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Optimization of multivariable functions

16.1 Local maximum and minimum values

17 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 prac_s2.3, line 19

    Figure from prac_s2.3, line 19

    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 R S T U(ii) f has a saddle pointP R S T U(iii)  the partial derivative fy is positiveP R S T U\begin{align*} \text{(i) }& \vnabla f\text{ is zero} & &\text{P R S T U} \\ \text{(ii) }& f\text{ has a saddle point} & &\text{P R S T U} \\ \text{(iii) }& \text{ the partial derivative }f_y\text{ is positive} & &\text{P R S T U} \\ \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, and 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 prac_s2.3, line 19

    Figure from prac_s2.3, line 19

Answer

(a) (i) TT, UU

(a) (ii) UU

(a) (iii) 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

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q2Stage 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 scales 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 prac_s2.3, line 205

Figure from prac_s2.3, line 205

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

(c) No. See the solutions.

Full solution
Q3Stage 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).

Hint

Remember a2<1a^2<1 means ∣a∣<1|a|<1, i.e. −1<a<1-1<a<1.

Answer

∣c∣>2|c|>2

Full solution
Q4Stage 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*}
Hint

Use the Second Derivative Test

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
Q5Stage 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.

Hint

Use the Second Derivative Test

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
Q6Stage 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.

Hint

Use the Second Derivative Test

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
Q7Stage 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*}
Hint

Use the second derivative test

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
Q8Stage 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).

Hint

Use the Second Derivative Test

Answer
criticalpoint\Atop{\text{critical}}{\text{point}}type
(0,0)(0,0)local max
(2,0)(2,0)saddle point
Full solution
Q9Stage 2Past exam · M200 2010A

  1. For the function z=f(x,y)=x3+3xy+3y2−6x−3y−6z = f (x, y) = x^3 + 3xy + 3y^2 - 6x - 3y - 6. Find and classify as [local maxima, local minima, or saddle points] all critical points of f(x,y)f(x, y).

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

Figure from prac_s2.3, line 674

Figure from prac_s2.3, line 674

Figure from prac_s2.3, line 674

Figure from prac_s2.3, line 674

Hint

When you're looking for critical points, remember you need both fx=0f_x=0 and fy=0f_y=0. So if it's hard to solve (say) fx=0f_x=0, then first solve fy=0f_y=0; then you can narrow your search of fx=0f_x=0.

Answer

(a)

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

(b)

(i)

Figure from prac_s2.3, line 707

Figure from prac_s2.3, line 707

(ii)

Figure from prac_s2.3, line 707

Figure from prac_s2.3, line 707

Full solution
Q10Stage 2Past exam · M200 2010D

Define 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 all critical points of f(x,y)f(x,y) as local maxima, local minima, or saddle points.

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
Q11Stage 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
Q12Stage 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
Q13Stage 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
Q14Stage 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.

Hint

“Explain your reasoning" is test-speak for “show your work."

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
Q15Stage 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

3Stage 3Application

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

Q16Stage 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
Q17Stage 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.

Hint

Check Example 16.1.11 in the text.

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

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From the UBC Math 100 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.