } 3x^2y=1 \\ f_y&=0 & &\quad\iff\quad &x(1-2x^2y)&=0 & &\quad\iff\quad & &x=0 \text{ or } 2x^2y=1 \end{alignat*}

    • If y=0y=0, we cannot have 2x2y=12x^2y=1, so we must have x=0x=0.

    • If 3x2y=13x^2y=1, we cannot have x=0x=0, so we must have 2x2y=12x^2y=1. Dividing gives 1=3x2y2x2y=321=\frac{3x^2y}{2x^2y}=\frac{3}{2} which is impossible.

    So the only critical point in the square is (0,0)(0,0). There f=0f=0.

  • Next, we look at the part of the boundary with x=0x=0. There f=0f=0.

  • Next, we look at the part of the boundary with y=0y=0. There f=0f=0.

  • Next, we look at the part of the boundary with x=1x=1. There f=y−y2f=y-y^2. As ddy(y−y2)=1−2y\diff{}{y}(y-y^2)=1-2y, the max and min of y−y2y-y^2 for 0≤y≤10\le y\le 1 must occur either at y=0y=0, where f=0f=0, or at y=12y=\half, where f=14f=\frac{1}{4}, or at y=1y=1, where f=0f=0.

  • Next, we look at the part of the boundary with y=1y=1. There f=x−x3f=x-x^3. As ddx(x−x3)=1−3x2\diff{}{x}(x-x^3)=1-3x^2, the max and min of x−x3x-x^3 for 0≤x≤10\le x\le 1 must occur either at x=0x=0, where f=0f=0, or at x=13x=\frac{1}{\sqrt{3}}, where f=233f=\frac{2}{3\sqrt{3}}, or at x=1x=1, where f=0f=0.

All together, we have the following candidates for max and min.

point(0,0)(0,0)x=0x=0y=0y=0(1,0)(1,0)(1,12)(1,\half)(1,1)(1,1)(0,1)(0,1)(13,1)(\frac{1}{\sqrt{3}},1)(1,1)(1,1)
value of ff0000000014\frac{1}{4}0000233\frac{2}{3\sqrt{3}}00
minminminminminminmaxmin

The largest and smallest values of ff in this table are

min=0max=233≈0.385\begin{equation*} \text{min}=0\qquad \text{max}=\frac{2}{3\sqrt{3}}\approx0.385 \end{equation*}
Q13Stage 2

The temperature at all points in the disc x2+y2≤1x^2+y^2\le 1 is given by T(x,y)=(x+y)e−x2−y2T(x,y)=(x+y)e^{-x^2-y^2}. Find the maximum and minimum temperatures at points of the disc.

Hint

One way to deal with the boundary x2+y2=1x^2+y^2=1 is to parametrize it by x=cos⁡θx=\cos\theta, y=sin⁡θy=\sin\theta, 0≤θ<2π0\le\theta<2\pi.

Answer

min=−1emax=1e\text{min}=-\frac{1}{\sqrt{e}}\qquad \text{max}=\frac{1}{\sqrt{e}}

Full solution

The specified temperature and its first order derivatives are

T(x,y)=(x+y)e−x2−y2Tx(x,y)=(1−2x2−2xy)e−x2−y2Ty(x,y)=(1−2xy−2y2)e−x2−y2\begin{align*} T(x,y)&=(x+y)e^{-x^2-y^2}\\ T_x(x,y)&=(1-2x^2-2xy)e^{-x^2-y^2}\\ T_y(x,y)&=(1-2xy-2y^2)e^{-x^2-y^2} \end{align*}

  • First, we find the critical points.

    Tx=0  ⟺  2x(x+y)=1Ty=0  ⟺  2y(x+y)=1\begin{alignat*}{5} T_x&=0 & &\quad\iff\quad & 2x(x+y)&=1 \\ T_y&=0 & &\quad\iff\quad & 2y(x+y)&=1 \end{alignat*}

    As x+yx+y may not vanish, this forces x=yx=y and then x=y=±12x=y=\pm\half. So the only critical points are (12,12)(\half,\half) and (−12,−12)(-\half,-\half).

  • On the boundary x=cos⁡θx=\cos\theta and y=sin⁡θy=\sin\theta, so T=(cos⁡θ+sin⁡θ)e−1T=(\cos\theta+\sin\theta)e^{-1}. This is a periodic function and so takes its max and min at zeroes of dTdθ=(−sin⁡θ+cos⁡θ)e−1\diff{T}{\theta}=\big(-\sin\theta+\cos\theta\big)e^{-1}. That is, when sin⁡θ=cos⁡θ\sin\theta=\cos\theta, which forces sin⁡θ=cos⁡θ=±12\sin\theta=\cos\theta=\pm\frac{1}{\sqrt{2}}.

All together, we have the following candidates for max and min.

point(12,12)(\half,\half)(−12,−12)(-\half,-\half)(12,12)(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})(−12,−12)(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})
value of TT1e≈0.61\frac{1}{\sqrt{e}}\approx0.61−1e-\frac{1}{\sqrt{e}}2e≈0.52\frac{\sqrt{2}}{e}\approx0.52−2e-\frac{\sqrt{2}}{e}
maxmin

The largest and smallest values of TT in this table are

min=−1emax=1e\begin{equation*} \text{min}=-\frac{1}{\sqrt{e}}\qquad \text{max}=\frac{1}{\sqrt{e}} \end{equation*}
Q14Stage 2Past exam · M200 2010A

The images below depict level sets f(x,y)=cf (x, y) = c of the functions in the list at heights c=0,0.1,0.2,…,1.9,2c = 0, 0.1, 0.2, \ldots , 1.9, 2. Label the pictures with the corresponding function and mark the critical points in each picture. (Note that in some cases, the critical points might not be drawn on the images already. In those cases you should add them to the picture.)

  1. f(x,y)=(x2+y2−1)(x−y)+1f(x, y) = (x^2 + y^2 - 1)(x - y) + 1

  2. f(x,y)=x2+y2f(x, y) = \sqrt{x^2 + y^2}

  3. f(x,y)=y(x+y)(x−y)+1f(x, y) = y(x + y)(x - y) + 1

  4. f(x,y)=x2+y2f(x, y) = x^2 + y^2

Figure from prob_s2.9, line 931

Figure from prob_s2.9, line 931

Figure from prob_s2.9, line 931

Figure from prob_s2.9, line 931

Figure from prob_s2.9, line 931

Figure from prob_s2.9, line 931

Figure from prob_s2.9, line 931

Figure from prob_s2.9, line 931

Answer

(i)

Figure from prob_s2.9, line 964

Figure from prob_s2.9, line 964

(ii)

Figure from prob_s2.9, line 964

Figure from prob_s2.9, line 964

(iii)

Figure from prob_s2.9, line 964

Figure from prob_s2.9, line 964

(iv)

Figure from prob_s2.9, line 964

Figure from prob_s2.9, line 964

Full solution

Both of the functions f(x,y)=x2+y2f(x, y) = \sqrt{x^2 + y^2} (i.e. (ii)) and f(x,y)=x2+y2f(x, y) = x^2 + y^2 (i.e. (iv)) are invariant under rotations around the (0,0)(0,0). So their level curves are circles centred on the origin. In polar coordinates x2+y2\sqrt{x^2 + y^2} is rr. So the sketched level curves of the function in (ii) are r=0,0.1,0.2,…,1.9,2r = 0, 0.1, 0.2, \ldots , 1.9, 2. They are equally spaced. So at this point, we know that the third picture goes with (iv) and the fourth picture goes with (ii).

Notice that the lines x=yx=y, x=−yx=-y and y=0y=0 are all level curves of the function f(x,y)=y(x+y)(x−y)+1f(x, y) = y(x + y)(x - y) + 1 (i.e. of (iii)) with f=1f=1. So the first picture goes with (iii). And the second picture goes with (i).

Here are the pictures with critical points marked on them. There are saddle points where level curves cross and there are local max's or min's at “bull's eyes”.

(i)

Figure from prob_s2.9, line 964

Figure from prob_s2.9, line 964

(ii)

Figure from prob_s2.9, line 964

Figure from prob_s2.9, line 964

(iii)

Figure from prob_s2.9, line 964

Figure from prob_s2.9, line 964

(iv)

Figure from prob_s2.9, line 964

Figure from prob_s2.9, line 964

Q15Stage 2Past exam · M200 2010D

Let the function

f(x,y)=x3+3xy+3y2−6x−3y−6\begin{equation*} f(x,y) = x^3+3xy+3y^2-6x-3y-6 \end{equation*}

Classify as [\big[local maxima, minima or saddle points]\big] all critical points of f(x,y)f(x,y).

Answer
criticalpoint\Atop{\text{critical}}{\text{point}}type
(32,−14)\big(\frac{3}{2},-\frac{1}{4}\big)local min
(−1,1)(-1,1)saddle point
Full solution

To find the critical points we will need the gradient of ff, and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

f=x3+3xy+3y2−6x−3y−6fx=3x2+3y−6fxx=6xfxy=3fy=3x+6y−3fyy=6fyx=3\begin{alignat*}{3} f&=x^3+3xy+3y^2-6x-3y-6 \\ f_x&=3x^2+3y-6 & f_{xx}&=6x \qquad & f_{xy}&= 3\\ f_y&=3x+6y-3 \qquad & f_{yy}&=6\qquad & f_{yx}&= 3 \end{alignat*}

(Of course, fxyf_{xy} and fyxf_{yx} have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

fx=3x2+3y−6=0fy=3x +6y−3=0\begin{alignat*}{3} f_x&=3x^2+3y-6&&=0 \tag{E1} \\ f_y&=3x\ +6y-3 &&= 0 \tag{E2} \end{alignat*}

Subtracting equation (E2) from twice equation (E1) gives

6x2−3x−9=0  ⟺  (2x−3)(3x+3)=0\begin{align*} 6x^2-3x-9=0 \iff (2x-3)(3x+3)=0 \end{align*}

So we must have either x=32x=\frac{3}{2} or x=−1x=-1.

  • If x=32x=\frac{3}{2}, (E2) reduces to 92+6y−3=0\frac{9}{2}+6y-3=0 so y=−14y=-\frac{1}{4}.

  • If x=−1x=-1, (E2) reduces to −3+6y−3=0-3+6y-3=0 so y=1y=1.

So there are two critical points: (32,−14)\big(\frac{3}{2},-\frac{1}{4}\big) and (−1,1)(-1,1).

The classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(32,−14)\big(\frac{3}{2},-\frac{1}{4}\big)(9)×(6)−(3)2>0(9)\times (6)-(3)^2> 099local min
(−1,1)(-1,1)(−6)×(6)−(3)2<0(-6)\times (6)-(3)^2<0saddle point
Q16Stage 2Past exam · M200 2011A

Let h(x,y)=y(4−x2−y2)h(x, y) = y(4 - x^2 - y^2).

  1. Find and classify the critical points of h(x,y)h(x, y) as local maxima, local minima or saddle points.

  2. Find the maximum and minimum values of h(x,y)h(x, y) on the disk x2+y2≤1x^2 + y^2 \le 1.

Answer

(a)

criticalpoint\Atop{\text{critical}}{\text{point}}type
(0,23)\left(0,\frac{2}{\sqrt{3}}\right)local max
(0,−23)\left(0,-\frac{2}{\sqrt{3}}\right)local min
(2,0)(2,0)saddle point
(−2,0)(-2,0)saddle point

(b) The maximum and minimum values of h(x,y)h(x,y) in x2+y2≤1x^2+y^2\le 1 are 33 (at (0,1)(0,1)) and −3-3 (at (0,−1)(0,-1)), respectively.

Full solution

(a) To find the critical points we will need the gradient of hh and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

h=y(4−x2−y2)hx=−2xyhxx=−2yhxy=−2xhy=4−x2−3y2hyy=−6yhyx=−2x\begin{alignat*}{3} h&=y(4 - x^2 - y^2) \\ h_x&=-2xy & h_{xx}&=-2y \qquad & h_{xy}&= -2x\\ h_y&=4-x^2-3y^2 \qquad & h_{yy}&=-6y\qquad & h_{yx}&= -2x \end{alignat*}

(Of course, hxyh_{xy} and hyxh_{yx} have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

hx=−2xy=0hy=4−x2−3y2=0\begin{equation*} h_x=-2xy=0 \qquad h_y=4-x^2-3y^2 = 0 \end{equation*}

The first equation is satisfied if at least one of x=0x=0, y=0y=0 are satisfied.

  • If x=0x=0, the second equation reduces to 4−3y2=04-3y^2=0, which is satisfied if y=±23y=\pm\frac{2}{\sqrt{3}}.

  • If y=0y=0, the second equation reduces to 4−x2=04-x^2=0 which is satisfied if x=±2x=\pm 2.

So there are four critical points: (0,23)\left(0,\frac{2}{\sqrt{3}}\right), (0,−23)\left(0,-\frac{2}{\sqrt{3}}\right), (2,0)(2,0), (−2,0)(-2,0).

The classification is

criticalpoint\Atop{\text{critical}}{\text{point}}hxxhyy−hxy2h_{xx}h_{yy}-h_{xy}^2hxxh_{xx}type
(0,23)\left(0,\frac{2}{\sqrt{3}}\right)(−43)×(−123)−(0)2>0\left(\frac{-4}{\sqrt{3}}\right)\times \left(-\frac{12}{\sqrt{3}}\right)-(0)^2> 0−43\frac{-4}{\sqrt{3}}local max
(0,−23)\left(0,-\frac{2}{\sqrt{3}}\right)(43)×(123)−(0)2>0\left(\frac{4}{\sqrt{3}}\right)\times \left(\frac{12}{\sqrt{3}}\right)-(0)^2>043\frac{4}{\sqrt{3}}local min
(2,0)(2,0)0×0−(−4)2<00\times 0-(-4)^2<0saddle point
(−2,0)(-2,0)0×0−(4)2<00\times 0-(4)^2<0saddle point

(b) The absolute max and min can occur either in the interior of the disk or on the boundary of the disk. The boundary of the disk is the circle x2+y2=1x^2+y^2=1.

  • Any absolute max or min in the interior of the disk must also be a local max or min and, since there are no singular points, must also be a critical point of hh. We found all of the critical points of hh in part (a). Since 2>12>1 and 23>1\frac{2}{\sqrt{3}}>1 none of the critical points are in the disk.

  • At each point of x2+y2=1x^2+y^2=1 we have h(x,y)=3yh(x,y)=3y with −1≤y≤1-1\le y\le 1. Clearly the maximum value is 33 (at (0,1)(0,1)) and the minimum value is −3-3 (at (0,−1)(0,-1)).

So all together, the maximum and minimum values of h(x,y)h(x,y) in x2+y2≤1x^2+y^2\le 1 are 33 (at (0,1)(0,1)) and −3-3 (at (0,−1)(0,-1)), respectively.

Q17Stage 2Past exam · M200 2012D

Find the absolute maximum and minimum values of the function f(x,y)=5+2x−x2−4y2f(x, y) = 5 + 2x - x^2 - 4y^2 on the rectangular region

R={ (x,y) ∣ −1≤x≤3, −1≤y≤1 }\begin{equation*} R = \Set{(x, y)}{ -1 \le x \le 3,\ -1 \le y \le 1} \end{equation*}
Answer

The minimum is −2-2 and the maximum is 66.

Full solution

The maximum and minimum must either occur at a critical point or on the boundary of RR.

  • The critical points are the solutions of

    0=fx(x,y)=2−2x0=fy(x,y)=−8y\begin{align*} 0&=f_x(x,y) = 2-2x \\ 0&=f_y(x,y) = -8y \\ \end{align*}

    So the only critical point is (1,0)(1,0).

  • On the side x=−1x=-1, −1≤y≤1-1\le y\le 1 of the boundary of RR

    f(−1,y)=2−4y2\begin{equation*} f(-1,y) = 2-4y^2 \end{equation*}

    This function decreases as ∣y∣|y| increases. So its maximum value on −1≤y≤1-1\le y\le 1 is achieved at y=0y=0 and its minimum value is achieved at y=±1y=\pm 1.

  • On the side x=3x=3, −1≤y≤1-1\le y\le 1 of the boundary of RR

    f(3,y)=2−4y2\begin{equation*} f(3,y) = 2-4y^2 \end{equation*}

    This function decreases as ∣y∣|y| increases. So its maximum value on −1≤y≤1-1\le y\le 1 is achieved at y=0y=0 and its minimum value is achieved at y=±1y=\pm 1.

  • On both sides y=±1y=\pm 1, −1≤x≤3-1\le x\le 3 of the boundary of RR

    f(x,±1)=1+2x−x2=2−(x−1)2\begin{equation*} f(x,\pm 1) = 1+2x-x^2 = 2 -(x-1)^2 \end{equation*}

    This function decreases as ∣x−1∣|x-1| increases. So its maximum value on −1≤x≤3-1\le x\le 3 is achieved at x=1x=1 and its minimum value is achieved at x=3x= 3 and x=−1x=-1 (both of whom are a distance 22 from x=1x=1).

So we have the following candidates for the locations of the min and max

point(1,0)(1,0)(−1,0)(-1,0)(1,±1)(1,\pm 1)(−1,±1)(-1,\pm 1)(3,0)(3,0)(3,±1)(3,\pm 1)
value of ff662222−2-222−2-2
maxminmin

So the minimum is −2-2 and the maximum is 66.

Q18Stage 2Past exam · M200 2013D

Find the minimum of the function h(x,y)=−4x−2y+6h(x,y) = -4x - 2y + 6 on the closed bounded domain defined by x2+y2≤1x^2 + y^2 \le 1.

Answer

6−256-2\sqrt{5}

Full solution

Since ∇h=<−4 , −2>\vnabla h = \llt -4\,,\,-2 \rgt is never zero, hh has no critical points and the minimum of hh on the disk x2+y2≤1x^2+y^2\le 1 must be taken on the boundary, x2+y2=1x^2+y^2=1, of the disk. To find the minimum on the boundary, we parametrize x2+y2≤1x^2+y^2\le 1 by x=cos⁡θx=\cos\theta, y=sin⁡θy=\sin\theta and find the minimum of

H(θ)=−4cos⁡θ−2sin⁡θ+6\begin{align*} H(\theta) = -4\cos\theta -2\sin\theta +6 \end{align*}

Since

0=H′(θ)=4sin⁡θ−2cos⁡θ  ⟹  x=cos⁡θ=2sin⁡θ=2y\begin{align*} 0=H'(\theta) = 4\sin\theta -2\cos\theta \implies x=\cos\theta = 2\sin\theta =2y \end{align*}

So

1=x2+y2=4y2+y2=5y2  ⟹  y=±15, x=±25\begin{align*} 1=x^2+y^2 = 4y^2 +y^2 =5 y^2 \implies y = \pm\frac{1}{\sqrt{5}},\ x = \pm\frac{2}{\sqrt{5}} \end{align*}

At these two points

h=6−4x−2y=6−10y=6∓105=6∓25\begin{align*} h = 6-4x-2y = 6 - 10y =6 \mp \frac{10}{\sqrt{5}} =6 \mp 2\sqrt{5} \end{align*}

The minimum is 6−256-2\sqrt{5}.

Q19Stage 2Past exam · M200 2014D

Let f(x,y)=xy(x+y−3)f(x,y) = xy(x + y - 3).

  1. Find all critical points of ff, and classify each one as a local maximum, a local minimum, or saddle point.

  2. Find the location and value of the absolute maximum and minimum of ff on the triangular region x≥0x \ge 0, y≥0y \ge 0, x+y≤8x + y \le 8.

Answer

(a) (0,0) and (3,0) and (0,3) are saddle points
(1,1) is a local min

(b) The minimum is −1-1 at (1,1)(1,1) and the maximum is 8080 at (4,4)(4,4).

Full solution

(a) Thinking a little way ahead, to find the critical points we will need the gradient of ff and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

f=xy(x+y−3)fx=2xy+y2−3yfxx=2yfxy=2x+2y−3fy=x2+2xy−3xfyy=2xfyx=2x+2y−3\begin{alignat*}{3} f&=xy(x + y - 3) \\ f_x&=2xy+y^2-3y & f_{xx}&=2y \qquad & f_{xy}&= 2x+2y-3\\ f_y&=x^2+2xy-3x \qquad & f_{yy}&=2x\qquad & f_{yx}&= 2x+2y-3 \end{alignat*}

(Of course, fxyf_{xy} and fyxf_{yx} have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

fx=y(2x+y−3)=0fy=x(x+2y−3)=0\begin{equation*} f_x=y(2x+y-3)=0 \qquad f_y=x(x+2y-3) = 0 \end{equation*}

The first equation is satisfied if at least one of y=0y=0, y=3−2xy=3-2x are satisfied.

  • If y=0y=0, the second equation reduces to x(x−3)=0x(x-3)=0, which is satisfied if either x=0x=0 or x=3x=3.

  • If y=3−2xy=3-2x, the second equation reduces to x(x+6−4x−3)=x(3−3x)=0x(x+6-4x-3)=x(3-3x)=0 which is satisfied if x=0x=0 or x=1x=1.

So there are four critical points: (0,0)(0,0), (3,0)(3,0), (0,3)(0,3), (1,1)(1,1).

The classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)0×0−(−3)2<00\times 0-(-3)^2< 0saddle point
(3,0)(3,0)0×6−(3)2<00\times 6-(3)^2<0saddle point
(0,3)(0,3)6×0−(3)2<06\times 0-(3)^2<0saddle point
(1,1)(1,1)2×2−(1)2>02\times 2-(1)^2>02local min

(b) The absolute max and min can occur either in the interior of the triangle or on the boundary of the triangle. The boundary of the triangle consists of the three line segments.

L1={ (x,y) ∣ x=0, 0≤y≤8 }L2={ (x,y) ∣ y=0, 0≤x≤8 }L3={ (x,y) ∣ x+y=8, 0≤x≤8 }\begin{align*} L_1 &= \Set{(x,y)}{x=0,\ 0\le y\le 8} \\ L_2 &= \Set{(x,y)}{y=0,\ 0\le x\le 8} \\ L_3 &= \Set{(x,y)}{x+y=8,\ 0\le x\le 8} \end{align*}

  • Any absolute max or min in the interior of the triangle must also be a local max or min and, since there are no singular points, must also be a critical point of ff. We found all of the critical points of ff in part (a). Only one of them, namely (1,1)(1,1) is in the interior of the triangle. (The other three critical points are all on the boundary of the triangle.) We have f(1,1)=−1f(1,1) = -1.

  • At each point of L1L_1 we have x=0x=0 and so f(x,y)=0f(x,y)=0.

  • At each point of L2L_2 we have y=0y=0 and so f(x,y)=0f(x,y)=0.

  • At each point of L3L_3 we have f(x,y)=x(8−x)(5)=40x−5x2=5[8x−x2]f(x,y)=x(8-x)(5)=40x-5x^2=5[8x-x^2] with 0≤x≤80\le x\le 8. As ddx(40x−5x2)=40−10x\diff{}{x}\big(40x-5x^2\big)= 40-10x, the max and min of 40x−5x240x-5x^2 on 0≤x≤80\le x\le 8 must be one of 5[8x−x2]x=0=05\big[8x-x^2\big]_{x=0}=0 or 5[8x−x2]x=8=05\big[8x-x^2\big]_{x=8}=0 or 5[8x−x2]x=4=805\big[8x-x^2\big]_{x=4}=80.

So all together, we have the following candidates for max and min, with the max and min indicated.

point(s)(1,1)(1,1)L1L_1L2L_2(0,8)(0,8)(8,0)(8,0)(4,4)(4,4)
value of ff−1-1000000008080
minmax

Figure from prob_s2.9, line 1365

Figure from prob_s2.9, line 1365

Q20Stage 2Past exam · M200 2015D

Find and classify the critical points of f(x,y)=3x2y+y3−3x2−3y2+4f(x,y) = 3x^2 y + y^3 - 3x^2 - 3y^2 + 4.

Answer

(0,0) is a local max

(0,2) is a local min

(1,1) and (-1,1) are saddle points

Full solution

Thinking a little way ahead, to find the critical points we will need the gradient of ff, and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

f=3x2y+y3−3x2−3y2+4fx=6xy−6xfxx=6y−6fxy=6xfy=3x2+3y2−6yfyy=6y−6fyx=6x\begin{alignat*}{3} f&=3x^2 y + y^3 - 3x^2 - 3y^2 + 4 \\ f_x&=6xy-6x & f_{xx}&=6y-6 \qquad & f_{xy}&= 6x\\ f_y&=3x^2+3y^2-6y \qquad & f_{yy}&=6y-6\qquad & f_{yx}&= 6x \end{alignat*}

(Of course, fxyf_{xy} and fyxf_{yx} have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

fx=6x(y−1)=0fy=3x2+3y2−6y=0\begin{equation*} f_x=6x(y-1)=0 \qquad f_y=3x^2+3y^2-6y = 0 \end{equation*}

The first equation is satisfied if at least one of x=0x=0, y=1y=1 are satisfied.

  • If x=0x=0, the second equation reduces to 3y2−6y=03y^2-6y=0, which is satisfied if either y=0y=0 or y=2y=2.

  • If y=1y=1, the second equation reduces to 3x2−3=03x^2 -3=0 which is satisfied if x=±1x=\pm 1.

So there are four critical points: (0,0)(0,0), (0,2)(0,2), (1,1)(1,1), (−1,1)(-1,1).

The classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)(−6)×(−6)−(0)2>0(-6)\times (-6)-(0)^2> 0−6-6local max
(0,2)(0,2)6×6−(0)2>06\times 6-(0)^2>06local min
(1,1)(1,1)0×0−(6)2<00\times 0-(6)^2<0saddle point
(−1,1)(-1,1)0×0−(−6)2<00\times 0-(-6)^2<0saddle point
Q21Stage 2Past exam · M200 2004A

Consider the function

f(x,y)=2x3−6xy+y2+4y\begin{equation*} f(x,y)=2x^3 - 6xy + y^2 +4y \end{equation*}

  1. Find and classify all of the critical points of f(x,y)f(x,y).

  2. Find the maximum and minimum values of f(x,y)f(x,y) in the triangle with vertices (1,0)(1,0), (0,1)(0,1) and (1,1)(1,1).

Answer

(a) (1,1)(1,1) is a saddle point and (2,4)(2,4) is a local min

(b) The min and max are 1927\frac{19}{27} and 55, respectively.

Full solution

(a) Since

f=2x3−6xy+y2+4yfx=6x2−6yfxx=12xfxy=−6fy=−6x+2y+4fyy=2\begin{alignat*}{5} f&=2x^3 - 6xy + y^2 +4y \\ f_x&=6x^2-6y & f_{xx}&=12x\qquad && f_{xy}&= -6\\ f_y&=-6x+2y+4\qquad & f_{yy}&=2 \end{alignat*}

the critical points are the solutions of

fx=0fy=0  ⟺  y=x2y−3x+2=0  ⟺  y=x2x2−3x+2=0  ⟺  y=x2x=1 or 2\begin{alignat*}{5} & & &f_x=0\qquad & &f_y=0 \\ &\iff\qquad& &y=x^2\qquad & &y-3x+2=0 \\ &\iff& &y=x^2 & &x^2-3x+2=0 \\ &\iff& &y=x^2 & &x=1\text{ or }2 \end{alignat*}

So, there are two critical points: (1,1), (2,4)(1,1),\ (2,4).

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(1,1)(1,1)12×2−(−6)2<012\times 2-(-6)^2<0saddle point
(2,4)(2,4)24×2−(−6)2>024\times 2-(-6)^2>02424local min

(b) There are no critical points in the interior of the allowed region, so both the maximum and the minimum occur only on the boundary. The boundary consists of the line segments (i) x=1x=1, 0≤y≤10\le y\le 1, (ii) y=1y=1, 0≤x≤10\le x\le 1 and (iii) y=1−xy=1-x, 0≤x≤10\le x\le 1.

Figure from prob_s2.9, line 1572

Figure from prob_s2.9, line 1572

  • First, we look at the part of the boundary with x=1x=1. There f=y2−2y+2f=y^2-2y+2. As ddy(y2−2y+2)=2y−2\diff{}{y}(y^2-2y+2)=2y-2 vanishes only at y=1y=1, the max and min of y2−2y+2y^2-2y+2 for 0≤y≤10\le y\le 1 must occur either at y=0y=0, where f=2f=2, or at y=1y=1, where f=1f=1.

  • Next, we look at the part of the boundary with y=1y=1. There f=2x3−6x+5f=2x^3-6x+5. As ddx(2x3−6x+5)=6x2−6\diff{}{x}(2x^3-6x+5)=6x^2-6, the max and min of 2x3−6x+52x^3-6x+5 for 0≤x≤10\le x\le 1 must occur either at x=0x=0, where f=5f=5, or at x=1x=1, where f=1f=1.

  • Next, we look at the part of the boundary with y=1−xy=1-x. There f=2x3−6x(1−x)+(1−x)2+4(1−x)=2x3+7x2−12x+5f=2x^3-6x(1-x) +(1-x)^2+4(1-x)=2x^3+7x^2-12x+5. As ddx(2x3+7x2−12x+5)=6x2+14x−12=2(3x2+7x−6)=2(3x−2)(x+3)\diff{}{x}(2x^3+7x^2-12x+5)=6x^2+14x-12 =2\big(3x^2+7x-6\big) =2(3x-2)(x+3), the max and min of 2x3+7x2−12x+52x^3+7x^2-12x+5 for 0≤x≤10\le x\le 1 must occur either at x=0x=0, where f=5f=5, or at x=1x=1, where f=2f=2, or at x=23x=\frac{2}{3}, where f=2(827)−6(23)(13)+19+43=16−36+3+3627=1927f=2(\frac{8}{27})-6(\frac{2}{3})(\frac{1}{3})+\frac{1}{9}+\frac{4}{3} =\frac{16-36+3+36}{27} =\frac{19}{27}.

So all together, we have the following candidates for max and min, with the max and min indicated.

point(1,0)(1,0)(1,1)(1,1)(0,1)(0,1)(23,13)\big(\frac{2}{3},\frac{1}{3}\big)
value of ff2211551927\frac{19}{27}
maxmin
Q22Stage 2Past exam · M200 2003D

Find all critical points of the function f(x,y)=x4+y4−4xy+2f(x,y)=x^4+y^4-4xy+2, and for each determine whether it is a local minimum, maximum or saddle point.

Answer

(0,0)(0,0) is a saddle point and ±(1,1)\pm(1,1) are local mins

Full solution

We have

f(x,y)=x4+y4−4xy+2fx(x,y)=4x3−4yfxx(x,y)=12x2fy(x,y)=4y3−4xfyy(x,y)=12y2fxy(x,y)=−4\begin{alignat*}{5} f(x,y)&=x^4+y^4-4xy+2\quad & f_x(x,y)&=4x^3-4y\quad & f_{xx}(x,y)&=12x^2 \\ & & f_y(x,y)&=4y^3-4x & f_{yy}(x,y)&=12y^2 \\ & & & &f_{xy}(x,y)&=-4 \end{alignat*}

At a critical point

fx(x,y)=fy(x,y)=0  ⟺  y=x3 and x=y3  ⟺  x=x9 and y=x3  ⟺  x(x8−1)=0, y=x3  ⟺  (x,y)=(0,0) or (1,1) or (−1,−1)\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff y=x^3\text{ and }x=y^3 \\ &\iff x=x^9\text{ and }y=x^3 \\ &\iff x(x^8-1)=0,\ y=x^3 \\ &\iff (x,y)=(0,0)\text{ or }(1,1)\text{ or }(-1,-1) \end{align*}

Here is a table giving the classification of each of the three critical points.

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)0×0−(−4)2<00\times 0-(-4)^2<0saddle point
(1,1)(1,1)12×12−(−4)2>012\times 12-(-4)^2>012local min
(−1,−1)(-1,-1)12×12−(−4)2>012\times 12-(-4)^2>012local min
Q23Stage 2Past exam · M200 2003A

Let

f(x,y)=xy(x+2y−6)\begin{equation*} f(x,y)=xy(x+2y-6) \end{equation*}

  1. Find every critical point of f(x,y)f(x,y) and classify each one.

  2. Let DD be the region in the plane between the hyperbola xy=4xy=4 and the line x+2y−6=0x+2y-6=0. Find the maximum and minimum values of f(x,y)f(x,y) on DD.

Answer

(a) (0,0)(0,0), (6,0)(6,0), (0,3)(0,3) are saddle points and (2,1)(2,1) is a local min

(b) The maximum value is 00 and the minimum value is 4(42−6)≈−1.374(4\sqrt{2}-6) \approx -1.37.

Full solution

(a) We have

f(x,y)=xy(x+2y−6)fx(x,y)=2xy+2y2−6yfxx(x,y)=2yfy(x,y)=x2+4xy−6xfyy(x,y)=4xfxy(x,y)=2x+4y−6\begin{alignat*}{5} f(x,y)&=xy(x+2y-6)\quad & f_x(x,y)&=2xy+2y^2-6y\quad & f_{xx}(x,y)&=2y \\ & & f_y(x,y)&=x^2+4xy-6x & f_{yy}(x,y)&=4x\\ & & & &f_{xy}(x,y)&=2x+4y-6 \end{alignat*}

At a critical point

fx(x,y)=fy(x,y)=0  ⟺  2y(x+y−3)=0 and x(x+4y−6)=0  ⟺  {y=0 or x+y=3} and {x=0 or x+4y=6}  ⟺  {x=y=0} or {y=0, x+4y=6} or {x+y=3, x=0} or {x+y=3, x+4y=6}  ⟺  (x,y)=(0,0) or (6,0) or (0,3) or (2,1)\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff 2y(x+y-3)=0\text{ and }x(x+4y-6)=0 \\ &\iff \{y=0\text{ or }x+y=3\}\text{ and }\{x=0\text{ or }x+4y=6\}\\ &\iff \{x=y=0\}\text{ or }\{y=0,\ x+4y=6\}\\ &\hskip0.5in\text{ or }\{x+y=3,\ x=0\} \text{ or }\{x+y=3,\ x+4y=6\}\\ &\iff (x,y)=(0,0)\text{ or }(6,0)\text{ or }(0,3)\text{ or }(2,1) \end{align*}

Here is a table giving the classification of each of the four critical points.

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)0×0−(−6)2<00\times 0-(-6)^2<0saddle point
(6,0)(6,0)0×24−62<00\times 24-6^2<0saddle point
(0,3)(0,3)6×0−62<06\times 0-6^2<0saddle point
(2,1)(2,1)2×8−22>02\times 8-2^2>02local min

(b) Observe that xy=4xy=4 and x+2y=6x+2y=6 intersect when x=6−2yx=6-2y and

(6−2y)y=4  ⟺  2y2−6y+4=0  ⟺  2(y−1)(y−2)=0  ⟺  (x,y)=(4,1) or (2,2)\begin{align*} (6-2y)y=4 &\iff 2y^2-6y+4=0 \iff 2(y-1)(y-2)=0 \\ &\iff (x,y) = (4,1)\text{ or }(2,2) \end{align*}

The shaded region in the sketch below is DD.

Figure from prob_s2.9, line 1739

Figure from prob_s2.9, line 1739

None of the critical points are in DD. So the max and min must occur at either (2,2)(2,2) or (4,1)(4,1) or on xy=4xy=4, 2<x<42<x<4 (in which case F(x)=f(x,4x)=4(x+8x−6)F(x)=f\big(x,\frac{4}{x}\big)=4\big(x+\frac{8}{x}-6) obeys F′(x)=4−32x2=0  ⟺  x=±22F'(x)=4-\frac{32}{x^2}=0\iff x=\pm2\sqrt{2}) or on x+2y=6x+2y=6, 2<x<42<x<4 (in which case f(x,y)f(x,y) is identically zero). So the min and max must occur at one of

(x,y)(x,y)f(x,y)f(x,y)
(2,2)(2,2)2×2(2+2×2−6)=02\times 2(2+2\times2-6)=0
(4,1)(4,1)4×1(4+2×1−6)=04\times 1(4+2\times 1-6)=0
(22,2)(2\sqrt{2},\sqrt{2})4(22+22−6)<04(2\sqrt{2}+2\sqrt{2}-6)<0

The maximum value is 00 and the minimum value is 4(42−6)≈−1.374(4\sqrt{2}-6) \approx -1.37.

Q24Stage 2Past exam · M200 2002D

Find all the critical points of the function

f(x,y)=x4+y4−4xy\begin{equation*} f(x,y)=x^4+y^4-4xy \end{equation*}

defined in the xyxy-plane. Classify each critical point as a local minimum, maximum or saddle point.

Answer

(0,0)(0,0) is a saddle point and ±(1,1)\pm(1,1) are local mins

Full solution

We have

f(x,y)=x4+y4−4xyfx(x,y)=4x3−4yfxx(x,y)=12x2fy(x,y)=4y3−4xfyy(x,y)=12y2fxy(x,y)=−4\begin{alignat*}{3} f(x,y)&=x^4+y^4-4xy\qquad & f_x(x,y)&=4x^3-4y\qquad & f_{xx}(x,y)&=12x^2\\ & & f_y(x,y)&=4y^3-4x & f_{yy}(x,y)&=12y^2 \\ & & & &f_{xy}(x,y)&=-4 \end{alignat*}

At a critical point

fx(x,y)=fy(x,y)=0  ⟺  y=x3 and x=y3  ⟺  x=x9 and y=x3  ⟺  x(x8−1)=0, y=x3  ⟺  (x,y)=(0,0) or (1,1) or (−1,−1)\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff y=x^3\text{ and }x=y^3 \iff x=x^9\text{ and }y=x^3 \\ &\iff x(x^8-1)=0,\ y=x^3\\ &\iff (x,y)=(0,0)\text{ or }(1,1)\text{ or }(-1,-1) \end{align*}

Here is a table giving the classification of each of the three critical points.

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)0×0−(−4)2<00\times 0-(-4)^2<0saddle point
(1,1)(1,1)12×12−(−4)2>012\times 12-(-4)^2>012local min
(−1,−1)(-1,-1)12×12−(−4)2>012\times 12-(-4)^2>012local min
Q25Stage 2Past exam · M200 2002A

A metal plate is in the form of a semi-circular disc bounded by the xx-axis and the upper half of x2+y2=4x^2+y^2=4. The temperature at the point (x,y)(x,y) is given by T(x,y)=ln⁡(1+x2+y2)−yT(x,y)=\ln\big(1+x^2+y^2\big)-y. Find the coldest point on the plate, explaining your steps carefully. (Note: ln⁡2≈0.693\ln 2\approx 0.693, ln⁡5≈1.609\ln 5\approx 1.609)

Answer

The coldest temperture is −0.391-0.391 and the coldest point is (0,2)(0,2).

Full solution

The coldest point must be either on the boundary of the plate or in the interior of the plate.

  • On the semi–circular part of the boundary 0≤y≤20\le y\le 2 and x2+y2=4x^2+y^2=4 so that T=ln⁡(1+x2+y2)−y=ln⁡5−yT=\ln\big(1+x^2+y^2\big)-y=\ln 5-y. The smallest value of ln⁡5−y\ln 5-y is taken when yy is as large as possible, i.e. when y=2y=2, and is ln⁡5−2≈−0.391\ln 5 -2\approx -0.391.

  • On the flat part of the boundary, y=0y=0 and −2≤x≤2-2\le x\le 2 so that T=ln⁡(1+x2+y2)−y=ln⁡(1+x2)T=\ln\big(1+x^2+y^2\big)-y=\ln\big(1+x^2\big). The smallest value of ln⁡(1+x2)\ln\big(1+x^2\big) is taken when xx is as small as possible, i.e. when x=0x=0, and is 00.

  • If the coldest point is in the interior of the plate, it must be at a critical point of T(x,y)T(x,y). Since

    Tx(x,y)=2x1+x2+y2Ty(x,y)=2y1+x2+y2−1\begin{align*} T_x(x,y)=\frac{2x}{1+x^2+y^2}\qquad T_y(x,y)=\frac{2y}{1+x^2+y^2}-1 \end{align*}

    a critical point must have x=0x=0 and 2y1+x2+y2−1=0\frac{2y}{1+x^2+y^2}-1=0, which is the case if and only if x=0x=0 and 2y−1−y2=02y-1-y^2=0. So the only critical point is x=0, y=1x=0,\ y=1, where T=ln⁡2−1≈−0.307T=\ln 2-1\approx -0.307.

Since −0.391<−0.307<0-0.391<-0.307<0, the coldest temperture is −0.391-0.391 and the coldest point is (0,2)(0,2).

Q26Stage 2Past exam · M200 2001D

Find all the critical points of the function

f(x,y)=x3+xy2−x\begin{equation*} f(x,y)=x^3+xy^2-x \end{equation*}

defined in the xyxy-plane. Classify each critical point as a local minimum, maximum or saddle point. Explain your reasoning.

Answer

(0,±1)(0,\pm 1) are saddle points, (13,0)\big(\frac{1}{\sqrt{3}},0\big) is a local min and (−13,0)\big(-\frac{1}{\sqrt{3}},0\big) is a local max

Full solution

We have

f(x,y)=x3+xy2−xfx(x,y)=3x2+y2−1fxx(x,y)=6xfy(x,y)=2xyfyy(x,y)=2xfxy(x,y)=2y\begin{alignat*}{3} f(x,y)&=x^3+xy^2-x\qquad & f_x(x,y)&=3x^2+y^2-1\qquad & f_{xx}(x,y)&=6x \\ & & f_y(x,y)&=2xy & f_{yy}(x,y)&=2x \\ & & & & f_{xy}(x,y)&=2y \end{alignat*}

At a critical point

fx(x,y)=fy(x,y)=0  ⟺  xy=0 and 3x2+y2=1  ⟺  {x=0 or y=0} and 3x2+y2=1  ⟺  (x,y)=(0,1) or (0,−1) or (13,0) or (−13,0)\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff xy=0\text{ and }3x^2+y^2=1 \\ &\iff \{x=0\text{ or }y=0\}\text{ and }3x^2+y^2=1 \\ &\iff (x,y)=(0,1)\text{ or }(0,-1)\text{ or }\left(\frac{1}{\sqrt{3}},0\right) \text{ or }\left(-\frac{1}{\sqrt{3}},0\right) \end{align*}

Here is a table giving the classification of each of the four critical points.

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,1)(0,1)0×0−22<00\times 0-2^2<0saddle point
(0,−1)(0,-1)0×0−(−2)2<00\times 0-(-2)^2<0saddle point
(13,0)\big(\frac{1}{\sqrt{3}},0\big)23×23−02>02\sqrt{3}\times\frac{2}{\sqrt{3}}-0^2>0232\sqrt{3}local min
(−13,0)\big(-\frac{1}{\sqrt{3}},0\big)−23×(−23)−02>0-2\sqrt{3}\times \big(-\frac{2}{\sqrt{3}}\big)-0^2>0−23-2\sqrt{3}local max
Q27Stage 2Past exam · M200 2001A

Consider the function g(x,y)=x2−10y−y2.g(x,y)=x^2-10y-y^2 .

  1. Find and classify all critical points of gg.

  2. Find the absolute extrema of gg on the bounded region given by

    x2+4y2≤16, y≤0\begin{equation*} x^2+4y^2\le 16,\ y\le 0 \end{equation*}
Answer

(a) (0,−5)(0,-5) is a saddle point

(b) The smallest value of gg is 00 at (0,0)(0,0) and the largest value is 2121 at (±23,−1)(\pm 2\sqrt{3},-1).

Full solution

(a) We have

g(x,y)=x2−10y−y2gx(x,y)=2xgxx(x,y)=2gy(x,y)=−10−2ygyy(x,y)=−2gxy(x,y)=0\begin{alignat*}{5} g(x,y)&=x^2-10y-y^2 \qquad& g_x(x,y)&=2x \qquad& g_{xx}(x,y)&=2\\ & & g_y(x,y)&=-10-2y\qquad & g_{yy}(x,y)&=-2 \\ & & & &g_{xy}(x,y)&=0 \end{alignat*}

At a critical point

gx(x,y)=gy(x,y)=0  ⟺  2x=0 and −10−2y=0  ⟺  (x,y)=(0,−5)\begin{align*} g_x(x,y)=g_y(x,y)=0 \iff 2x=0\text{ and }-10-2y=0 \iff (x,y)=(0,-5) \end{align*}

Since gxx(0,−5)gyy(0,−5)−gxy(0,−5)2=2×(−2)−02<0g_{xx}(0,-5)g_{yy}(0,-5)-g_{xy}(0,-5)^2=2\times(-2)-0^2<0, the critical point is a saddle point.

(b) The extrema must be either on the boundary of the region or in the interior of the region.

  • On the semi-elliptical part of the boundary −2≤y≤0-2\le y\le 0 and x2+4y2=16x^2+4y^2=16 so that g=x2−10y−y2=16−10y−5y2=21−5(y+1)2g=x^2-10y-y^2=16-10y-5y^2=21-5(y+1)^2. This has a minimum value of 16 (at y=0,−2y=0,-2) and a maximum value of 21 (at y=−1y=-1). You could also come to this conclusion by checking the critical point of 16−10y−5y216-10y-5y^2 (i.e. solving ddy(16−10y−5y2)=0\diff{}{y}(16-10y-5y^2)=0) and checking the end points of the allowed interval (namely y=0y=0 and y=−2y=-2).

  • On the flat part of the boundary y=0y=0 and −4≤x≤4-4\le x\le 4 so that g=x2g=x^2.
    The smallest value is taken when x=0x=0 and is 00 and the largest value is taken when x=±4x=\pm 4 and is 1616.

  • If an extremum is in the interior of the plate, it must be at a critical point of g(x,y)g(x,y). The only critical point is not in the prescribed region.

Here is a table giving all candidates for extrema:

(x,y)(x,y)g(x,y)g(x,y)
(0,−2)(0,-2)1616
(±4,0)(\pm 4,0)1616
(±12,−1)(\pm \sqrt{12},-1)2121
(0,0)(0,0)00

From the table the smallest value of gg is 00 at (0,0)(0,0) and the largest value is 2121 at (±23,−1)(\pm 2\sqrt{3},-1).

Q28Stage 2Past exam · M200 2000D

Find and classify all critical points of

f(x,y)=x3−3xy2−3x2−3y2f(x,y)=x^3-3xy^2-3x^2-3y^2
Answer

(−1,±3)(-1,\pm\sqrt{3}) and (2,0)(2,0) are saddle points and (0,0)(0,0) is a local max.

Full solution

We have

f(x,y)=x3−3xy2−3x2−3y2fx(x,y)=3x2−3y2−6xfxx(x,y)=6x−6fy(x,y)=−6xy−6yfyy(x,y)=−6x−6fxy(x,y)=−6y\begin{alignat*}{5} f(x,y)&=x^3-3xy^2-3x^2-3y^2\qquad & f_x(x,y)&=3x^2-3y^2-6x\qquad & f_{xx}(x,y)&=6x-6 \\ & & f_y(x,y)&=-6xy-6y & f_{yy}(x,y)&=-6x-6 \\ & & & &f_{xy}(x,y)&=-6y \end{alignat*}

At a critical point

fx(x,y)=fy(x,y)=0  ⟺  3(x2−y2−2x)=0 and −6y(x+1)=0  ⟺  {x=−1 or y=0} and x2−y2−2x=0  ⟺  (x,y)=(−1,3) or (−1,−3) or (0,0) or (2,0)\begin{align*} f_x(x,y)=f_y(x,y)=0 &\iff 3(x^2-y^2-2x)=0\text{ and }-6y(x+1)=0 \\ &\iff \{x=-1\text{ or }y=0\}\text{ and }x^2-y^2-2x=0 \\ &\iff (x,y)=(-1,\sqrt{3})\text{ or }(-1,-\sqrt{3})\text{ or }(0,0) \text{ or }(2,0) \end{align*}

Here is a table giving the classification of each of the four critical points.

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)(−6)×(−6)−02>0(-6)\times(-6)-0^2>0−6-6local max
(2,0)(2,0)6×(−18)−02<06\times(-18)-0^2<0saddle point
(−1,3)(-1,\sqrt{3})(−12)×0−(−63)2<0(-12)\times0-(-6\sqrt{3})^2<0saddle point
(−1,−3)(-1,-\sqrt{3})(−12)×0−(63)2<0(-12)\times 0-(6\sqrt{3})^2<0saddle point
Q29Stage 2Past exam · M200 2000A

Find the maximum value of

f(x,y)=xye−(x2+y2)/2\begin{equation*} f(x, y) = xye^{-(x^2 + y^2) / 2} \end{equation*}

on the quarter-circle D={ (x,y) ∣ x2+y2≤4, x≥0, y≥0 }D = \Set{(x, y)} { x^2 + y^2 \le 4,\ x\ge0,\ y\ge 0}.

Answer

e−1≈0.368e^{-1}\approx0.368

Full solution

The maximum must be either on the boundary of DD or in the interior of DD.

  • On the circular part of the boundary r=2r=2, 0≤θ≤π20\le\theta\le\frac{\pi}{2} (in polar coordinates) so that f=r2cos⁡θsin⁡θe−r2/2=2sin⁡(2θ)e−2f=r^2\cos\theta\sin\theta e^{-r^2/2}=2\sin(2\theta)e^{-2}. This has a maximum value of 2e−22e^{-2} at θ=π4\theta=\frac{\pi}{4} or x=y=2x=y=\sqrt{2}.

  • On the two flat parts of the boundary x=0x=0 or y=0y=0 so that f=0f=0.

  • If the maximum is in the interior of DD, it must be at a critical point of f(x,y)f(x,y). Since

    fx(x,y)=e−(x2+y2)/2[y−x2y]fy(x,y)=e−(x2+y2)/2[x−xy2]\begin{align*} f_x(x,y)=e^{-(x^2 + y^2) / 2}\big[y-x^2y\big]\qquad f_y(x,y)=e^{-(x^2 + y^2) / 2}\big[x-xy^2\big] \end{align*}

    (x,y)(x,y) is a critical point if and only if

    y(1−x2)=0 and x(1−y2)=0  ⟺  {y=0 or x=1 or x=−1} and {x=0 or y=1 or y=−1}\begin{align*} &y(1-x^2)=0\text{ and }x(1-y^2)=0 \\ &\iff \{y=0\text{ or }x=1\text{ or }x=-1\}\text{ and } \{x=0\text{ or }y=1\text{ or }y=-1\} \end{align*}

    There are two critical points with x,y≥0x,y\ge 0, namely (0,0)(0,0) and (1,1)(1,1). The first of these is on the boundary of DD and the second is in the interior of DD.

Here is a table giving all candidates for the maximum:

(x,y)(x,y)g(x,y)g(x,y)
(2,2)(\sqrt{2},\sqrt{2})2e−2≈0.2712e^{-2}\approx0.271
(x,0)(x,0)00
(0,y)(0,y)00
(1,1)(1,1)e−1≈0.368e^{-1}\approx0.368

Since e>2e>2, we have that 2e−2=e−12e<e−12e^{-2}=e^{-1}\frac{2}{e}<e^{-1} and the largest value is e−1e^{-1}.

Q30Stage 2

Equal angle bends are made at equal distances from the two ends of a 100 metre long fence, so that the resulting three segment fence can be placed along an existing wall to make an enclosure of trapezoidal shape. What is the largest possible area for such an enclosure?

Hint

Suppose that the bends are made a distance xx from the ends of the fence and that the bends are through an angle θ\theta. Draw a sketch of the enclosure and figure out its area, as a function of xx and θ\theta.

Answer

25003\frac{2500}{\sqrt{3}}

Full solution

Suppose that the bends are made a distance xx from the ends of the fence and that the bends are through an angle θ\theta. Here is a sketch of the enclosure.

Figure from prob_s2.9, line 2185

Figure from prob_s2.9, line 2185

It consists of a rectangle, with side lengths 100−2x100-2x and xsin⁡θx\sin\theta, together with two triangles, each of height xsin⁡θx\sin\theta and base length xcos⁡θx\cos\theta. So the enclosure has area

A(x,θ)=(100−2x)xsin⁡θ+2⋅12⋅xsin⁡θ⋅xcos⁡θ=(100x−2x2)sin⁡θ+12x2sin⁡(2θ)\begin{align*} A(x,\theta)&=(100-2x)x\sin\theta+2\cdot\half\cdot x\sin\theta\cdot x\cos\theta\\ &=(100x-2x^2)\sin\theta+\half x^2\sin(2\theta) \end{align*}

The maximize the area, we need to solve

0=Ax=(100−4x)sin⁡θ+xsin⁡(2θ)  ⟹  (100−4x)+2xcos⁡θ=00=Aθ=(100x−2x2)cos⁡θ+x2cos⁡(2θ)  ⟹  (100−2x)cos⁡θ+xcos⁡(2θ)=0\begin{alignat*}{5} 0=A_x&=(100-4x)\sin\theta+x\sin(2\theta) & &\implies & (100-4x)+2x\cos\theta&=0\cr 0=A_\theta&=(100x-2x^2)\cos\theta+x^2\cos(2\theta)\quad & &\implies\quad & (100-2x)\cos\theta+x\cos(2\theta)&=0 \end{alignat*}

Here we have used that the fence of maximum area cannot have sin⁡θ=0\sin\theta=0 or x=0x=0, because in either of these two cases, the area enclosed will be zero. The first equation forces cos⁡θ=−100−4x2x\cos\theta=-\frac{100-4x}{2x} and hence cos⁡(2θ)=2cos⁡2θ−1=(100−4x)22x2−1\cos(2\theta)=2\cos^2\theta-1=\frac{(100-4x)^2}{2x^2}-1. Substituting these into the second equation gives

−(100−2x)100−4x2x+x[(100−4x)22x2−1]=0  ⟹  −(100−2x)(100−4x)+(100−4x)2−2x2=0  ⟹  6x2−200x=0  ⟹  x=1003cos⁡θ=−−100/3200/3=12θ=60∘A=(1001003−2100232)32+1210023232=25003\begin{alignat*}{5} & & -(100-2x)\frac{100-4x}{2x}+x\Big[\frac{(100-4x)^2}{2x^2}-1\Big]&=0 \\ &\implies & -(100-2x)(100-4x)+(100-4x)^2-2x^2&=0 \\ &\implies & 6x^2-200x&=0 \\ &\implies & x=\frac{100}{3} \quad\cos\theta=-\frac{-100/3}{200/3}=\frac{1}{2}\quad \theta&=60^\circ\\ & & A= \left(100\frac{100}{3}-2\frac{100^2}{3^2}\right) \frac{\sqrt{3}}{2}+\frac{1}{2} \frac{100^2}{3^2}\frac{\sqrt{3}}{2} &=\frac{2500}{\sqrt{3}} \end{alignat*}
Q31Stage 2

Find the most economical shape of a rectangular box that has a fixed volume VV and that has no top.

Hint

Suppose that the box has side lengths xx, yy and zz.

Answer

The box has dimensions (2V)1/3×(2V)1/3×2−2/3V1/3(2V)^{1/3}\times(2V)^{1/3}\times 2^{-2/3}V^{1/3}.

Full solution

Suppose that the box has side lengths xx, yy and zz. Here is a sketch.

Figure from prob_s2.9, line 2238

Figure from prob_s2.9, line 2238

Because the box has to have volume VV we need that V=xyzV=xyz. We wish to minimize the area A=xy+2yz+2xzA=xy+2yz+2xz of the four sides and bottom. Substituting in z=Vxyz=\frac{V}{xy},

A=xy+2Vx+2VyAx=y−2Vx2Ay=x−2Vy2\begin{align*} A&=xy+2\frac{V}{x}+2\frac{V}{y}\\ A_x&=y-2\frac{V}{x^2}\\ A_y&=x-2\frac{V}{y^2} \end{align*}

To minimize, we want Ax=Ay=0A_x=A_y=0, which is the case when yx2=2V, xy2=2Vyx^2=2V,\ xy^2=2V. This forces yx2=xy2yx^2=xy^2. Since V=xyzV=xyz is nonzero, neither xx nor yy may be zero. So x=y=(2V)1/3x=y=(2V)^{1/3}, z=2−2/3V1/3z=2^{-2/3}V^{1/3}.

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q32Stage 3Past exam · M200 2009D

The temperature T(x,y)T(x,y) at a point of the xyxy–plane is given by

T(x,y)=20−4x2−y2\begin{equation*} T(x,y) = 20 - 4x^2 - y^2 \end{equation*}

  1. Find the maximum and minimum values of T(x,y)T(x,y) on the disk DD defined by x2+y2≤4x^2 + y^2 \le 4.

  2. Suppose an ant lives on the disk DD. If the ant is initially at point (1,1)(1, 1), in which direction should it move so as to increase its temperature as quickly as possible?

  3. Suppose that the ant moves at a velocity v=<−2,−1>\vv = \llt -2, -1\rgt. What is its rate of increase of temperature as it passes through (1,1)(1, 1)?

  4. Suppose the ant is constrained to stay on the curve y=2−x2y = 2 - x^2. Where should the ant go if it wants to be as warm as possible?

Answer

(a) The maximum and minimum values of T(x,y)T(x,y) in x2+y2≤4x^2+y^2\le 4 are 2020 (at (0,0)(0,0)) and 44 (at (±2,0)(\pm 2,0)), respectively.

(b) 117<−4,−1>\frac{1}{\sqrt{17}}\llt -4,-1\rgt (c) 1818 (d) (0,2)(0,2)

Full solution

(a) The maximum and minimum can occur either in the interior of the disk or on the boundary of the disk. The boundary of the disk is the circle x2+y2=4x^2+y^2=4.

  • Any absolute max or min in the interior of the disk must also be a local max or min and, since there are no singular points, must also be a critical point of hh. Since Tx=−8xT_x=-8x and Ty=−2yT_y=-2y, the only critical point is (x,y)=(0,0)(x,y)=(0,0), where T=20T=20. Since 4x2+y2≥04x^2+y^2\ge 0, we have T(x,y)=20−4x2−y2≤20T(x,y)=20-4x^2-y^2\le 20. So the maximum value of TT (even in R2\bbbr^2) is 2020.

  • At each point of x2+y2=4x^2+y^2=4 we have T(x,y)=20−4x2−y2=20−4x2−(4−x2)=16−3x2T(x,y)=20-4x^2-y^2=20 -4x^2-(4-x^2)=16-3x^2 with −2≤x≤2-2\le x\le 2. So TT is a minimum when x2x^2 is a maximum. Thus the minimum value of TT on the disk is 16−3(±2)2=416-3(\pm 2)^2=4.

So all together, the maximum and minimum values of T(x,y)T(x,y) in x2+y2≤4x^2+y^2\le 4 are 2020 (at (0,0)(0,0)) and 44 (at (±2,0)(\pm 2,0)), respectively.

(b) To increase its temperature as quickly as possible, the ant should move in the direction of the temperature gradient ∇T(1,1)=<−8x,−2y>∣(x,y)=(1,1)=<−8,−2>\vnabla T(1,1)=\llt -8x,-2y\rgt\Big|_{(x,y)=(1,1)}=\llt -8,-2\rgt. A unit vector in that direction is 117<−4,−1>\frac{1}{\sqrt{17}}\llt -4,-1\rgt.

(c) The ant's rate f increase of temperature (per unit time) is

∇T(1,1)⋅<−2,−1>=<−8,−2>⋅<−2,−1>=18\begin{align*} \vnabla T(1,1)\cdot\llt -2,-1\rgt =\llt -8,-2\rgt\cdot\llt -2,-1\rgt =18 \end{align*}

(d) We are being asked to find the (x,y)=(x,2−x2)(x,y)=(x,2-x^2) which maximizes

T(x,2−x2)=20−4x2−(2−x2)2=16−x4\begin{align*} T\big(x,2-x^2\big) =20 -4x^2-\big(2-x^2\big)^2 = 16-x^4 \end{align*}

The maximum of 16−x416-x^4 is obviously 1616 at x=0x=0. So the ant should go to (0,2−02)=(0,2)\big(0,2-0^2\big)=(0,2).

Q33Stage 3Past exam · M200 2014A

Consider the function

f(x,y)=3kx2y+y3−3x2−3y2+4\begin{equation*} f (x,y) = 3kx^2 y + y^3 - 3x^2 - 3y^2 + 4 \end{equation*}

where k>0k > 0 is a constant. Find and classify all critical points of f(x,y)f(x,y) as local minima, local maxima, saddle points or points of indeterminate type. Carefully distinguish the cases k<12k < \frac{1}{2}, k=12k = \frac{1}{2} and k>12k > \frac{1}{2}.

Answer

*Case k<12k<\frac{1}{2}: *

criticalpoint\Atop{\text{critical}}{\text{point}}type
(0,0)(0,0)local max
(0,2)(0,2)saddle point

*Case k=12k=\frac{1}{2}: *

criticalpoint\Atop{\text{critical}}{\text{point}}type
(0,0)(0,0)local max
(0,2)(0,2)unknown

*Case k>12k>\frac{1}{2}: *

criticalpoint\Atop{\text{critical}}{\text{point}}type
(0,0)(0,0)local max
(0,2)(0,2)local min
(1k3(2k−1) , 1k)\left(\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)saddle point
(−1k3(2k−1) , 1k)\left(-\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)saddle point
Full solution

To find the critical points we will need the gradient of ff and to apply the second derivative test of Theorem 2.9.16 in the CLP-3 text we will need all second order partial derivatives. So we need all partial derivatives of ff up to order two. Here they are.

f=3kx2y+y3−3x2−3y2+4fx=6kxy−6xfxx=6ky−6fxy=6kxfy=3kx2+3y2−6yfyy=6y−6fyx=6kx\begin{alignat*}{3} f&=3kx^2 y + y^3 - 3x^2 - 3y^2 + 4 \\ f_x&=6kxy-6x & f_{xx}&=6ky-6 \qquad & f_{xy}&= 6kx\\ f_y&=3kx^2+3y^2-6y \qquad & f_{yy}&=6y-6\qquad & f_{yx}&= 6kx \end{alignat*}

(Of course, fxyf_{xy} and fyxf_{yx} have to be the same. It is still useful to compute both, as a way to catch some mechanical errors.)

The critical points are the solutions of

fx=6x(ky−1)=0fy=3kx2+3y2−6y=0\begin{equation*} f_x=6x(ky-1)=0 \qquad f_y=3kx^2+3y^2-6y = 0 \end{equation*}

The first equation is satisfied if at least one of x=0x=0, y=1 ⁣/ky=\nicefrac{1}{k} are satisfied. (Recall that k>0k>0.)

  • If x=0x=0, the second equation reduces to 3y(y−2)=03y(y-2)=0, which is satisfied if either y=0y=0 or y=2y=2.

  • If y=1 ⁣/ky=\nicefrac{1}{k}, the second equation reduces to 3kx2+3k2−6k=3kx2+3k2(1−2k)=03kx^2+\frac{3}{k^2}-\frac{6}{k}=3kx^2+\frac{3}{k^2}(1-2k)=0.

*Case k<12k<\frac{1}{2}: * If k<12k<\frac{1}{2}, then 3k2(1−2k)>0\frac{3}{k^2}(1-2k)>0 and the equation 3kx2+3k2(1−2k)=03kx^2+\frac{3}{k^2}(1-2k)=0 has no real solutions. In this case there are two critical points: (0,0)(0,0), (0,2)(0,2) and the classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)(−6)×(−6)−(0)2>0(-6)\times(-6)-(0)^2> 0−6-6local max
(0,2)(0,2)(12k−6)×6−(0)2<0(12k-6)\times 6-(0)^2<0saddle point

*Case k=12k=\frac{1}{2}: * If k=12k=\frac{1}{2}, then 3k2(1−2k)=0\frac{3}{k^2}(1-2k)=0 and the equation 3kx2+3k2(1−2k)=03kx^2+\frac{3}{k^2}(1-2k)=0 reduces to 3kx2=03kx^2=0 which has as its only solution x=0x=0. We have already seen this third critical point, x=0x=0, y=1 ⁣/k=2y=\nicefrac{1}{k}=2. So there are again two critical points: (0,0)(0,0), (0,2)(0,2) and the classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)(−6)×(−6)−(0)2>0(-6)\times(-6)-(0)^2> 0−6-6local max
(0,2)(0,2)(12k−6)×6−(0)2=0(12k-6)\times 6-(0)^2=0unknown

*Case k>12k>\frac{1}{2}: * If k>12k>\frac{1}{2}, then 3k2(1−2k)<0\frac{3}{k^2}(1-2k)<0 and the equation 3kx2+3k2(1−2k)=03kx^2+\frac{3}{k^2}(1-2k)=0 reduces to 3kx2=3k2(2k−1)3kx^2=\frac{3}{k^2}(2k-1) which has two solutions, namely x=±1k3(2k−1)x=\pm\sqrt{\frac{1}{k^3}(2k-1)}.
So there are four critical points: (0,0)(0,0), (0,2)(0,2), (1k3(2k−1) , 1k)\left(\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right) and (−1k3(2k−1) , 1k)\left(-\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right) and the classification is

criticalpoint\Atop{\text{critical}}{\text{point}}fxxfyy−fxy2f_{xx}f_{yy}-f_{xy}^2fxxf_{xx}type
(0,0)(0,0)(−6)×(−6)−(0)2>0(-6)\times(-6)-(0)^2> 0−6-6local max
(0,2)(0,2)(12k−6)×6−(0)2>0(12k-6)\times 6-(0)^2>012k−6>012k-6>0local min
(1k3(2k−1) , 1k)\left(\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)(6−6)×(6k−6)−(>0)2<0(6-6)\times (\frac{6}{k}-6)-(> 0)^2<0saddle point
(−1k3(2k−1) , 1k)\left(-\sqrt{\frac{1}{k^3}(2k-1)}\,,\,\frac{1}{k}\right)(6−6)×(6k−6)−(<0)2<0(6-6)\times (\frac{6}{k}-6)-(< 0)^2<0saddle point
Q34Stage 3Past exam · M200 2002A

  1. Show that the function f(x,y)=2x+4y+1xyf(x,y)=2x+4y+\frac{1}{xy} has exactly one critical point in the first quadrant x>0x>0, y>0y>0, and find its value at that point.

  2. Use the second derivative test to classify the critical point in part (a).

  3. Hence explain why the inequality 2x+4y+1xy≥62x+4y+\frac{1}{xy}\ge 6 is valid for all positive real numbers xx and yy.

Answer

(a) x=1x=1, y=12y=\half, f(1,12)=6f\big(1,\half\big)=6 (b) local minimum

(c) As xx or yy tends to infinity (with the other at least zero), 2x+4y2x+4y tends to +∞+\infty. As (x,y)(x,y) tends to any point on the first quadrant part of the xx- and yy–axes, 1xy\frac{1}{xy} tends to +∞+\infty.
Hence as xx or yy tends to the boundary of the first quadrant (counting infinity as part of the boundary), f(x,y)f(x,y) tends to +∞+\infty. As a result (1,12)\big(1,\half\big) is a global (and not just local) minimum for ff in the first quadrant. Hence f(x,y)≥f(1,12)=6f(x,y)\ge f\big(1,\half\big)=6 for all x,y>0x,y>0.

Full solution

(a) For x,y>0x,y>0,

fx=2−1x2y=0  ⟺  y=12x2fy=4−1xy2=0\begin{align*} f_x&=2-\frac{1}{x^2y}=0\iff y=\frac{1}{2x^2}\cr f_y&=4-\frac{1}{xy^2}=0 \end{align*}

Substituting y=12x2y=\frac{1}{2x^2}, from the first equation, into the second gives 4−4x3=04-4x^3=0 which forces x=1x=1, y=12y=\half. At x=1x=1, y=12y=\half,

f(1,12)=2+2+2=6\begin{equation*} f\big(1,\half\big)=2+2+2=6 \end{equation*}

(b) The second derivatives are

fxx(x,y)=2x3yfxy(x,y)=1x2y2fyy(x,y)=2xy3\begin{equation*} f_{xx}(x,y)=\frac{2}{x^3y}\qquad f_{xy}(x,y)=\frac{1}{x^2y^2}\qquad f_{yy}(x,y)=\frac{2}{xy^3} \end{equation*}

In particular

fxx(1,12)=4fxy(1,12)=4fyy(1,12)=16\begin{equation*} f_{xx}\big(1,\half\big)=4\qquad f_{xy}\big(1,\half\big)=4\qquad f_{yy}\big(1,\half\big)=16 \end{equation*}

Since fxx(1,12)fyy(1,12)−fxy(1,12)2=4×16−42=48>0f_{xx}\big(1,\half\big)f_{yy}\big(1,\half\big)- f_{xy}\big(1,\half\big)^2=4\times 16-4^2=48>0 and fxx(1,12)=4>0f_{xx}\big(1,\half\big)=4>0, the point (1,12)\big(1,\half\big) is a local minimum.

(c) As xx or yy tends to infinity (with the other at least zero), 2x+4y2x+4y tends to +∞+\infty. As (x,y)(x,y) tends to any point on the first quadrant part of the xx- and yy–axes, 1xy\frac{1}{xy} tends to +∞+\infty.
Hence as xx or yy tends to the boundary of the first quadrant (counting infinity as part of the boundary), f(x,y)f(x,y) tends to +∞+\infty. As a result (1,12)\big(1,\half\big) is a global (and not just local) minimum for ff in the first quadrant. Hence f(x,y)≥f(1,12)=6f(x,y)\ge f\big(1,\half\big)=6 for all x,y>0x,y>0.

Q35Stage 3

An experiment yields data points (xi,yi), i=1,2,⋯ ,n.(x_i,y_i),\ i=1,2,\cdots,n. We wish to find the straight line y=mx+by=mx+b which “best" fits the data. The definition of “best" is “minimizes the root mean square error", i.e. minimizes ∑i=1n(mxi+b−yi)2\sum_{i=1}^n (mx_i+b-y_i)^2. Find mm and bb.

Answer

m=nSxy−SxSynSx2−Sx2m=\frac{nS_{xy}-S_xS_y}{nS_{x^2} -S_x^2} and b=SySx2−SxSxynSx2−Sx2b=\frac{S_yS_{x^2}-S_xS_{xy}}{nS_{x^2} -S_x^2} where Sy=∑i=1nyiS_y=\smsum\limits_{i=1}^n y_i, Sx2=∑i=1nxi2S_{x^2}=\smsum\limits_{i=1}^n x^2_i and Sxy=∑i=1nxiyiS_{xy}=\smsum\limits_{i=1}^n x_iy_i.

Full solution

We wish to choose mm and bb so as to minimize the (square of the) rms error E(m,b)=∑i=1n(mxi+b−yi)2E(m,b)=\sum\limits_{i=1}^n (mx_i+b-y_i)^2.

0=∂E∂m=∑i=1n2(mxi+b−yi)xi=m[∑i=1n2xi2]+b[∑i=1n2xi]−[∑i=1n2xiyi]0=∂E∂b=∑i=1n2(mxi+b−yi)=m[∑i=1n2xi]+b[∑i=1n2]−[∑i=1n2yi]\begin{alignat*}{5} 0&=\pdiff{E}{m}&&=\smsum_{i=1}^n 2(mx_i+b-y_i)x_i &&=m\Big[\smsum_{i=1}^n 2x^2_i\Big]+b\Big[\smsum_{i=1}^n 2x_i\Big] -\Big[\smsum_{i=1}^n 2x_iy_i\Big]\\ 0&=\pdiff{E}{b}&&=\smsum_{i=1}^n 2(mx_i+b-y_i) &&=m\Big[\smsum_{i=1}^n 2x_i\Big]+b\Big[\smsum_{i=1}^n 2\Big] -\Big[\smsum_{i=1}^n 2y_i\Big] \end{alignat*}

There are a lot of symbols in those two equations. But remember that only two of them, namely mm and bb, are unknowns. All of the xix_i's and yiy_i's are given data. We can make the equations look a lot less imposing if we define Sx=∑i=1nxiS_x=\smsum_{i=1}^n x_i, Sy=∑i=1nyiS_y=\smsum_{i=1}^n y_i, Sx2=∑i=1nxi2S_{x^2}=\smsum_{i=1}^n x^2_i and Sxy=∑i=1nxiyiS_{xy}=\smsum_{i=1}^n x_iy_i. In terms of this notation, the two equations are (after dividing by two)

Sx2 m+Sx b=SxySx m+n b=Sy\begin{align*} S_{x^2}\, m+S_x\, b&=S_{xy} \tag{\rm 1}\\ S_{x}\,m+n\,b&=S_{y} \tag{\rm 2} \end{align*}

This is a system of two linear equations in two unknowns. One way (This procedure is probably not the most efficient one. But it has the advantage that it always works, it does not require any ingenuity on the part of the solver, and it generalizes easily to larger linear systems of equations.) to solve them, is to use one of the two equations to solve for one of the two unknowns in terms of the other unknown. For example, equation (2) gives that

b=1n(Sy−Sx m)\begin{equation*} b=\frac{1}{n}\big(S_y-S_x\,m\big) \end{equation*}

If we now substitute this into equation (1) we get

Sx2 m+Sxn(Sy−Sx m)=Sxy  ⟹  (Sx2−Sx2n)m=Sxy−SxSyn\begin{equation*} S_{x^2}\, m+\frac{S_x}{n}\big(S_y-S_x\,m\big)=S_{xy} \implies \left(S_{x^2}-\frac{S_x^2}{n}\right)m = S_{xy} -\frac{S_xS_y}{n} \end{equation*}

which is a single equation in the single unkown mm. We can easily solve it for mm. It tells us that

m=nSxy−SxSynSx2−Sx2\begin{equation*} m=\frac{nS_{xy}-S_xS_y}{nS_{x^2} -S_x^2} \end{equation*}

Then substituting this back into b=1n(Sy−Sx m)b=\frac{1}{n}\big(S_y-S_x\,m\big) gives us

b=Syn−Sxn(nSxy−SxSynSx2−Sx2)=SySx2−SxSxynSx2−Sx2\begin{equation*} b=\frac{S_y}{n} -\frac{S_x}{n}\left(\frac{nS_{xy}-S_xS_y}{nS_{x^2} -S_x^2}\right) =\frac{S_yS_{x^2}-S_xS_{xy}}{nS_{x^2} -S_x^2} \end{equation*}

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From the UBC Math 200 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.