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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
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
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

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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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

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
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
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
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.

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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.