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Partial Derivatives

2.10 Lagrange Multipliers

30 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 2010A

  1. Does the function f(x,y)=x2+y2f(x, y) = x^2 +y^2 have a maximum or a minimum on the curve xy=1xy = 1? Explain.

  2. Find all maxima and minima of f(x,y)f(x, y) on the curve xy=1xy = 1.

Answer

(a) ff does not have a maximum. It does have a minimum.

(b) The minima are at ±(1,1)\pm (1,1), where ff takes the value 22.

Full solution

(a) f(x,y)=x2+y2f(x, y) = x^2 +y^2 is the square of the distance from the point (x,y)(x,y) to the origin. There are points on the curve xy=1xy=1 that have either xx or yy arbitrarily large and so whose distance from the origin is arbitrarily large. So ff has no maximum on the curve. On the other hand ff will have a minimum, achieved at the points of xy=1xy=1 that are closest to the origin.

(b) On the curve xy=1xy=1 we have y=1xy=\frac{1}{x} and hence f=x2+1x2f=x^2+\frac{1}{x^2}. As

ddx(x2+1x2)=2x−2x3=2x3(x4−1)\begin{align*} \diff{}{x}\left(x^2+\frac{1}{x^2}\right) =2x-\frac{2}{x^3} =\frac{2}{x^3}(x^4-1) \end{align*}

and as no point of the curve has x=0x=0, the minimum is achieved when x=±1x=\pm 1. So the minima are at ±(1,1)\pm (1,1), where ff takes the value 22.

Q2Stage 1

The surface SS is given by the equation g(x,y,z)=0g(x,y,z)=0. You are walking on SS measuring the function f(x,y,z)f(x,y,z) as you go. You are currently at the point (x0,y0,z0)(x_0,y_0,z_0) where ff takes its largest value on SS, and are walking in the direction d≠0\vd\ne\vZero. Because you are walking on SS, the vector d\vd is tangent to SS at (x0,y0,z0)(x_0,y_0,z_0).

  1. What is the directional derivative of ff at (x0,y0,z0)(x_0,y_0,z_0) in the direction d\vd? Do not use the method of Lagrange multipliers.

  2. What is the directional derivative of ff at (x0,y0,z0)(x_0,y_0,z_0) in the direction d\vd? This time use the method of Lagrange multipliers.

Hint

(a) The function ff decreases, or at least does not increase, as you leave (x0,y0,z0)(x_0,y_0,z_0) in the direction d\vd.
The function ff also decreases, or at least does not increase, as you leave (x0,y0,z0)(x_0,y_0,z_0) in the direction −d-\vd.

Answer

(a), (b) 00

Full solution

(a) As you leave (x0,y0,z0)(x_0,y_0,z_0) walking in the direction d≠0\vd\ne\vZero, ff has to be decreasing, or at least not increasing, because ff takes its largest value on SS at (x0,y0,z0)(x_0,y_0,z_0). So the directional derivative

Dd/∣d∣f(x0,y0,z0)=∇f(x0,y0,z0)⋅d∣d∣≤0\begin{equation*} D_{\vd/|\vd|}f(x_0,y_0,z_0)=\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|}\le 0 \tag{E1}\end{equation*}

As you leave (x0,y0,z0)(x_0,y_0,z_0) walking in the direction −d≠0-\vd\ne\vZero, ff also has to be decreasing, or at least not increasing, because ff still takes its largest value on SS at (x0,y0,z0)(x_0,y_0,z_0). So the directional derivative

D−d/∣d∣f(x0,y0,z0)=−∇f(x0,y0,z0)⋅d∣d∣≤0\begin{equation*} D_{-\vd/|\vd|}f(x_0,y_0,z_0)=-\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|}\le 0 \tag{E2}\end{equation*}

(E1) and (E2) can both be true only if the directional derivative

Dd/∣d∣f(x0,y0,z0)=∇f(x0,y0,z0)⋅d∣d∣=0\begin{equation*} D_{\vd/|\vd|}f(x_0,y_0,z_0)=\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|} = 0 \end{equation*}

(b) By Definition 2.7.5 in the CLP-3 text, the directional derivative is

Dd/∣d∣f(x0,y0,z0)=∇f(x0,y0,z0)⋅d∣d∣\begin{equation*} D_{\vd/|\vd|}f(x_0,y_0,z_0)=\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|} \end{equation*}

  • As (x0,y0,z0)(x_0,y_0,z_0) is a local maximum for ff on SS, the method of Lagrange multipliers, Theorem 2.10.2 in the CLP-3 text, gives that ∇f(x0,y0,z0)=λ∇g(x0,y0,z0)\vnabla f(x_0,y_0,z_0) =\la\vnabla g(x_0,y_0,z_0) for some λ\la.

  • By Theorem 2.5.5, the vector ∇g(x0,y0,z0)\vnabla g(x_0,y_0,z_0) is perpendicular to the surface SS at (x0,y0,z0)(x_0,y_0,z_0), and, in particular, is perpendicular to the vector d\vd, which after all is tangent to the surface SS at (x0,y0,z0)(x_0,y_0,z_0).

So ∇g(x0,y0,z0)⋅d=0\vnabla g(x_0,y_0,z_0)\cdot\vd=0 and the directional derivative

Dd/∣d∣f(x0,y0,z0)=∇f(x0,y0,z0)⋅d∣d∣=0\begin{equation*} D_{\vd/|\vd|}f(x_0,y_0,z_0)=\vnabla f(x_0,y_0,z_0)\cdot\frac{\vd}{|\vd|}=0 \end{equation*}

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q3Stage 2

Find the maximum and minimum values of the function
f(x,y,z)=x+y−zf(x,y,z)=x+y-z on the sphere x2+y2+z2=1x^2+y^2+z^2=1.

Answer

The max is f=3f=\sqrt{3} and the min is f=−3f=-\sqrt{3}.

Full solution

We are to find the maximum and minimum of f(x,y,z)=x+y−zf(x,y,z)=x+y-z subject to the constraint g(x,y,z)=x2+y2+z2−1=0g(x,y,z) = x^2+y^2+z^2 -1=0. According to the method of Lagrange multipliers, we need to find all solutions to

fx=1=2λx=λgx  ⟹  x=12λfy=1=2λy=λgy  ⟹  y=12λfz=−1=2λz=λgz  ⟹  z=−12λx2+y2+z2=1  ⟹  3(12λ)2=1  ⟹  λ=±32\begin{alignat*}{7} f_x = 1 = 2\la x &= \la g_x \quad&&\implies\quad && x=\frac{1}{2\la} \tag{E1} \\ f_y = 1 = 2\la y &= \la g_y \quad&&\implies\quad && y=\frac{1}{2\la}\tag{E2} \\ f_z = -1 = 2\la z &= \la g_z \quad&&\implies\quad && z=-\frac{1}{2\la} \tag{E3} \\ x^2+y^2+z^2&=1 \quad&&\implies\quad &&3\left(\frac{1}{2\la}\right)^2=1 \quad&&\implies\quad & \la&=\pm\frac{\sqrt{3}}{2}\tag{E4} \end{alignat*}

Thus the critical points are (−13,−13,13)\big(-\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\big), where f=−3f=-\sqrt{3} and (13,13,−13)\big(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}}\big), where f=3f=\sqrt{3}. So, the max is f=3f=\sqrt{3} and the min is f=−3f=-\sqrt{3}.

Q4Stage 2

Find a, ba,\ b and cc so that the volume 4π3abc\frac{4\pi}{3} abc of an ellipsoid x2a2+y2b2+z2c2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 passing through the point (1,2,1)(1,2,1) is as small as possible.

Hint

The ellipsoid x2a2+y2b2+z2c2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 passes through the point (1,2,1)(1,2,1) if and only if 1a2+4b2+1c2=1\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2}=1.

Answer

a=c=3, b=23a=c=\sqrt{3},\ b=2\sqrt{3}.

Full solution

The ellipsoid x2a2+y2b2+z2c2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1 passes through the point (1,2,1)(1,2,1) if and only if 1a2+4b2+1c2=1\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2}=1. We are to minimize f(a,b,c)=43πabcf(a,b,c)=\frac{4}{3}\pi abc subject to the constraint that g(a,b,c)=1a2+4b2+1c2−1=0g(a,b,c) = \frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2} -1=0. According to the method of Lagrange multipliers, we need to find all solutions to

fa=43πbc=−2λa3=λga  ⟹  32πλ=−a3bcfb=43πac=−8λb3=λgb  ⟹  32πλ=−14ab3cfc=43πab=−2λc3=λgc  ⟹  32πλ=−abc31a2+4b2+1c2=1\begin{alignat*}{7} f_a = \frac{4}{3}\pi bc = -\frac{2\la}{a^3} &= \la g_a \quad&&\implies\quad && \frac{3}{2\pi}\la=-a^3bc \tag{E1} \\ f_b = \frac{4}{3}\pi ac = - \frac{8\la}{b^3} &= \la g_b \quad&&\implies\quad && \frac{3}{2\pi}\la=-\frac{1}{4}ab^3c \tag{E2} \\ f_c = \frac{4}{3}\pi ab = -\frac{2\la}{c^3} &= \la g_c \quad&&\implies\quad && \frac{3}{2\pi}\la=-abc^3 \tag{E3} \\ \frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2}&=1 \tag{E4} \end{alignat*}

The equations −32πλ=a3bc=14ab3c-\frac{3}{2\pi}\la=a^3bc=\frac{1}{4}ab^3c force b=2ab=2a (since we want a,b,c>0a,b,c>0). The equations −32πλ=a3bc=abc3-\frac{3}{2\pi}\la=a^3bc=abc^3 force a=ca=c. Hence, by (E4),

1=1a2+4b2+1c2=3a2  ⟹  a=c=3, b=23\begin{align*} 1=\frac{1}{a^2}+\frac{4}{b^2}+\frac{1}{c^2}=\frac{3}{a^2} \implies a=c=\sqrt{3},\ b=2\sqrt{3} \end{align*}
Q5Stage 2Past exam · M200 2005D

Use the Method of Lagrange Multipliers to find the minimum value of z=x2+y2z = x^2 + y^2 subject to x2y=1x^2 y = 1. At which point or points does the minimum occur?

Answer

The minimum value is 213+2−23=3223=3432^{\frac{1}{3}} + 2^{-\frac{2}{3}} =\frac{3}{2}\sqrt[3]{2} =\frac{3}{\sqrt[3]{4}} at (±216 , 2−13)\big(\pm 2^{\frac{1}{6}}\,,\, 2^{-\frac{1}{3}}\big).

Full solution

So we are to minimize f(x,y)=x2+y2f(x,y) = x^2+y^2 subject to the constraint g(x,y)=x2y−1=0g(x,y) = x^2 y -1=0. According to the method of Lagrange multipliers, we need to find all solutions to

fx=2x=2λxy=λgxfy=2y=λx2=λgyx2y=1\begin{align*} f_x = 2x &=2 \la xy = \la g_x \tag{E1} \\ f_y = 2y &= \la x^2 = \la g_y \tag{E2} \\ x^2y&=1 \tag{E3} \end{align*}

  • Equation (E1), 2x(1−λy)=02x(1-\la y)=0, gives that either x=0x=0 or λy=1\la y=1.

  • But substituting x=0x=0 in (E3) gives 0=10=1, which is impossible.

  • Also note that λ=0\la=0 is impossible, since substituting λ=0\la=0 in (E1) and (E2) gives x=y=0x=y=0, which violates (E3).

  • So y=1λy=\frac{1}{\la}.

  • Substituting y=1λy=\frac{1}{\la} into (E2) gives 2λ=λx2\frac{2}{\la} = \la x^2 or x2=2λ2x^2=\frac{2}{\la^2}. So x=±2λx=\pm\frac{\sqrt{2}}{\la}.

  • Substituting y=1λy=\frac{1}{\la}, x=±2λx=\pm\frac{\sqrt{2}}{\la} into (E3) gives 2λ3=1\frac{2}{\la^3} =1 or λ3=2\la^3 = 2 or λ=23\la= \sqrt[3]{2}.

  • λ=21/3\la= 2^{1/3} gives x=±212−13=±216x=\pm 2^{\frac{1}{2}-\frac{1}{3}}=\pm 2^{\frac{1}{6}} and y=2−13y= 2^{-\frac{1}{3}}.

So the two critical points are (216 , 2−13)\big(2^{\frac{1}{6}}\,,\,2^{-\frac{1}{3}}\big) and (−216 , 2−13)\big(-2^{\frac{1}{6}}\,,\,2^{-\frac{1}{3}}\big). For both of these critical points,

x2+y2=213+2−23=213+12213=3223=343\begin{equation*} x^2+y^2= 2^{\frac{1}{3}} + 2^{-\frac{2}{3}} = 2^{\frac{1}{3}} + \frac{1}{2}2^{\frac{1}{3}} =\frac{3}{2}\sqrt[3]{2} =\frac{3}{\sqrt[3]{4}} \end{equation*}
Q6Stage 2Past exam · M200 2006A

Use the Method of Lagrange Multipliers to find the radius of the base and the height of a right circular cylinder of maximum volume which can be fit inside the unit sphere x2+y2+z2=1x^2 + y^2 + z^2 = 1.

Answer

radius=23\text{radius}=\sqrt{\frac{2}{3}} and height=23\text{height}=\frac{2}{\sqrt{3}}.

Full solution

Let rr and hh denote the radius and height, respectively, of the cylinder. We can always choose our coordinate system so that the axis of the cylinder is parallel to the zz–axis.

  • If the axis of the cylinder does not lie exactly on the zz–axis, we can enlarge the cylinder sideways. (See the figure on the left below. It shows the y=0y=0 cross–section of the cylinder.) So we can assume that the axis of the cylinder lies on the zz–axis

  • If the top and/or the bottom of the cylinder does not touch the sphere x2+y2+z2=1x^2+y^2+z^2=1, we can enlarge the cylinder vertically. (See the central figure below.)

  • So we may assume that the cylinder is

    { (x,y,z) ∣ x2+y2≤r2, −h/2≤z≤h/2 }\begin{equation*} \Set{(x,y,z)}{x^2+y^2\le r^2,\ -h/2\le z\le h/2} \end{equation*}

    with r2+(h/2)2=1r^2+(h/2)^2=1. See the figure on the right below.

Figure from prob_s2.10, line 286

Figure from prob_s2.10, line 286

Figure from prob_s2.10, line 286

Figure from prob_s2.10, line 286

Figure from prob_s2.10, line 286

Figure from prob_s2.10, line 286

So we are to maximize the volume, f(r,h)=πr2hf(r,h) = \pi r^2 h, of the cylinder subject to the constraint g(r,h)=r2+h24−1=0g(r,h) = r^2+ \frac{h^2}{4} -1=0. According to the method of Lagrange multipliers, we need to find all solutions to

fr=2πrh=2λr=λgrfh=πr2=λh2=λghr2+h24=1\begin{align*} f_r = 2\pi r h &=2 \la r = \la g_r \tag{E1} \\ f_h = \pi r^2 &= \la \frac{h}{2} = \la g_h \tag{E2} \\ r^2+ \frac{h^2}{4}&=1 \tag{E3} \end{align*}

Equation (E1), 2r(πh−λ)=02r(\pi h-\la)=0, gives that either r=0r=0 or λ=πh\la=\pi h. Clearly r=0r=0 cannot give the maximum volume, so λ=πh\la=\pi h. Substituting λ=πh\la=\pi h into (E2) gives

πr2=12πh2  ⟹  r2=h22\begin{equation*} \pi r^2 = \frac{1}{2}\pi h^2 \implies r^2=\frac{h^2}{2} \end{equation*}

Substituting r2=h22r^2=\frac{h^2}{2} into (E3) gives

h22+h24=1  ⟹  h2=43\begin{align*} \frac{h^2}{2} + \frac{h^2}{4} =1 \implies h^2 =\frac{4}{3} \end{align*}

Clearly both rr and hh have to be positive, so h=23h=\frac{2}{\sqrt{3}} and r=23r=\sqrt{\frac{2}{3}}.

Q7Stage 2Past exam · M200 2006D

Use the method of Lagrange Multipliers to find the maximum and minimum values of

f(x,y)=xy\begin{equation*} f(x, y) = xy \end{equation*}

subject to the constraint

x2+2y2=1.\begin{equation*} x^2 + 2y^2 = 1. \end{equation*}
Answer

The maximum and minimum values of ff are 122\frac{1}{2\sqrt{2}} and −122-\frac{1}{2\sqrt{2}}, respectively.

Full solution

For this problem the objective function is f(x,y)=xyf(x,y) = xy and the constraint function is g(x,y)=x2+2y2−1g(x,y)=x^2 + 2y^2 - 1. To apply the method of Lagrange multipliers we need ∇f\vnabla f and ∇g\vnabla g. So we start by computing the first order derivatives of these functions.

fx=yfy=xgx=2xgy=4y\begin{equation*} f_x=y\qquad f_y=x\qquad g_x=2x\qquad g_y=4y \end{equation*}

So, according to the method of Lagrange multipliers, we need to find all solutions to

y=λ(2x)x=λ(4y)x2+2y2−1=0\begin{align*} y&=\la (2x) \tag{E1}\\ x&=\la (4y) \tag{E2}\\ x^2+2y^2-1&=0 \tag{E3} \end{align*}

First observe that none of xx, yy, λ\la can be zero, because if at least one of them is zero, then (E1) and (E2) force x=y=0x=y=0, which violates (E3). Dividing (E1) by (E2) gives yx=x2y\frac{y}{x} = \frac{x}{2y} so that x2=2y2x^2=2y^2 or x=±2 yx=\pm \sqrt{2}\,y. Then (E3) gives

2y2+2y2=1  ⟺  y=±12\begin{align*} 2y^2+2y^2=1 \iff y=\pm\frac{1}{2} \end{align*}

The method of Lagrange multipliers, Theorem 2.10.2 in the CLP-3 text, gives that the only possible locations of the maximum and minimum of the function ff are (±12,±12)\left(\pm\frac{1}{\sqrt{2}},\pm\frac{1}{2}\right). So the maximum and minimum values of ff are 122\frac{1}{2\sqrt{2}} and −122-\frac{1}{2\sqrt{2}}, respectively.

Q8Stage 2Past exam · M200 2008A

Find the maximum and minimum values of f(x,y)=x2+y2f(x,y) = x^2 + y^2 subject to the constraint x4+y4=1x^4 + y^4 = 1.

Answer

min=1=1, max=2=\sqrt{2}.

Full solution

This is a constrained optimization problem with the objective function being f(x,y)=x2+y2f(x,y) = x^2 + y^2 and the constraint function being g(x,y)=x4+y4−1g(x,y) =x^4 + y^4 - 1. By Theorem 2.10.2 in the CLP-3 text, any minimum or maximum (x,y)(x,y) must obey the Lagrange multiplier equations

fx=2x=4λx3=λgxfy=2y=4λy3=λgyx4+y4=1\begin{align*} f_x = 2x &=4 \la x^3 = \la g_x \tag{E1} \\ f_y = 2y &=4 \la y^3 = \la g_y \tag{E2} \\ x^4 + y^4 &= 1 \tag{E3} \end{align*}

for some real number λ\la. By equation (E1), 2x(1−2λx2)=02x(1-2\la x^2)=0, which is obeyed if and only if at least one of x=0x=0, 2λx2=12\la x^2=1 is obeyed. Similarly, by equation (E2), 2y(1−2λy2)=02y(1-2\la y^2)=0, which is obeyed if and only if at least one of y=0y=0, 2λy2=12\la y^2=1 is obeyed.

  • If x=0x=0, (E3) reduces to y4=1y^4=1 or y=±1y=\pm 1. At both (0,±1)\big(0,\pm 1\big) we have f(0,±1)=1f\big(0,\pm1\big)=1.

  • If y=0y=0, (E3) reduces to x4=1x^4=1 or x=±1x=\pm 1. At both (±1,0)\big(\pm 1,0\big) we have f(±1,0)=1f\big(\pm1,0\big)=1.

  • If both xx and yy are nonzero, we have x2=12λ=y2x^2=\frac{1}{2\lambda}=y^2. Then (E3) reduces to

    2x4=1\begin{align*} 2x^4=1 \end{align*}

    so that x2=y2=12x^2=y^2=\frac{1}{\sqrt{2}} and x=±2−1/4x=\pm 2^{-1/4}, y=±2−1/4y=\pm 2^{-1/4}. At all four of these points, we have f=2f=\sqrt{2}.

So the minimum value of ff on x4+y4=1x^4+y^4=1 is 11 and the maximum value of ff on x4+y4=1x^4+y^4=1 is 2\sqrt{2}.

Q9Stage 2Past exam · M200 2008D

Use Lagrange multipliers to find the points on the sphere z2+x2+y2−2y−10=0z^2 + x^2 + y^2 - 2y - 10 = 0 closest to and farthest from the point (1,−2,1)(1, -2, 1).

Answer

(1,−2,1)(1,-2,1) is the closest point. (−1,4,−1)(-1,4,-1) is the farthest point.

Full solution

The function f(x,y,z)=(x−1)2+(y+2)2+(z−1)2f(x,y,z)=(x-1)^2+(y+2)^2+(z-1)^2 gives the square of the distance from the point (x,y,z)(x,y,z) to the point (1,−2,1)(1,-2,1). So it suffices to find the (x,y,z)(x,y,z) which minimizes f(x,y,z)=(x−1)2+(y+2)2+(z−1)2f(x,y,z)=(x-1)^2+(y+2)^2+(z-1)^2 subject to the constraint g(x,y,z)=z2+x2+y2−2y−10=0g(x,y,z) = z^2 + x^2 + y^2 - 2y - 10=0. By Theorem 2.10.2 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the Lagrange multiplier equations

fx=2(x−1)=2λx=λgxfy=2(y+2)=2λ(y−1)=λgyfz=2(z−1)=2λz=λgzz2+x2+y2−2y=10\begin{align*} f_x = 2(x-1) &= 2 \la x = \la g_x \tag{E1} \\ f_y = 2(y+2) &=2 \la (y-1) = \la g_y \tag{E2} \\ f_z = 2(z-1) &= 2\la z = \la g_z \tag{E3} \\ z^2 + x^2 + y^2 - 2y &= 10 \tag{E4} \end{align*}

for some real number λ\la. Now

(E1)  ⟹  x=11−λ(E2)  ⟹  y=−2+λ1−λ(E3)  ⟹  z=11−λ\begin{align*} \text{(E1)} &\implies x=\frac{1}{1-\la} \\ \text{(E2)} &\implies y=-\frac{2+\la}{1-\la} \\ \text{(E3)} &\implies z=\frac{1}{1-\la} \end{align*}

(Note that λ\la cannot be 11, because if it were (E1) would reduce to −2=0-2=0.) Substituting these into (E4), and using that

y−2=−2+λ1−λ−2−2λ1−λ=−4−λ1−λ\begin{equation*} y-2=-\frac{2+\la}{1-\la} -\frac{2-2\la}{1-\la}=-\frac{4-\la}{1-\la} \end{equation*}

gives

1(1−λ)2+1(1−λ)2+2+λ1−λ 4−λ1−λ=10  ⟺  2+(2+λ)(4−λ)=10(1−λ)2  ⟺  11λ2−22λ=0  ⟺  λ=0 or λ=2\begin{align*} &\frac{1}{{(1-\la)}^2}+\frac{1}{{(1-\la)}^2} +\frac{2+\la}{1-\la}\ \frac{4-\la}{1-\la} = 10 \\ \iff & 2+(2+\la)(4-\la) = 10 (1-\la)^2 \\ \iff & 11\la^2 -22\la =0 \\ \iff & \la=0\text{ or }\la=2 \end{align*}

When λ=0\la=0, we have (x,y,z)=(1,−2,1)(x,y,z) = (1,-2,1) (nasty!), which gives distance zero and so is certainly the closest point. When λ=2\la=2, we have (x,y,z)=(−1,4,−1)(x,y,z) = (-1,4,-1), which does not give distance zero and so is certainly the farthest point.

Q10Stage 2Past exam · M200 2009A

Use Lagrange multipliers to find the maximum and minimum values of the function f(x,y,z)=x2+y2−120z2f(x,y,z) = x^2 + y^2 -\frac{1}{20} z^2 on the curve of intersection of the plane x+2y+z=10x + 2y + z = 10 and the paraboloid x2+y2−z=0x^2 + y^2 - z = 0.

Answer

The maximum is 55 and the minimum is 00.

Full solution

We are to maximize and minimize f(x,y,z)=x2+y2−120z2f(x,y,z)=x^2 + y^2 -\frac{1}{20} z^2 subject to the constraints g(x,y,z)=x+2y+z−10=0g(x,y,z)=x + 2y + z - 10=0 and h(x,y,z)=x2+y2−z=0h(x,y,z) = x^2 + y^2 - z=0. By Theorem 2.10.8 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the double Lagrange multiplier equations

fx=2x=λ+2μx=λgx+μhxfy=2y=2λ+2μy=λgy+μhyfz=−z10=λ−μ=λgz+μhzx+2y+z=10x2+y2−z=0\begin{align*} f_x = 2x &=\la + 2 \mu x = \la g_x +\mu h_x\tag{E1} \\ f_y = 2y &=2\la + 2 \mu y = \la g_y +\mu h_y\tag{E2} \\ f_z = -\frac{z}{10} &=\la - \mu = \la g_z +\mu h_z\tag{E3} \\ x + 2y + z &= 10\tag{E4} \\ x^2 + y^2 - z &= 0 \tag{E5} \end{align*}

for some real numbers λ\la and μ\mu.

Equation (E1) gives 2(1−μ)x=λ2(1-\mu)x=\la and equation (E2) gives (1−μ)y=λ(1-\mu)y=\la. So

2(1−μ)x=(1−μ)y  ⟹  (1−μ)(2x−y)=0\begin{align*} 2(1-\mu)x=(1-\mu)y \implies (1-\mu)(2x-y)=0 \end{align*}

So at least one of μ=1\mu=1 and y=2xy=2x must be true.

  • If μ=1\mu=1, equations (E1) and (E2) both reduce to λ=0\la=0 and then the remaining equations reduce to

    −z10=−1x+2y+z=10x2+y2−z=0\begin{align*} -\frac{z}{10} &=-1\tag{E3} \\ x + 2y + z &= 10\tag{E4} \\ x^2 + y^2 - z &= 0 \tag{E5} \end{align*}

    Then (E3) implies z=10z=10, and (E4) in turn implies x+2y+10=10x+2y+10=10 so that x=−2yx=-2y. Finally, substituting z=10z=10 and x=−2yx=-2y into (E5) gives

    4y2+y2−10=0  ⟺  5y2=10  ⟺  y=±2\begin{align*} 4y^2+y^2-10=0 \iff 5y^2=10 \iff y=\pm\sqrt{2} \end{align*}

  • If y=2xy=2x, equations (E4) and (E5) reduce to

    5x+z=105x2−z=0\begin{align*} 5x + z &= 10\tag{E4} \\ 5x^2 - z &= 0 \tag{E5} \end{align*}

    Substituting z=5x2z=5x^2, from (E5), into (E4) gives

    5x2+5x−10=0  ⟺  x2+x−2=0  ⟺  (x+2)(x−1)=0\begin{align*} 5x^2+5x-10=0 \iff x^2+x-2=0 \iff (x+2)(x-1)=0 \end{align*}

    So we have either x=−2x=-2, y=2x=−4y=2x=-4, z=5x2=20z=5x^2=20 or x=1x=1, y=2x=2y=2x=2, z=5x2=5z=5x^2=5. (In both cases, we could now solve (E1) and (E3) for λ\la and μ\mu, but we don't care what the values of λ\la and μ\mu are.)

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

point(−22,2,10)(-2\sqrt{2},\sqrt{2}, 10)(22,−2,10)(2\sqrt{2},-\sqrt{2}, 10)(−2,−4,20)(-2,-4,20)(1,2,5)(1,2,5)
value of ff8+2−58+2-58+2−58+2-54+16−204+16-201+4−25201+4-\frac{25}{20}
maxmaxmin

So the maximum is 55 and the minimum is 00.

Q11Stage 2Past exam · M200 2010D

Find the point P=(x,y,z)P = (x, y, z) (with xx, yy and z>0z> 0) on the surface x3y2z=63x^3 y^2 z = 6 \sqrt{3} that is closest to the origin.

Answer

(3 , 2 , 1)\big(\sqrt{3}\,,\,\sqrt{2}\,,\,1\big)

Full solution

The function f(x,y,z)=x2+y2+z2f(x,y,z)=x^2+y^2+z^2 gives the square of the distance from the point (x,y,z)(x,y,z) to the origin. So it suffices to find the (x,y,z)(x,y,z) (in the first octant) which minimizes f(x,y,z)=x2+y2+z2f(x,y,z)=x^2+y^2+z^2 subject to the constraint g(x,y,z)=x3y2z−63=0g(x,y,z) = x^3y^2z -6\sqrt{3}=0. To start, we'll find the minimizers in all of R3\bbbr^3. By Theorem 2.10.2 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the Lagrange multiplier equations

fx=2x=3λx2y2z=λgxfy=2y=2λx3yz=λgyfz=2z=λx3y2=λgzx3y2z=63\begin{align*} f_x = 2x &= 3 \la x^2y^2z = \la g_x \tag{E1} \\ f_y = 2y &=2 \la x^3 yz = \la g_y \tag{E2} \\ f_z = 2z &= \la x^3y^2 = \la g_z \tag{E3} \\ x^3y^2z &= 6\sqrt{3} \tag{E4} \end{align*}

for some real number λ\la.

Multiplying (E1) by 2x2x, (E2) by 3y3y, and (E3) by 6z6z gives

4x2=6λx3y2z6y2=6λx3y2z12z2=6λx3y2z\begin{align*} 4x^2 &= 6 \la x^3y^2z \tag{E1'} \\ 6y^2 &= 6 \la x^3y^2z \tag{E2'} \\ 12z^2&= 6 \la x^3y^2z \tag{E3'} \end{align*}

The three right hand sides are all identical. So the three left hand sides must all be equal.

4x2=6y2=12z2  ⟺  x=±3 z, y=±2 z\begin{equation*} 4x^2=6y^2=12z^2 \iff x=\pm\sqrt{3}\, z,\ y=\pm\sqrt{2}\, z \end{equation*}

Equation (E4) forces xx and zz to have the same sign. So we must have x=3 zx=\sqrt{3}\,z and y=±2 zy=\pm \sqrt{2}\,z. Substituting this into (E4) gives

(3 z)3(±2 z)2z=63  ⟺  z6=1  ⟺  z=±1\begin{align*} \big(\sqrt{3}\,z\big)^3 \big(\pm \sqrt{2}\,z\big)^2 z=6\sqrt{3} \iff z^6=1 \iff z=\pm 1 \end{align*}

So our minimizer (in all of R3\bbbr^3) must be one of (3 , ±2 , 1)\big(\sqrt{3}\,,\,\pm\sqrt{2}\,,\,1\big) or (−3 , ±2 , −1)\big(-\sqrt{3}\,,\,\pm\sqrt{2}\,,\,-1\big). All of these points give exactly the same value of ff (namely 3+2+1=63+2+1=6). That is all four points are a distance 6\sqrt{6} from the origin and all other points on x3y2z=63x^3y^2z=6\sqrt{3} have distance from the origin strictly greater than 6\sqrt{6}. So the first octant point on x3y2z=63x^3y^2z=6\sqrt{3} that is closest to the origin is (3 , 2 , 1)\big(\sqrt{3}\,,\,\sqrt{2}\,,\,1\big).

Q12Stage 2Past exam · M200 2011A

Find the maximum value of f(x,y,z)=xyzf (x, y, z) = xyz on the ellipsoid

g(x,y,z)=x2+xy+y2+3z2=9\begin{equation*} g(x, y, z) = x^2 + xy + y^2 + 3z^2 = 9 \end{equation*}

Specify all points at which this maximum value occurs.

Answer

The maximum is 66 and is achieved at (6 , −6 , −1)\big(\sqrt{6}\,,\,-\sqrt{6}\,,\,-1\big) and (−6 , 6 , −1)\big(-\sqrt{6}\,,\,\sqrt{6}\,,\,-1\big).

Full solution

This is a constrained optimization problem with the objective function being

f(x,y,z)=xyz\begin{equation*} f(x,y,z) = xyz \end{equation*}

and the constraint function being

G(x,y,z)=x2+xy+y2+3z2−9\begin{equation*} G(x,y,z) =x^2 + xy + y^2 + 3z^2 - 9 \end{equation*}

By Theorem 2.10.2 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the Lagrange multiplier equations

fx=yz=λ(2x+y)=λGxfy=xz=λ(2y+x)=λGyfz=xy=6λz=λGzx2+xy+y2+3z2=9\begin{align*} f_x = yz &= \la (2x+y) = \la G_x \tag{E1} \\ f_y = xz &= \la (2y+x) = \la G_y \tag{E2} \\ f_z = xy &= 6\la z = \la G_z \tag{E3} \\ x^2 + xy + y^2 + 3z^2 &= 9 \tag{E4} \end{align*}

for some real number λ\la.

  • If λ=0\la=0, then, by (E1), yz=0yz=0 so that f(x,y,z)=xyz=0f(x,y,z)=xyz=0. This cannot possibly be the maximum value of ff because there are points (x,y,z)(x,y,z) on g(x,y,z)=9g(x,y,z)=9 (for example x=y=1x=y=1, z=2z=\sqrt{2}) with f(x,y,z)>0f(x,y,z)>0.

  • If λ≠0\la\ne 0, then multiplying (E1) by xx, (E2) by yy, and (E3) by zz gives

    xyz=λ(2x2+xy)=λ(2y2+xy)=6λz2  ⟹  2x2+xy=2y2+xy=6z2  ⟹  x=±y, z2=16(2x2+xy)\begin{align*} xyz = \la (2x^2+xy) = \la(2y^2 +xy) =6\la z^2 &\implies 2x^2+xy =2y^2 +xy =6z^2 \\ &\implies x=\pm y,\ z^2=\frac{1}{6}(2x^2+xy) \end{align*}

    • If x=yx=y, then z2=x22z^2=\frac{x^2}{2} and, by (E4)

      x2+x2+x2+32x2=9  ⟹  x2=2  ⟹  x=y=±2, z=±1\begin{align*} x^2+x^2+x^2 +\frac{3}{2}x^2=9 \implies x^2=2 \implies x=y=\pm \sqrt{2},\ z=\pm 1 \end{align*}

      For these points

      f(x,y,z)=2z={2if z=1−2if z=−1\begin{equation*} f(x,y,z)=2z=\begin{cases} 2&\text{if }z=1 \\ -2&\text{if }z=-1 \end{cases} \end{equation*}

    • If x=−yx=-y, then z2=x26z^2=\frac{x^2}{6} and, by (E4)

      x2−x2+x2+x22=9  ⟹  x2=6  ⟹  x=−y=±6, z=±1\begin{align*} x^2-x^2+x^2 +\frac{x^2}{2}=9 \implies x^2=6 \implies x=-y=\pm \sqrt{6},\ z=\pm 1 \end{align*}

      For these points

      f(x,y,z)=−6z={−6if z=16if z=−1\begin{equation*} f(x,y,z)=-6z=\begin{cases} -6&\text{if }z=1 \\ 6&\text{if }z=-1 \end{cases} \end{equation*}

So the maximum is 66 and is achieved at (6 , −6 , −1)\big(\sqrt{6}\,,\,-\sqrt{6}\,,\,-1\big) and (−6 , 6 , −1)\big(-\sqrt{6}\,,\,\sqrt{6}\,,\,-1\big).

Q13Stage 2Past exam · M200 2011D

Find the radius of the largest sphere centred at the origin that can be inscribed inside (that is, enclosed inside) the ellipsoid

2(x+1)2+y2+2(z−1)2=8\begin{equation*} 2(x+1)^2 + y^2 + 2(z-1)^2 =8 \end{equation*}
Answer

6−42≈0.59\sqrt{6-4\sqrt{2}}\approx 0.59

Full solution

In order for a sphere of radius rr centred on the origin to be enclosed in the ellipsoid, every point of the ellipsoid must be at least a distance rr from the origin. So the largest allowed rr is the distance from the origin to the nearest point on the ellipsoid.

We have to minimize f(x,y,z)=x2+y2+z2f(x,y,z)=x^2+y^2+z^2 subject to the constraint g(x,y,z)=2(x+1)2+y2+2(z−1)2−8g(x,y,z) = 2(x+1)^2 + y^2 + 2(z-1)^2 -8. By Theorem 2.10.2 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the Lagrange multiplier equations

fx=2x=4λ(x+1)=λgxfy=2y=2λy=λgyfz=2z=4λ(z−1)=λgz2(x+1)2+y2+2(z−1)2=8\begin{align*} f_x = 2x &=4 \la (x+1) = \la g_x \tag{E1} \\ f_y = 2y &=2 \la y = \la g_y \tag{E2} \\ f_z = 2z &=4 \la (z-1) = \la g_z \tag{E3} \\ 2(x+1)^2 + y^2 + 2(z-1)^2 &= 8 \tag{E4} \end{align*}

for some real number λ\la.

By equation (E2), 2y(1−λ)=02y(1-\la)=0, which is obeyed if and only if at least one of y=0y=0, λ=1\la=1 is obeyed.

  • If y=0y=0, the remaining equations reduce to

    x=2λ(x+1)z=2λ(z−1)(x+1)2+(z−1)2=4\begin{align*} x &=2 \la (x+1) \tag{E1} \\ z &=2 \la (z-1) \tag{E3} \\ (x+1)^2 + (z-1)^2 &= 4 \tag{E4} \end{align*}

    Note that 2λ2\la cannot be 11 — if it were, (E1) would reduce to 0=10=1. So equation (E1) gives

    x=2λ1−2λorx+1=11−2λ\begin{align*} x = \frac{2\la}{1-2\la}\qquad\text{or}\qquad x+1 = \frac{1}{1-2\la} \end{align*}

    Equation (E3) gives

    z=−2λ1−2λorz−1=−11−2λ\begin{align*} z = -\frac{2\la}{1-2\la}\qquad\text{or}\qquad z-1 = -\frac{1}{1-2\la} \end{align*}

    Substituting x+1=11−2λx+1 = \frac{1}{1-2\la} and z−1=−11−2λz-1 = -\frac{1}{1-2\la} into (E4) gives

    1(1−2λ)2+1(1−2λ)2=4  ⟺  1(1−2λ)2=2  ⟺  11−2λ=±2\begin{align*} \frac{1}{(1-2\la)^2} + \frac{1}{(1-2\la)^2} =4 &\iff \frac{1}{(1-2\la)^2} = 2 \\ &\iff \frac{1}{1-2\la} =\pm\sqrt{2} \end{align*}

    So we now have two candidates for the location of the max and min, namely (x,y,z)=(−1+2,0,1−2)(x,y,z) = \big(-1 + \sqrt{2}, 0, 1-\sqrt{2}\big) and (x,y,z)=(−1−2,0,1+2)(x,y,z) = \big(-1 - \sqrt{2}, 0, 1+\sqrt{2}\big).

  • If λ=1\la=1, the remaining equations reduce to

    x=2(x+1)z=2(z−1)2(x+1)2+y2+2(z−1)2=8\begin{align*} x &=2 (x+1) \tag{E1} \\ z &=2 (z-1) \tag{E3} \\ 2(x+1)^2 + y^2 + 2(z-1)^2 &= 8 \tag{E4} \end{align*}

    Equation (E1) gives x=−2x=-2 and equation (E3) gives z=2z=2. Substituting these into (E4) gives

    2+y2+2=8  ⟺  y2=4  ⟺  y=±2\begin{align*} 2+y^2+2=8 \iff y^2=4 \iff y=\pm 2 \end{align*}

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

point(−1+2,0,1−2)\big(-1 + \sqrt{2}, 0, 1-\sqrt{2}\big)(−1−2,0,1+2)\big(-1 - \sqrt{2}, 0, 1+\sqrt{2}\big)(−2,2,2)(-2,2,2)(−2,−2,2)(-2,-2,2)
value of ff2(3−22)2\big(3-2\sqrt{2}\big)2(3+22)2\big(3+2\sqrt{2}\big)12121212
minmaxmax

Recalling that f(x,y,z)f(x,y,z) is the square of the distance from (x,y,z)(x,y,z) to the origin, the maximum allowed radius for the enclosed sphere is 6−42≈0.59\sqrt{6-4\sqrt{2}}\approx 0.59.

Q14Stage 2Past exam · M200 2012a

Let CC be the intersection of the plane x+y+z=2x + y + z = 2 and the sphere x2+y2+z2=2x^2 + y^2 + z^2 = 2.

  1. Use Lagrange multipliers to find the maximum value of f(x,y,z)=zf(x, y, z) = z on CC.

  2. What are the coordinates of the lowest point on CC?

Answer

(a) 43\frac{4}{3} (b) (1,1)(1,1)

Full solution

(a) We are to maximize f(x,y,z)=zf(x,y,z)=z subject to the constraints g(x,y,z)=x+y+z−2=0g(x,y,z)=x+y+z-2=0 and h(x,y,z)=x2+y2+z2−2=0h(x,y,z) = x^2 + y^2 + z^2 -2=0. By Theorem 2.10.8 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the double Lagrange multiplier equations

fx=0=λ+2μx=λgx+μhxfy=0=λ+2μy=λgy+μhyfz=1=λ+2μz=λgz+μhzx+y+z=2x2+y2+z2=2\begin{align*} f_x = 0 &=\la + 2 \mu x = \la g_x +\mu h_x\tag{E1} \\ f_y = 0 &=\la + 2 \mu y = \la g_y +\mu h_y\tag{E2} \\ f_z = 1 &=\la + 2 \mu z = \la g_z +\mu h_z\tag{E3} \\ x + y + z &= 2\tag{E4} \\ x^2 + y^2 + z^2 &= 2 \tag{E5} \end{align*}

for some real numbers λ\la and μ\mu. Subtracting (E2) from (E1) gives 2μ(x−y)=02\mu(x-y)=0. So at least one of μ=0\mu=0 and y=xy=x must be true.

  • If μ=0\mu=0, equations (E1) and (E3) reduce to λ=0\la=0 and λ=1\la=1, which is impossible. So μ≠0\mu\ne 0.

  • If y=xy=x, equations (E2) through (E5) reduce to

    λ+2μx=0λ+2μz=12x+z=22x2+z2=2\begin{align*} \la + 2 \mu x &= 0 \tag{E2} \\ \la + 2 \mu z &= 1\tag{E3} \\ 2x + z &= 2\tag{E4} \\ 2x^2 + z^2 &= 2 \tag{E5} \end{align*}

    By (E4), x=2−z2x=\frac{2-z}{2}. Substituting this into (E5) gives

    2(2−z)24+z2=2  ⟺  (2−z)2+2z2=4  ⟺  3z2−4z=0  ⟺  z=0, 43\begin{align*} 2\frac{(2-z)^2}{4} +z^2 =2 &\iff (2-z)^2 +2z^2 = 4 \iff 3z^2-4z=0 \\ &\iff z = 0,\ \frac{4}{3} \end{align*}

The maximum zz is thus 43\frac{4}{3}.

(b) Presumably the “lowest point” is the point with the minimal zz–coordinate. By our work in part (a), we have that the minimal value of zz on CC is 00. We have also already seen in part (a) that y=xy=x. When z=0z=0, (E4) reduces to 2x=22x=2. So the desired point is (1,1)(1,1).

Q15Stage 2Past exam · M200 2012D

  1. Use Lagrange multipliers to find the extreme values of

    f(x,y,z)=(x−2)2+(y+2)2+(z−4)2\begin{equation*} f (x, y, z) = (x - 2)^2 + (y + 2)^2 + (z - 4)^2 \end{equation*}

    on the sphere x2+y2+z2=6x^2 + y^2 + z^2 = 6.

  2. Find the point on the sphere x2+y2+z2=6x^2 + y^2 + z^2 = 6 that is farthest from the point (2,−2,4)(2, -2, 4).

Answer

(a) The min is 66 and the max is 5454. (b) (−1,1,−2)(-1,1,-2)

Full solution

(a) This is a constrained optimization problem with the objective function being f(x,y,z)=(x−2)2+(y+2)2+(z−4)2f(x,y,z) = (x - 2)^2 + (y + 2)^2 + (z - 4)^2 and the constraint function being g(x,y,z)=x2+y2+z2−6g(x,y,z) =x^2 + y^2 + z^2 - 6. By Theorem 2.10.2 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the Lagrange multiplier equations

fx=2(x−2)=2λx=λgxfy=2(y+2)=2λy=λgyfz=2(z−4)=2λz=λgzx2+y2+z2=6\begin{align*} f_x = 2(x-2) &=2 \la x = \la g_x \tag{E1} \\ f_y = 2(y+2) &=2 \la y = \la g_y \tag{E2} \\ f_z = 2(z-4) &=2 \la z = \la g_z \tag{E3} \\ x^2 + y^2 + z^2 &= 6 \tag{E4} \end{align*}

for some real number λ\la. Simplifying

x−2=λxy+2=λyz−4=λzx2+y2+z2=6\begin{align*} x-2 &= \la x \tag{E1} \\ y+2 &= \la y \tag{E2} \\ z-4 &= \la z \tag{E2} \\ x^2 + y^2 + z^2 &= 6 \tag{E4} \end{align*}

Note that we cannot have λ=1\la=1, because then (E1) would reduce to −2=0-2=0. Substituting x=21−λx=\frac{2}{1-\la}, from (E1), and y=−21−λy=\frac{-2}{1-\la}, from (E2), and z=41−λz=\frac{4}{1-\la}, from (E3), into (E4) gives

4(1−λ)2+4(1−λ)2+16(1−λ)2=6  ⟺  (1−λ)2=4  ⟺  1−λ=±2\begin{align*} \frac{4}{(1-\la)^2} + \frac{4}{(1-\la)^2} + \frac{16}{(1-\la)^2} =6 \iff (1-\la)^2=4 \iff 1-\la =\pm 2 \end{align*}

and hence

(x,y,z)=±(2,−2,4)2=±(1,−1,2)\begin{equation*} (x,y,z) = \pm \frac{(2,-2,4)}{2}= \pm (1,-1,2) \end{equation*}

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

point(1,−1,2)(1,-1,2)−(1,−1,2)-(1,-1,2)
value of ff665454
minmax

So the minimum is 66 and the maximum is 5454.

(b) f(x,y,z)f(x,y,z) is the square of the distance from (x,y,z)(x,y,z) to (2,−2,4)(2,-2,4). So the point on the sphere x2+y2+z2=6x^2 + y^2 + z^2 = 6 that is farthest from the point (2,−2,4)(2, -2, 4) is the point from part (a) that maximizes ff, which is (−1,1,−2)(-1,1,-2).

Q16Stage 2Past exam · M200 2013D

  1. Find the minimum of the function

    f(x,y,z)=(x−2)2+(y−1)2+z2\begin{equation*} f(x,y,z) = (x-2)^2 + (y-1)^2 + z^2 \end{equation*}

    subject to the constraint x2+y2+z2=1x^2 + y^2 + z^2 = 1, using the method of Lagrange multipliers.

  2. Give a geometric interpretation of this problem.

Answer

(a) (5−1)2=6−25\big(\sqrt{5}-1\big)^2=6-2\sqrt{5}

(b) The minimum of ff subject to the constraint x2+y2+z2=1x^2+y^2+z^2=1 is the square of the distance from (2,1,0)(2,1,0) to the point on the sphere x2+y2+z2=1x^2+y^2+z^2=1 that is nearest (2,1,0)(2,1,0).

Full solution

(a) This is a constrained optimization problem with the objective function being f(x,y,z)=(x−2)2+(y−1)2+z2f(x,y,z) = (x-2)^2 + (y-1)^2 + z^2 and the constraint function being g(x,y,z)=x2+y2+z2−1g(x,y,z) =x^2 + y^2 + z^2 - 1. By Theorem 2.10.2 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the Lagrange multiplier equations

fx=2(x−2)=2λx=λgxfy=2(y−1)=2λy=λgyfz=2z=2λz=λgzx2+y2+z2=1\begin{align*} f_x = 2(x-2) &=2 \la x = \la g_x \tag{E1} \\ f_y = 2(y-1) &=2 \la y = \la g_y \tag{E2} \\ f_z = 2z &=2 \la z = \la g_z \tag{E3} \\ x^2 + y^2 + z^2 &= 1 \tag{E4} \end{align*}

for some real number λ\la. By equation (E3), 2z(1−λ)=02z(1-\la)=0, which is obeyed if and only if at least one of z=0z=0, λ=1\la=1 is obeyed.

  • If z=0z=0 and λ≠1\la\ne 1, the remaining equations reduce to

    x−2=λxy−1=λyx2+y2=1\begin{align*} x-2 &= \la x \tag{E1} \\ y-1 &= \la y \tag{E2} \\ x^2 + y^2 &= 1 \tag{E4} \end{align*}

    Substituting x=21−λx=\frac{2}{1-\la}, from (E1), and y=11−λy=\frac{1}{1-\la}, from (E2), into (E3) gives

    4(1−λ)2+1(1−λ)2=1  ⟺  (1−λ)2=5  ⟺  1−λ=±5\begin{align*} \frac{4}{(1-\la)^2} + \frac{1}{(1-\la)^2} =1 \iff (1-\la)^2=5 \iff 1-\la =\pm\sqrt{5} \end{align*}

    and hence

    (x,y,z)=±15(2,1,0)\begin{equation*} (x,y,z) = \pm\frac{1}{\sqrt{5}}(2,1,0) \end{equation*}

    To aid in the evaluation of f(x,y,z)f(x,y,z) at these points note that, at these points,

    x−2=λx=2λ1−λ,y−1=λy=λ1−λ  ⟹  f(x,y,z)=4λ2(1−λ)2+λ2(1−λ)2=5λ2(1−λ)2=λ2=(1∓5)2\begin{align*} &x-2=\la x = \frac{2\la}{1-\la},\qquad y-1=\la y = \frac{\la}{1-\la} \\ &\implies f(x,y,z) =\frac{4\la^2}{(1-\la)^2} + \frac{\la^2}{(1-\la)^2} =\frac{5\la^2}{(1-\la)^2} =\la^2 =\big(1\mp\sqrt{5}\big)^2 \end{align*}

  • If λ=1\la=1, the remaining equations reduce to

    x−2=xy−1=yx2+y2+z2=1\begin{align*} x-2 &=x \tag{E1} \\ y -1 &=y \tag{E2} \\ x^2 +y^2 + z^2 &= 1 \tag{E4} \end{align*}

    Since −2≠0-2\ne 0 and −1≠0-1\ne 0, neither (E1) nor (E2) has any solution.

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

point15(2,1,0)\frac{1}{\sqrt{5}}(2,1,0)−15(2,1,0)-\frac{1}{\sqrt{5}}(2,1,0)
value of ff(1−5)2\big(1-\sqrt{5}\big)^2(1+5)2\big(1+\sqrt{5}\big)^2
minmax

So the minimum is (5−1)2=6−25\big(\sqrt{5}-1\big)^2=6-2\sqrt{5}.

(b) The function f(x,y,z)=(x−2)2+(y−1)2+z2f(x,y,z) = (x-2)^2 + (y-1)^2 + z^2 is the square of the distance from the point (x,y,z)(x,y,z) to the point (2,1,0)(2,1,0). So the minimum of ff subject to the constraint x2+y2+z2=1x^2+y^2+z^2=1 is the square of the distance from (2,1,0)(2,1,0) to the point on the sphere x2+y2+z2=1x^2+y^2+z^2=1 that is nearest (2,1,0)(2,1,0).

Q17Stage 2Past exam · M200 2015D

Use Lagrange multipliers to find the minimum and maximum values of (x+z)ey(x + z)e^y subject to x2+y2+z2=6x^2 + y^2 + z^2 = 6.

Answer

The maximum value is 2e22e^2 and the minimum value is −2e2-2 e^2.

Full solution

For this problem the objective function is f(x,y,z)=(x+z)eyf(x,y,z) = (x + z)e^y and the constraint function is g(x,y,z)=x2+y2+z2−6g(x,y,z)=x^2 + y^2 + z^2 -6. To apply the method of Lagrange multipliers we need ∇f\vnabla f and ∇g\vnabla g. So we start by computing the first order derivatives of these functions.

fx=eyfy=(x+z)eyfz=eygx=2xgy=2ygz=2z\begin{equation*} f_x=e^y\qquad f_y=(x+z)e^y\qquad f_z=e^y\qquad g_x=2x\qquad g_y=2y\qquad g_z=2z \end{equation*}

So, according to the method of Lagrange multipliers, we need to find all solutions to

ey=λ(2x)(x+z)ey=λ(2y)ey=λ(2z)x2+y2+z2−6=0\begin{align*} e^y&=\la (2x) \tag{E1}\\ (x+z)e^y&=\la (2y) \tag{E2}\\ e^y&=\la (2z) \tag{E3}\\ x^2+y^2+z^2-6&=0 \tag{E4} \end{align*}

First notice that, since ey≠0e^y\ne 0, equation (E1) guarantees that λ≠0\la\ne 0 and x≠0x\ne 0 and equation (E3) guarantees that z≠0z\ne 0 too.

  • So dividing (E1) by (E3) gives xz=1\frac{x}{z}=1 and hence x=zx=z.

  • Then subbing x=zx=z into (E2) gives 2zey=λ(2y)2z e^y = \la(2y). Dividing this equation by (E3) gives 2z=yz2z=\frac{y}{z} or y=2z2y=2z^2.

  • Then subbing x=zx=z and y=2z2y=2z^2 into (E4) gives

    z2+4z4+z2−6=0  ⟺  4z4+2z2−6=0  ⟺  (2z2+3)(2z2−2)=0\begin{align*} z^2+4z^4+z^2-6=0 \iff 4z^4 +2z^2 -6 = 0 \iff (2z^2+3)(2z^2-2) =0 \end{align*}

  • As 2z2+3>02z^2+3>0, we must have 2z2−2=02z^2-2=0 or z=±1z=\pm 1.

Recalling that x=zx=z and y=2z2y=2z^2, the method of Lagrange multipliers, Theorem 2.10.2 in the CLP-3 text, gives that the only possible locations of the maximum and minimum of the function ff are (1,2,1)(1,2,1) and (−1,2,−1)(-1,2,-1). To complete the problem, we only have to compute ff at those points.

point(1,2,1)(1,2,1)(−1,2,−1)(-1,2,-1)
value of ff2e22e^2−2e2-2e^2
maxmin

Hence the maximum value of (x+z)ey(x + z)e^y on x2+y2+z2=6x^2 + y^2 + z^2 = 6 is 2e22e^2 and the minimum value is −2e2-2 e^2.

Q18Stage 2Past exam · M200 2004A

Find the points on the ellipse 2x2+4xy+5y2=302x^2 + 4xy + 5y^2 = 30 which are closest to and farthest from the origin.

Answer

The farthest points are ±6(−2,1)\pm\sqrt{6}(-2,1). The nearest points are ±(1,2)\pm(1,2).

Full solution

Let (x,y)(x,y) be a point on 2x2+4xy+5y2=302x^2 + 4xy + 5y^2 = 30. We wish to maximize and minimize x2+y2x^2+y^2 subject to 2x2+4xy+5y2=302x^2 + 4xy + 5y^2 = 30. Define L(x,y,λ)=x2+y2−λ(2x2+4xy+5y2−30)L(x,y,\la)=x^2+y^2-\la(2x^2 + 4xy + 5y^2 - 30). Then

0=Lx=2x−λ(4x+4y)  ⟹  (1−2λ)x−2λy=00=Ly=2y−λ(4x+10y)  ⟹  −2λx+(1−5λ)y=00=Lλ=2x2+4xy+5y2−30\begin{alignat*}{5} 0&=L_x=2x-\la (4x+4y)\qquad&&\implies\qquad &(1-2\la)x-2\la y&=0\tag{1} \\ 0&=L_y=2y-\la (4x+10y)\qquad&&\implies& -2\la x+(1-5\la)y&=0\tag{2} \\ 0&=L_\la=2x^2 + 4xy + 5y^2 - 30 \end{alignat*}

Note that λ\la cannot be zero because if it is, (1) forces x=0x=0 and (2) forces y=0y=0, but (0,0)(0,0) is not on the ellipse. So equation (1) gives y=1−2λ2λxy=\frac{1-2\la}{2\la}x. Substituting this into equation (2) gives −2λx+(1−5λ)(1−2λ)2λx=0-2\la x+\frac{(1-5\la)(1-2\la)}{2\la}x=0. To get a nonzero (x,y)(x,y) we need

−2λ+(1−5λ)(1−2λ)2λ=0  ⟺  0=−4λ2+(1−5λ)(1−2λ)=6λ2−7λ+1=(6λ−1)(λ−1)\begin{equation*} -2\la +\frac{(1-5\la)(1-2\la)}{2\la}=0\iff 0=-4\la^2+(1-5\la)(1-2\la) =6\la^2-7\la+1=(6\la-1)(\la-1) \end{equation*}

So λ\la must be either 11 or 16\frac{1}{6}. Substituting these into either (1) or (2) gives

λ=1  ⟹  −x−2y=0  ⟹  x=−2y  ⟹  8y2−8y2+5y2=30  ⟹  y=±6λ=16  ⟹  23x−13y=0  ⟹  y=2x  ⟹  2x2+8x2+20x2=30  ⟹  x=±1\begin{alignat*}{7} \la&=1&&\implies -x-2y&=0 &&\implies x&=-2y &&\implies 8y^2-8y^2+5y^2&=30 &&\implies y&=\pm \sqrt{6}\\ \la&=\frac{1}{6}&&\implies \frac{2}{3} x-\frac{1}{3} y&=0 &&\implies y&=2x &&\implies 2x^2+8x^2+20x^2&=30 &&\implies x&=\pm 1 \end{alignat*}

The farthest points are ±6(−2,1)\pm\sqrt{6}(-2,1). The nearest points are ±(1,2)\pm(1,2).

Q19Stage 2

Find the ends of the major and minor axes of the ellipse 3x2−2xy+3y2=43x^2-2xy+3y^2=4.

Hint

The ends of the major axes are the points on the ellipse which are farthest from the origin. The ends of the minor axes are the points on the ellipse which are closest to the origin.

Answer

The ends of the minor axes are ±(12,−12)\pm\big(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\big). The ends of the major axes are ±(1,1)\pm(1,1).

Full solution

Let (x,y)(x,y) be a point on 3x2−2xy+3y2=43x^2-2xy+3y^2=4. This point is at the end of a major axis when it maximizes its distance from the centre, (0,0)(0,0), of the ellipse. It is at the end of a minor axis when it minimizes its distance from (0,0)(0,0). So we wish to maximize and minimize f(x,y)=x2+y2f(x,y)=x^2+y^2 subject to the constraint g(x,y)=3x2−2xy+3y2−4=0g(x,y)=3x^2-2xy+3y^2-4=0. According to the method of Lagrange multipliers, we need to find all solutions to

fx=2x=λ(6x−2y)−=λgx  ⟹  (1−3λ)x+λy=0fy=2y=λ(−2x+6y)=λgy  ⟹  λx+(1−3λ)y=03x2−2xy+3y2=4\begin{alignat*}{5} f_x = 2x = \la (6x-2y)\phantom{-} &= \la g_x \quad&&\implies\quad & (1-3\la)x+\la y&=0 \tag{E1} \\ f_y = 2y = \la (-2x+6y) &= \la g_y \quad&&\implies\quad & \la x+(1-3\la)y&=0\tag{E2} \\ 3x^2-2xy+3y^2&=4 \tag{E3} \end{alignat*}

To start, let's concentrate on the first two equations. Pretend for a couple of minutes, that we already know the value of λ\la and are trying to find xx and yy. The system of equations (1−3λ)x+λy=0(1-3\la)x+\la y=0, λx+(1−3λ)y=0\la x+(1-3\la)y=0 has one obvious solution. Namely x=y=0x=y=0. But this solution is not acceptable because it does not satisfy the equation of the ellipse. If you have already taken a linear algebra course, you know that a system of two linear homogeneous equations in two unknowns has a nonzero solution if and only if the determinant of the matrix of coefficients is zero. (You use this when you find eigenvalues and eigenvectors.) For the equations of interest, this is

det⁡[1−3λλλ1−3λ]=(1−3λ)2−λ2=(1−2λ)(1−4λ)=0  ⟹  λ=12,14\begin{align*} \det\left[\begin{matrix}1-3\la&\la\\ \la&1-3\la\end{matrix}\right] =(1-3\la)^2-\la^2 =(1-2\la)(1-4\la)=0\implies\la=\frac{1}{2},\frac{1}{4} \end{align*}

Even if you have not already taken a linear algebra course, you also come to this conclusion directly when you try to solve the equations. Note that λ\la cannot be zero because if it is, (E1) forces x=0x=0 and (E2) forces y=0y=0. So equation (E1) gives y=−1−3λλxy=-\frac{1-3\la}{\la}x. Substituting this into equation (E2) gives λx−(1−3λ)2λx=0\la x-\frac{(1-3\la)^2}{\la}x=0. To get a nonzero (x,y)(x,y) we need

λ−(1−3λ)2λ=0  ⟺  λ2−(1−3λ)2=0\begin{equation*} \la -\frac{(1-3\la)^2}{\la}=0\iff \la^2-(1-3\la)^2=0 \end{equation*}

By either of these two methods, we now know that λ\la must be either 12\frac{1}{2} or 14\frac{1}{4}. Substituting these into either (E1) or (E2) and then using (E3) gives

λ=12  ⟹  −12x+12y=0  ⟹   x=y  ⟹  3x2−2x2+3x2=4  ⟹   x=±1λ=14  ⟹  14x+14y=0  ⟹   x=−y  ⟹  3x2+2x2+3x2=4  ⟹   x=±12\begin{alignat*}{9} \la&=\frac{1}{2}&&\implies& -\frac{1}{2} x+\frac{1}{2} y&=0 &&\implies\ & x&=y &&\implies 3x^2-2x^2+3x^2&=4 &&\implies\ x&=\pm 1\\ \la&=\frac{1}{4}&&\implies& \frac{1}{4} x+\frac{1}{4} y&=0 &&\implies\ & x&=-y &&\implies 3x^2+2x^2+3x^2&=4 &&\implies\ x&=\pm \frac{1}{\sqrt{2}} \end{alignat*}

The ends of the minor axes are ±(12,−12)\pm\big(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\big). The ends of the major axes are ±(1,1)\pm(1,1).

Q20Stage 2Past exam · M200 2003D

A closed rectangular box with a volume of 96 cubic meters is to be constructed of two materials. The material for the top costs twice as much per square meter as that for the sides and bottom. Use the method of Lagrange multipliers to find the dimensions of the least expensive box.

Answer

x=y=4, z=6 metersx=y=4,\ z=6\ \text{meters}

Full solution

Let the box have dimensions x×y×zx\times y\times z. Use units of money so that the sides and bottom cost one unit per square meter and the top costs two units per square meter. Then the top costs 2xy2xy, the bottom costs xyxy and the four sides cost 2xz+2yz2xz+2yz. We are to find the xx, yy and zz that minimize the cost f(x,y,z)=2xy+xy+2xz+2yzf(x,y,z)=2xy +xy +2xz+2yz subject to the constraint that g(x,y,z)=xyz−96=0g(x,y,z)=xyz-96=0. By the method of Lagrange multipliers (Theorem 2.10.2 in the CLP-3 text), the minimizing xx, yy, zz must obey

fx=3y+2z=λyz=λgxfy=3x+2z=λxz=λgyfz=2x+2y=λxy=λgz   xyz−96=0\begin{alignat*}{3} f_x&=3y+2z&&=\la yz=\la g_x \\ f_y&=3x+2z&&=\la xz=\la g_y \\ f_z&=2x+2y&&=\la xy=\la g_z \\ &\ \ \ xyz-96&&=0 \end{alignat*}

Multiplying the first equation by xx, the second equation by yy and the third equation by zz and then substituting in xyz=96xyz=96 gives

3xy+2xz=96λ3xy+2yz=96λ2xz+2yz=96λ\begin{align*} 3xy+2xz&=96\la \\ 3xy+2yz&=96\la \\ 2xz+2yz&=96\la \end{align*}

Subtracting the second equation from the first gives 2z(x−y)=02z(x-y)=0. Since z=0z=0 is impossible, we must have x=yx=y. Substituting this in,

3x2+2xz=96λ4xz=96λ\begin{equation*} 3x^2+2xz=96\la\qquad 4xz=96\la \end{equation*}

Subtracting,

3x2−2xz=0  ⟹  z=32x  ⟹  96=xyz=32x3  ⟹  x3=64  ⟹  x=y=4, z=6 meters\begin{align*} 3x^2-2xz=0&\implies z=\frac{3}{2}x \implies 96=xyz=\frac{3}{2}x^3 \implies x^3=64 \\ &\implies x=y=4,\ z=6\ \text{meters} \end{align*}
Q21Stage 2Past exam · M200 2003A

Consider the unit sphere

S={ (x,y,z) ∣ x2+y2+z2=1 }\begin{equation*} S=\Set{(x,y,z)}{x^2+y^2+z^2=1} \end{equation*}

in R3\bbbr^3. Assume that the temperature at a point (x,y,z)(x,y,z) of SS is

T(x,y,z)=40xy2z\begin{equation*} T(x,y,z)=40xy^2z \end{equation*}

Find the hottest and coldest temperatures on SS.

Answer

The hottest temperature is +5+5 and the coldest temperature is −5-5.

Full solution

We are to find the xx, yy and zz that minimize the temperature T(x,y,z)=40xy2zT(x,y,z)=40xy^2z subject to the constraint that g(x,y,z)=x2+y2+z2−1=0g(x,y,z)=x^2+y^2+z^2-1=0. By the method of Lagrange multipliers (Theorem 2.10.2 in the CLP-3 text), the minimizing xx, yy, zz must obey

Tx=40y2z=λ(2x)=λgxTy=80xyz=λ(2y)=λgyTz=40xy2=λ(2z)=λgzx2+y2+z2−1=0\begin{alignat*}{3} T_x&=40y^2z&&=\la(2x)=\la g_x \\ T_y&=80xyz&&=\la(2y)=\la g_y \\ T_z&=40xy^2&&=\la(2z)=\la g_z \\ &\hskip-0.5in x^2+y^2+z^2-1&&=0 \end{alignat*}

Multiplying the first equation by xx, the second equation by y/2y/2 and the third equation by zz gives

40xy2z=2x2λ40xy2z=y2λ40xy2z=2z2λ\begin{align*} 40xy^2z&=2x^2\la \\ 40xy^2z&=y^2\la \\ 40xy^2z&=2z^2\la \end{align*}

Hence we must have

2x2λ=y2λ=2z2λ\begin{equation*} 2x^2\la=y^2\la=2z^2\la \end{equation*}

  • If λ=0\la=0, then 40y2z=0, 80xyz=0, 40xy2=040y^2z=0,\ 80xyz=0,\ 40xy^2=0 which is possible only if at least one of x,y,zx,y,z is zero so that T(x,y,z)=0T(x,y,z)=0.

  • If λ≠0\la\ne 0, then

    2x2=y2=2z2  ⟹  1=x2+y2+z2=x2+2x2+x2=4x2  ⟹  x=±12, y2=12, z=±12  ⟹  T=40(±12)12(±12)=±5\begin{align*} 2x^2=y^2=2z^2 &\implies 1=x^2+y^2+z^2=x^2+2x^2+x^2=4x^2 \\ &\implies x=\pm\half,\ y^2=\half,\ z=\pm \half \\ &\implies T=40\big(\pm\half)\half\big(\pm\half)=\pm 5 \end{align*}

(The sign of xx and zz need not be the same.) So the hottest temperature is +5+5 and the coldest temperature is −5-5.

Q22Stage 2Past exam · M200 2002A

Find the dimensions of the box of maximum volume which has its faces parallel to the coordinate planes and which is contained inside the region 0≤z≤48−4x2−3y20\le z\le 48-4x^2-3y^2.

Figure from prob_s2.10, line 1458

Figure from prob_s2.10, line 1458

Answer

23×4×242\sqrt{3} \times 4\times 24

Full solution

The optimal box will have vertices (±x,±y,0)(\pm x,\pm y, 0), (±x,±y, z)(\pm x,\pm y,\ z) with x,y,z>0x,y,z>0 and z=48−4x2−3y2z=48-4x^2-3y^2. (If the lower vertices are not in the xyxy–plane, the volume of the box can be increased by lowering the bottom of the box to the xyxy–plane. If any of the four upper vertices are not on the hemisphere, the volume of the box can be increased by moving the upper vertices outwards to the hemisphere.)
The volume of this box will be (2x)(2y)z(2x)(2y)z. So we are to find the xx, yy and zz that maximize the volume f(x,y,z)=4xyzf(x,y,z)=4xyz subject to the constraint that g(x,y,z)=48−4x2−3y2−z=0g(x,y,z)=48-4x^2-3y^2-z =0. By the method of Lagrange multipliers (Theorem 2.10.2 in the CLP-3 text), the minimizing xx, yy, zz must obey

fx=4yz=−8λx=λgxfy=4xz=−6λy=λgyfz=4xy=−λ=λgz48−4x2−3y2−z=0\begin{alignat*}{3} f_x&=4yz=-8\la x&&=\la g_x\cr f_y&=4xz=-6\la y&&=\la g_y\cr f_z&=4xy=-\la&&=\la g_z\cr &\hskip-0.2in 48-4x^2-3y^2-z&&=0 \end{alignat*}

Multiplying the first equation by xx, the second equation by yy and the third equation by zz gives

4xyz=−8λx24xyz=−6λy24xyz=−λz\begin{align*} 4xyz&=-8\la x^2 \\ 4xyz&=-6\la y^2 \\ 4xyz&=-\la z \end{align*}

This forces 8λx2=6λy2=λz8\la x^2=6\la y^2=\la z. Since λ\la cannot be zero (because that would force 4xyz=04xyz=0), this in turn gives 8x2=6y2=z8 x^2=6 y^2= z. Substituting in to the fourth equation gives

48−z2−z2−z=0  ⟹  2z=48  ⟹  z=24, 8x2=24, 6y2=24\begin{equation*} 48 -\frac{z}{2}-\frac{z}{2}-z=0\implies 2z=48\implies z=24,\ 8x^2=24,\ 6y^2=24 \end{equation*}

The dimensions of the box of biggest volume are 2x=232x=2\sqrt{3} by 2y=42y=4 by z=24z=24.

Q23Stage 2Past exam · M200 2001D

A rectangular bin is to be made of a wooden base and heavy cardboard with no top. If wood is three times more expensive than cardboard, find the dimensions of the cheapest bin which has a volume of 12m312{\rm m}^3.

Answer

2m×2m×3m2{\rm m}\times 2{\rm m}\times 3{\rm m}

Full solution

Use units of money for which cardboard costs one unit per square meter. Then, if the bin has dimensions x×y×zx\times y\times z, it costs 3xy+2xz+2yz3xy+2xz+2yz. We are to find the xx, yy and zz that minimize the cost f(x,y,z)=3xy+2xz+2yzf(x,y,z)=3xy+2xz+2yz subject to the constraint that g(x,y,z)=xyz−12=0g(x,y,z)=xyz-12=0. By the method of Lagrange multipliers (Theorem 2.10.2 in the CLP-3 text), the minimizing xx, yy, zz must obey

fx=3y+2z=λyz=λgxfy=3x+2z=λxz=λgyfz=2x+2y=λxy=λgz   xyz−12=0\begin{alignat*}{3} f_x&=3y+2z&&=\la yz=\la g_x \\ f_y&=3x+2z&&=\la xz=\la g_y \\ f_z&=2x+2y&&=\la xy=\la g_z \\ &\ \ \ xyz-12&&=0 \end{alignat*}

Multiplying the first equation by xx, the second equation by yy and the third equation by zz and then substituting in xyz=12xyz=12 gives

3xy+2xz=12λ3xy+2yz=12λ2xz+2yz=12λ\begin{align*} 3xy+2xz&=12\la \\ 3xy+2yz&=12\la \\ 2xz+2yz&=12\la \end{align*}

Subtracting the second equation from the first gives 2z(x−y)=02z(x-y)=0. Since z=0z=0 is impossible, we must have x=yx=y. Substituting this in

3x2+2xz=12λ4xz=12λ\begin{equation*} 3x^2+2xz=12\la\qquad 4xz=12\la \end{equation*}

Subtracting

3x2−2xz=0  ⟹  z=32x  ⟹  12=xyz=32x3  ⟹  x3=8  ⟹  x=y=2, z=3 meters\begin{align*} 3x^2-2xz=0\implies z=\frac{3}{2}x &\implies 12=xyz=\frac{3}{2}x^3 \implies x^3=8 \\ &\implies x=y=2,\ z=3\ {\rm meters} \end{align*}
Q24Stage 2Past exam · M200 2000D

A closed rectangular box having a volume of 44 cubic metres is to be built with material that costs $8 per square metre for the sides but $12 per square metre for the top and bottom. Find the least expensive dimensions for the box.

Answer

233m×233m×32/3m\frac{2}{\sqrt[3]{3}}{\rm m}\times\frac{2}{\sqrt[3]{3}}{\rm m} \times 3^{2/3}{\rm m}

Full solution

If the box has dimensions x×y×zx\times y\times z, it costs 24xy+16xz+16yz24xy+16xz+16yz. We are to find the xx, yy and zz that minimize the cost f(x,y,z)=24xy+16xz+16yzf(x,y,z)=24xy+16xz+16yz subject to the constraint that g(x,y,z)=xyz−4=0g(x,y,z)=xyz-4=0. By the method of Lagrange multipliers (Theorem 2.10.2 in the CLP-3 text), the minimizing xx, yy, zz must obey

fx=24y+16z=λyz=λgxfy=24x+16z=λxz=λgyfz=16x+16y=λxy=λgz      xyz−4=0\begin{alignat*}{3} f_x&=24y+16z&&=\la yz=\la g_x \\ f_y&=24x+16z&&=\la xz=\la g_y \\ f_z&=16x+16y&&=\la xy=\la g_z \\ &\ \ \ \ \ \ xyz-4&&=0 \end{alignat*}

Multiplying the first equation by xx, the second equation by yy and the third equation by zz and then substituting in xyz=4xyz=4 gives

24xy+16xz=4λ24xy+16yz=4λ16xz+16yz=4λ\begin{align*} 24xy+16xz&=4\la \\ 24xy+16yz&=4\la \\ 16xz+16yz&=4\la \end{align*}

Subtracting the second equation from the first gives 16z(x−y)=016z(x-y)=0. Since z=0z=0 is impossible, we must have x=yx=y. Subbing this in

24x2+16xz=4λ32xz=4λ\begin{equation*} 24x^2+16xz=4\la\qquad 32xz=4\la \end{equation*}

Subtracting

24x2−16xz=0  ⟹  z=32x  ⟹  4=xyz=32x3  ⟹  x3=83  ⟹  x=y=233, z=32/3metres\begin{align*} 24x^2-16xz=0&\implies z=\frac{3}{2}x \implies 4=xyz=\frac{3}{2}x^3 \implies x^3=\frac{8}{3} \\ &\implies x=y=\frac{2}{\sqrt[3]{3}},\ z=3^{2/3}{\rm metres} \end{align*}
Q25Stage 2Past exam · M200 2000A

Suppose that aa, bb, cc are all greater than zero and let DD be the pyramid bounded by the plane ax+by+cz=1ax+by+cz=1 and the 3 coordinate planes. Use the method of Lagrange multipliers to find the largest possible volume of DD if the plane ax+by+cz=1ax + by + cz = 1 is required to pass through the point (1,2,3)(1, 2, 3). (The volume of a pyramid is equal to one-third of the area of its base times the height.)

Answer

a=13a=\frac{1}{3}, b=16b=\frac{1}{6}, c=19c=\frac{1}{9}, max volume=27= 27

Full solution

The vertices of the pyramid are (0,0,0)(0,0,0), (1a,0,0)\big(\frac{1}{a},0,0\big), (0,1b,0)\big(0, \frac{1}{b},0\big) and (0,0,1c)\big(0,0,\frac{1}{c}\big). So the base of the pyramid is a triangle of area 121a1b\half\frac{1}{a}\frac{1}{b} and the height of the pyramid is 1c\frac{1}{c}. So the volume of the pyramid is 16abc\frac{1}{6abc}. The plane passes through (1,2,3)(1,2,3) if and only if a+2b+3c=1a+2b+3c=1. Thus we are to find the aa, bb and cc that maximize the volume f(a,b,c)=16abcf(a,b,c)=\frac{1}{6abc} subject to the constraint that g(a,b,c)=a+2b+3c−1=0g(a,b,c)=a+2b+3c-1=0. By the method of Lagrange multipliers (Theorem 2.10.2 in the CLP-3 text), the maximizing aa, bb, cc must obey

fa=−16a2bc=λ=λga  ⟺  6λa2bc=−1fb=−16ab2c=2λ=λgb  ⟺  6λab2c=−12fc=−16abc2=3λ=λgc  ⟺  6λabc2=−13a+2b+3c=1\begin{alignat*}{5} f_a&=-\frac{1}{6a^2bc}&&=\la&=\la g_a\quad &&\iff\quad 6\la a^2bc&=-1 \\ f_b&=-\frac{1}{6ab^2c}&&=2\la&=\la g_b\quad&&\iff\quad 6\la ab^2c&=-\half \\ f_c&=-\frac{1}{6abc^2}&&=3\la&=\la g_c\quad &&\iff\quad 6\la abc^2&=-\frac{1}{3}\\ &a+2b+3c&&=1 \end{alignat*}

Dividing the first two equations gives ab=2\frac{a}{b}=2 and dividing the first equation by the third gives ac=3\frac{a}{c}=3. Substituting b=12ab=\half a and c=13ac=\frac{1}{3} a in to the final equation gives

a+2b+3c=3a=1  ⟹  a=13, b=16, c=19\begin{equation*} a+2b+3c=3a=1 \implies a=\frac{1}{3},\ b=\frac{1}{6},\ c=\frac{1}{9} \end{equation*}

and the maximum volume is 3×6×96=27\frac{3\times 6\times 9}{6}=27.

3Stage 3Application

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

Q26Stage 3Past exam · M200 2009D

Use Lagrange multipliers to find the minimum distance from the origin to all points on the intersection of the curves

g(x,y,z)=x−z−4=0and h(x,y,z)=x+y+z−3=0\begin{align*} g(x,y,z) &= x-z-4=0 \\ \text{and } h(x,y,z) &= x+y+z-3=0 \end{align*}
Answer

11\sqrt{11}

Full solution

We'll find the minimum distance2^2 and then take the square root. That is, we'll find the minimum of f(x,y,z)=x2+y2+z2f(x,y,z)=x^2+y^2+z^2 subject to the constraints g(x,y,z)=x−z−4=0g(x,y,z)=x-z-4=0 and h(x,y,z)=x+y+z−3=0h(x,y,z) = x + y + z -3=0. By Theorem 2.10.8 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the double Lagrange multiplier equations

fx=2x=λ+μ=λgx+μhxfy=2y=μ=λgy+μhyfz=2z=−λ+μ=λgz+μhzx−z=4x+y+z=3\begin{align*} f_x = 2x &=\la + \mu = \la g_x +\mu h_x\tag{E1} \\ f_y = 2y &= \mu = \la g_y +\mu h_y\tag{E2} \\ f_z = 2z &=-\la + \mu = \la g_z +\mu h_z\tag{E3} \\ x - z &= 4\tag{E4} \\ x + y + z &= 3 \tag{E5} \end{align*}

for some real numbers λ\la and μ\mu. Adding (E1)and (E3) and then subtracting 2 times (E2) gives

2x−4y+2z=0orx−2y+z=0\begin{equation*} 2x-4y+2z=0\qquad\text{or}\qquad x-2y+z=0 \tag{E6}\end{equation*}

Substituting x=4+zx=4+z (from (E4)) into (E5) and (E6) gives

y+2z=−1−2y+2z=−4\begin{align*} y+2z&=-1 \tag{E5'}\\ -2y+2z&=-4 \tag{E6'} \end{align*}

Substituing y=−1−2zy=-1-2z (from (E5')) into (E6') gives

6z=−6  ⟹  z=−1  ⟹  y=−1−2(−1)=1  ⟹  x=4+(−1)=3\begin{align*} 6z =-6 \implies z=-1 \implies y=-1-2(-1)=1 \implies x=4+(-1)=3 \end{align*}

So the closest point is (3,1,−1)(3,1,-1) and the minimum distance is 32+12+(−1)2=11\sqrt{3^2+1^2+(-1)^2}=\sqrt{11}.

Q27Stage 3Past exam · M200 2014A

Find the largest and smallest values of

f(x,y,z)=6x+y2+xz\begin{equation*} f(x,y,z) = 6x + y^2 + xz \end{equation*}

on the sphere x2+y2+z2=36x^2 + y^2 + z^2 = 36. Determine all points at which these values occur.

Answer

The min is −273-27\sqrt{3} at (−33,0,3)(-3\sqrt{3},0,3) and the max is 4848 at (4,±4,2)(4,\pm 4,2).

Full solution

Solution 1:
This is a constrained optimization problem with objective function f(x,y,z)=6x+y2+xzf(x,y,z) = 6 x +y^2 +xz and constraint function g(x,y,z)=x2+y2+z2−36g(x,y,z) =x^2+y^2+z^2-36. By Theorem 2.10.2 in the CLP-3 text, any local minimum or maximum (x,y,z)(x,y,z) must obey the Lagrange multiplier equations

fx=6+z=2λx=λgxfy=2y=2λy=λgyfz=x=2λz=λgzx2+y2+z2=36\begin{align*} f_x = 6+z &=2 \la x = \la g_x \tag{E1} \\ f_y = 2y &=2 \la y = \la g_y \tag{E2} \\ f_z = x &=2 \la z = \la g_z \tag{E3} \\ x^2 + y^2 + z^2 &= 36 \tag{E4} \end{align*}

for some real number λ\la. By equation (E2), y(1−λ)=0y(1-\la)=0, which is obeyed if and only if at least one of y=0y=0, λ=1\la=1 is obeyed.

  • If y=0y=0, the remaining equations reduce to

    6+z=2λxx=2λzx2+z2=36\begin{align*} 6+z &=2 \la x \tag{E1} \\ x &=2 \la z \tag{E3} \\ x^2 + z^2 &= 36 \tag{E4} \end{align*}

    Substituting (E3) into (E1) gives 6+z=4λ2z6 + z = 4\la^ 2 z, which forces 4λ2≠14\la^2\ne 1 (since 6≠06\ne 0) and gives z=64λ2−1z = \frac{6}{4\la^2-1} and then x=12λ4λ2−1x=\frac{12\la}{4\la^2-1}. Substituting this into (E4) gives

    144λ2(4λ2−1)2+36(4λ2−1)2=364λ2(4λ2−1)2+1(4λ2−1)2=14λ2+1=(4λ2−1)2\begin{align*} \frac{144\la^2}{{(4\la^2-1)}^2} +\frac{36}{{(4\la^2-1)}^2}&=36 \\ \frac{4\la^2}{{(4\la^2-1)}^2} +\frac{1}{{(4\la^2-1)}^2}&=1 \\ 4\la^2+1 &= {(4\la^2-1)}^2 \end{align*}

    Write μ=4λ2\mu=4\la^2. Then this last equation is

    μ+1=μ2−2μ+1  ⟺  μ2−3μ=0  ⟺  μ=0,3\begin{align*} \mu+1 = \mu^2 -2\mu +1 &\iff \mu^2-3\mu =0 \\ &\iff \mu=0,3 \end{align*}

    When μ=0\mu=0, we have z=6μ−1=−6z=\frac{6}{\mu-1}=-6 and x=0x=0 (by (E4)). When μ=3\mu=3, we have z=6μ−1=3z=\frac{6}{\mu-1}=3 and then x=±27=±33x=\pm \sqrt{27} =\pm 3\sqrt{3} (by (E4)).

  • If λ=1\la=1, the remaining equations reduce to

    6+z=2xx=2zx2+y2+z2=36\begin{align*} 6+z &=2 x \tag{E1} \\ x &=2 z \tag{E3} \\ x^2 +y^2 + z^2 &= 36 \tag{E4} \end{align*}

    Substituting (E3) into (E1) gives 6+z=4z6+z=4z and hence z=2z=2. Then (E3) gives x=4x=4 and (E4) gives 42+y2+22=364^2 + y^2 + 2^2 =36 or y2=16y^2=16 or y=±4y=\pm 4.

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

point(0,0,−6)(0,0, -6)(33,0,3)(3\sqrt{3},0,3)(−33,0,3)(-3\sqrt{3},0,3)(4,4,2)(4,4,2)(4,−4,2)(4,-4,2)
value of ff0027327\sqrt{3}−273-27\sqrt{3}48484848
minmaxmax

Solution 2:
On the sphere we have y2=36−x2−z2y^2=36-x^2-z^2 and hence f=36+6x+xz−x2−z2f= 36 + 6x +xz -x^2-z^2 and x2+z2≤36x^2+z^2\le 36. So it suffices to find the max and min of h(x,z)=36+6x+xz−x2−z2h(x,z)= 36 + 6x +xz -x^2-z^2 on the disk D={ (x,z) ∣ x2+z2≤36 }D=\Set{(x,z)}{x^2+z^2\le 36}.

  • If a max or min occurs at an interior point (x,z)(x,z) of DD, then (x,z)(x,z) must be a critical point of hh and hence must obey

    hx=6+z−2x=0hz=x−2z=0\begin{align*} h_x = 6+z-2x&=0 \\ h_z = x-2z=0 \end{align*}

    Substituting x=2zx=2z into the first equation gives 6−3z=06-3z=0 and hence z=2z=2 and x=4x=4.

  • If a max or min occurs a point (x,z)(x,z) on the boundary of DD, we have x2+z2=36x^2+z^2=36 and hence x=±36−z2x=\pm\sqrt{36-z^2} and h=6x+zx=±(6+z)36−z2h=6x+zx=\pm(6+z)\sqrt{36-z^2} with −6≤z≤6-6\le z\le 6. So the max or min can occur either when z=−6z=-6 or z=+6z=+6 or at a zz obeying

    0=ddz[(6+z)36−z2]=36−z2−z(6+z)36−z2\begin{align*} 0=\diff{}{z}\big[(6+z)\sqrt{36-z^2}\big] =\sqrt{36-z^2} - \frac{z(6+z)}{\sqrt{36-z^2}} \end{align*}

    or equivalently

    36−z2−z(6+z)=02z2+6z−36=0z2+3z−18=0(z+6)(z−3)=0\begin{align*} 36-z^2-z(6+z)&=0 \\ 2z^2 +6z -36 &=0 \\ z^2 +3z -18 &=0 \\ (z+6)(z-3)&=0 \end{align*}

    So the max or min can occur either when z=±6z=\pm 6 or z=3z=3.

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

point(0,0,±6)(0,0, \pm 6)(33,0,3)(3\sqrt{3},0,3)(−33,0,3)(-3\sqrt{3},0,3)(4,4,2)(4,4,2)(4,−4,2)(4,-4,2)
value of ff0027327\sqrt{3}−273-27\sqrt{3}48484848
minmaxmax
Q28Stage 3Past exam · M200 2016D

The temperature in the plane is given by T(x,y)=ey(x2+y2)T(x,y) = e^y\big(x^2+y^2\big).

    1. Give the system of equations that must be solved in order to find the warmest and coolest point on the circle x2+y2=100x^2+y^2=100 by the method of Lagrange multipliers.

    2. Find the warmest and coolest points on the circle by solving that system.

    1. Give the system of equations that must be solved in order to find the critical points of T(x,y)T(x,y).

    2. Find the critical points by solving that system.

  3. Find the coolest point on the solid disc x2+y2≤100x^2+y^2\le 100.

Answer

(a) (i)

2x ey=λ(2x)ey(x2+y2+2y)=λ(2y)x2+y2=100\begin{align*} 2x\,e^y &=\la (2x) \\ e^y\big(x^2+y^2+2y\big) &=\la (2y) \\ x^2+y^2&=100 \end{align*}

(a) (ii) The warmest point is (0,10)(0,10) and the coolest point is (0,−10)(0,-10).

(b) (i)

2x ey=0ey(x2+y2+2y)=0\begin{align*} 2x\,e^y &=0 \\ e^y\big(x^2+y^2+2y\big) &=0 \end{align*}

(b) (ii) (0,0)(0,0) and (0,−2)(0,-2)

(c) (0,0)(0,0)

Full solution

By way of preparation, we have

∂T∂x(x,y)=2x ey∂T∂y(x,y)=ey(x2+y2+2y)\begin{align*} \pdiff{T}{x}(x,y) = 2x\,e^y\qquad \pdiff{T}{y}(x,y) = e^y\big(x^2+y^2+2y\big) \end{align*}

(a) (i) For this problem the objective function is T(x,y)=ey(x2+y2)T(x,y) = e^y\big(x^2+y^2\big) and the constraint function is g(x,y)=x2+y2−100g(x,y)=x^2 + y^2 - 100. According to the method of Lagrange multipliers, Theorem 2.10.2 in the CLP-3 text, we need to find all solutions to

Tx=2x ey=λ(2x)=λgxTy=ey(x2+y2+2y)=λ(2y)=λgyx2+y2=100\begin{align*} T_x = 2x\,e^y &=\la (2x) = \la g_x \tag{E1}\\ T_y = e^y\big(x^2+y^2+2y\big) &=\la (2y) = \la g_y \tag{E2}\\ x^2+y^2&=100 \tag{E3} \end{align*}

(a) (ii) According to equation (E1), 2x(ey−λ)=02x(e^y-\la)=0. This condition is satisfied if and only if at least one of x=0x=0, λ=ey\la=e^y is obeyed.

  • If x=0x=0, then equation (E3) reduces to y2=100y^2=100, which is obeyed if y=±10y=\pm 10. Equation (E2) then gives the corresponding values for λ\la, which we don't need.

  • If λ=ey\la=e^y, then equation (E2) reduces to

    ey(x2+y2+2y)=(2y)ey  ⟺  ey(x2+y2)=0\begin{equation*} e^y\big(x^2+y^2+2y\big) = (2y)e^y \iff e^y\big(x^2+y^2\big)=0 \end{equation*}

    which conflicts with (E3). So we can't have λ=ey\la=e^y.

So the only possible locations of the maximum and minimum of the function TT are (0,10)(0,10) and (0,−10)(0,-10). To complete the problem, we only have to compute TT at those points.

point(0,10)(0,10)(0,−10)(0,-10)
value of TT100e10100 e^{10}100e−10100 e^{-10}
maxmin

Hence the maximum value of T(x,y)=ey(x2+y2)T(x,y) = e^y\big(x^2+y^2\big) on x2+y2=100x^2 + y^2 = 100 is 100e10100 e^{10} at (0,10)(0,10) and the minimum value is 100e−10100 e^{-10} at (0,−10)(0,-10).

We remark that, on x2+y2=100x^2+y^2=100, the objective function T(x,y)=ey(x2+y2)=100eyT(x,y) = e^y\big(x^2+y^2\big) = 100 e^y. So of course the maximum value of TT is achieved when yy is a maximum, i.e. when y=10y=10, and the minimum value of TT is achieved when yy is a minimum, i.e. when y=−10y=-10.

(b) (i) By definition, the point (x,y)(x,y) is a critical point of T(x,y)T(x,y) if ane only if

Tx=2x ey=0Ty=ey(x2+y2+2y)=0\begin{align*} T_x = 2x\,e^y &=0 \tag{E1}\\ T_y = e^y\big(x^2+y^2+2y\big) &=0 \tag{E2} \end{align*}

(b) (ii) Equation (E1) forces x=0x=0. When x=0x=0, equation (E2) reduces to

ey(y2+2y)=0  ⟺  y(y+2)=0  ⟺  y=0 or y=−2\begin{align*} e^y\big(y^2+2y\big) =0 \iff y(y+2)=0 \iff y=0\text{ or }y=-2 \end{align*}

So there are two critical points, namely (0,0)(0,0) and (0,−2)(0,-2).

(c) Note that T(x,y)=ey(x2+y2)≥0T(x,y) = e^y\big(x^2+y^2\big)\ge 0 on all of R2\bbbr^2. As T(x,y)=0T(x,y)=0 only at (0,0)(0,0), it is obvious that (0,0)(0,0) is the coolest point.

In case you didn't notice that, here is a more conventional solution.

The coolest point on the solid disc x2+y2≤100x^2+y^2\le 100 must either be on the boundary, x2+y2=100x^2+y^2= 100, of the disc or be in the interior, x2+y2<100x^2+y^2 < 100, of the disc.

In part (a) (ii) we found that the coolest point on the boundary is (0,−10)(0,-10), where T=100e−10T=100 e^{-10}.

If the coolest point is in the interior, it must be a critical point and so must be either (0,0)(0,0), where T=0T=0, or (0,−2)(0,-2), where T=4e−2T= 4e^{-2}.

So the coolest point is (0,0)(0,0).

Q29Stage 3Past exam · M200 2002D

  1. By finding the points of tangency, determine the values of cc for which x+y+z=cx+y+z=c is a tangent plane to the surface 4x2+4y2+z2=964x^2+4y^2+z^2=96.

  2. Use the method of Lagrange Multipliers to determine the absolute maximum and minimum values of the function f(x,y,z)=x+y+zf(x,y,z)=x+y+z along the surface g(x,y,z)=4x2+4y2+z2=96g(x,y,z)=4x^2+4y^2+z^2=96.

  3. Why do you get the same answers in (a) and (b)?

Answer

(a) c=±12c=\pm 12 (b) ±12\pm 12

(c) The level surfaces of x+y+zx+y+z are planes with equation of the form x+y+z=cx+y+z=c. To find the largest (smallest) value of x+y+zx+y+z on 4x2+4y2+z2=964x^2+4y^2+z^2=96 we keep increasing (decreasing) cc until we get to the largest (smallest) value of cc for which the plane x+y+z=cx+y+z=c intersects 4x2+4y2+z2=964x^2+4y^2+z^2=96. For this value of cc, x+y+z=cx+y+z=c is tangent to 4x2+4y2+z2=964x^2+4y^2+z^2=96.

Full solution

(a) A normal vector to F(x,y,z)=4x2+4y2+z2=96F(x,y,z)=4x^2+4y^2+z^2=96 at (x0,y0,z0)(x_0,y_0,z_0) is ∇F(x0,y0,z0)=<8x0,8y0,2z0>\vnabla F(x_0,y_0,z_0)=\llt 8x_0,8y_0,2z_0\rgt. (Note that this normal vector is never the zero vector because (0,0,0)(0,0,0) is not on the surface.) So the tangent plane to 4x2+4y2+z2=964x^2+4y^2+z^2=96 at (x0,y0,z0)(x_0,y_0,z_0) is

8x0(x−x0)+8y0(y−y0)+2z0(z−z0)=0or8x0x+8y0y+2z0z=8x02+8y02+2z02\begin{equation*} 8x_0(x-x_0)+8y_0(y-y_0)+2z_0(z-z_0)=0\quad\text{or}\quad 8x_0x+8y_0y+2z_0z=8x_0^2+8y_0^2+2z_0^2 \end{equation*}

This plane is of the form x+y+z=cx+y+z=c if and only if 8x0=8y0=2z08x_0=8y_0=2z_0. A point (x0,y0,z0)(x_0,y_0,z_0) with 8x0=8y0=2z08x_0=8y_0=2z_0 is on the surface 4x2+4y2+z2=964x^2+4y^2+z^2=96 if and only if

4x02+4y02+z02=4x02+4x02+(4x0)2=96  ⟺  24x02=96  ⟺  x02=4  ⟺  x0=±2\begin{equation*} 4x_0^2+4y_0^2+z_0^2=4x_0^2+4x_0^2+(4x_0)^2=96 \iff 24 x_0^2=96 \iff x_0^2=4 \iff x_0=\pm 2 \end{equation*}

When x0=±2x_0=\pm 2, we have y0=±2y_0=\pm 2 and z0=±8z_0=\pm 8 (upper signs go together and lower signs go together) so that the tangent plane 8x0x+8y0y+2z0z=8x02+8y02+2z028x_0x+8y_0y+2z_0z=8x_0^2+8y_0^2+2z_0^2 is

8(±2)x+8(±2)y+2(±8)z=8(±2)2+8(±2)2+2(±8)2or±x±y±z=2+2+8orx+y+z=∓12  ⟹  c=±12\begin{align*} 8(\pm 2)x+8(\pm 2)y+2(\pm 8)z=8(\pm 2)^2+8(\pm 2)^2+2(\pm 8)^2 &\quad\text{or}\quad \pm x \pm y \pm z=2+2+8\\ &\quad\text{or}\quad x + y + z=\mp 12\\ &\implies c=\pm 12 \end{align*}

(b) We are to find the xx, yy and zz that minimize or maximize f(x,y,z)=x+y+zf(x,y,z)=x+y+z subject to the constraint that g(x,y,z)=4x2+4y2+z2−96=0g(x,y,z)=4x^2+4y^2+z^2-96=0. By the method of Lagrange multipliers (Theorem 2.10.2 in the CLP-3 text), the minimizing/maximizing xx, yy, zz must obey

fx=1=λ(8x)=λgxfy=1=λ(8y)=λgyfz=1=λ(2z)=λgz4x2+4y2+z2−96=0\begin{alignat*}{3} f_x&=1=\la(8x)&&=\la g_x\\ f_y&=1=\la(8y)&&=\la g_y\\ f_z&=1=\la(2z)&&=\la g_z\\ &\hskip-0.6in 4x^2+4y^2+z^2-96&&=0 \end{alignat*}

The first three equations give

x=18λy=18λz=12λwith λ≠0\begin{equation*} x=\frac{1}{8\la}\qquad y=\frac{1}{8\la}\qquad z=\frac{1}{2\la} \qquad\text{with }\la\ne 0 \end{equation*}

Substituting this into the fourth equation gives

4(18λ)2+4(18λ)2+(12λ)2=96  ⟺  (116+116+14)1λ2=96  ⟺  λ2=38196=18×32  ⟺  λ=±116\begin{align*} 4\left(\frac{1}{8\la}\right)^2+4\left(\frac{1}{8\la}\right)^2 +\left(\frac{1}{2\la}\right)^2=96 &\iff \left(\frac{1}{16}+\frac{1}{16}+\frac{1}{4}\right)\frac{1}{\la^2}=96 \\ &\iff \la^2=\frac{3}{8}\frac{1}{96}=\frac{1}{8\times 32} \\ &\iff \la=\pm \frac{1}{16} \end{align*}

Hence x=±2x=\pm 2, y=±2y=\pm 2 and z=±8z=\pm 8 so that the largest and smallest values of x+y+zx+y+z on 4x2+4y2+z2−964x^2+4y^2+z^2-96 are ±2±2±8\pm 2\pm2 \pm 8 or ±12\pm 12.

(c) The level surfaces of x+y+zx+y+z are planes with equation of the form x+y+z=cx+y+z=c. To find the largest (smallest) value of x+y+zx+y+z on 4x2+4y2+z2=964x^2+4y^2+z^2=96 we keep increasing (decreasing) cc until we get to the largest (smallest) value of cc for which the plane x+y+z=cx+y+z=c intersects 4x2+4y2+z2=964x^2+4y^2+z^2=96. For this value of cc, x+y+z=cx+y+z=c is tangent to 4x2+4y2+z2=964x^2+4y^2+z^2=96.

Q30Stage 3

Let f(x,y)f(x,y) have continuous partial derivatives. Consider the problem of finding local minima and maxima of f(x,y)f(x,y) on the curve xy=1xy=1.

  • Define g(x,y)=xy−1g(x,y) = xy -1. According to the method of Lagrange multipliers, if (x,y)(x,y) is a local minimum or maximum of f(x,y)f(x,y) on the curve xy=1xy=1, then there is a real number λ\la such that

    ∇f(x,y)=λ∇g(x,y),g(x,y)=0\begin{equation*} \vnabla f(x,y) =\la \vnabla g(x,y),\quad g(x,y)=0 \tag{E1}\end{equation*}

  • On the curve xy=1xy=1, we have y=1xy=\frac{1}{x} and f(x,y)=f(x,1x)f(x,y) =f\big(x,\frac{1}{x}\big). Define F(x)=f(x,1x)F(x)=f\big(x,\frac{1}{x}\big). If x≠0x\ne 0 is a local minimum or maximum of F(x)F(x), we have that

    F′(x)=0\begin{equation*} F'(x)=0 \tag{E2}\end{equation*}

Show that (E1) is equivalent to (E2), in the sense that

there is a λ such that (x,y,λ) obeys (E1)if and only ifx≠0 obeys (E2) and y=1 ⁣/x.\begin{align*} &\text{there is a }\la\text{ such that }(x,y,\la)\text{ obeys (E1)}\\ &\hskip-0.5in\text{if and only if}\\ &x\ne 0\text{ obeys (E2) and }y=\nicefrac{1}{x}\text{.} \end{align*}
Answer

See the solution.

Full solution

Note that if (x,y)(x,y) obeys g(x,y)=xy−1=0g(x,y)=xy-1=0, then xx is necessarily nonzero. So we may assume that x≠0x\ne 0. Then

There is a λ such that (x,y,λ) obeys (E1)  ⟺  there is a λ such that fx(x,y)=λgx(x,y),fy(x,y)=λgy(x,y),g(x,y)=0  ⟺  there is a λ such that fx(x,y)=λy,fy(x,y)=λx,xy=1  ⟺  there is a λ such that 1yfx(x,y)=1xfy(x,y)=λ,xy=1  ⟺  1yfx(x,y)=1xfy(x,y),xy=1  ⟺  xfx( ⁣x,1x)=1xfy( ⁣x,1x),y=1x  ⟺  F′(x)=ddxf( ⁣x,1x)=fx( ⁣x,1x)−1x2fy( ⁣x,1x)=0,y=1x\begin{align*} &\text{There is a }\la\text{ such that }(x,y,\la)\text{ obeys (E1)} \\ &\hskip0.2in \iff \text{there is a }\la\text{ such that } f_x(x,y)=\la g_x(x,y),\quad f_y(x,y)=\la g_y(x,y),\quad g(x,y)=0 \\ &\hskip0.2in \iff \text{there is a }\la\text{ such that } f_x(x,y)=\la y,\quad f_y(x,y)=\la x,\quad xy=1 \\ &\hskip0.2in \iff \text{there is a }\la\text{ such that } \frac{1}{y}f_x(x,y)= \frac{1}{x} f_y(x,y)=\la ,\quad xy=1 \\ &\hskip0.2in \iff \frac{1}{y}f_x(x,y)= \frac{1}{x} f_y(x,y) ,\quad xy=1 \\ &\hskip0.2in \iff xf_x\Big(\!x,\frac{1}{x}\Big)= \frac{1}{x} f_y\Big(\!x,\frac{1}{x}\Big) ,\quad y=\frac{1}{x} \\ &\hskip0.2in \iff F'(x) = \diff{}{x} f\Big(\!x,\frac{1}{x}\Big) = f_x\Big(\!x,\frac{1}{x}\Big) -\frac{1}{x^2}f_y\Big(\!x,\frac{1}{x}\Big) =0,\quad y=\frac{1}{x} \end{align*}

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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.