MA261-A Calculus III

2006 Fall

Homework 8 Solutions

Due 10/30/2006 8:00AM

11.6 #8 Letf (x; y) = y lnx.

(a) Find the gradient of f .(b) Evaluate the gradient at the point P (1;�3).(c) Find the rate of change of f at P in the direction of the vector u =

�45; 35

�.

[Solution](a) The gradient is

rf (x; y) =�@f

@x;@f

@y

�=Dyx; lnx

E.

(b) The gradient at the point P (1;�3) is

rf (1;�3) =��31; ln 1

�= h�3; 0i .

(c) The directional derivative is

Duf (x; y) = rf (x; y) � u.At the point P (1;�3), we have

Duf (1;�3) = rf (1;�3) ���45;3

5

�= h�3; 0i �

��45;3

5

�=12

5.

�1

2

11.6 #12 Find the directional derivative of the function

f (x; y) = ln�x2 + y2

�at the point (2; 1) in the direction of the vector v = h�1; 2i.[Solution]First, we need an unit vector of v. The unit vector u = v

jvj =h�1;2ip(�1)2+22

=D� 1p

5; 2p

5

E.

Second, the gradient of f is

rf (x; y) =�@f

@x;@f

@y

�=

�2x

x2 + y2;

2y

x2 + y2

�.

Thus, the directional derivative at the point (2; 1) is

Duf (1;�3) = rf (2; 1) ��� 1p

5;2p5

�=

�4

5;2

5

���� 1p

5;2p5

�= 0.

�11.6 #18 Find the directional derivative of the function

f (x; y; z) = x2 + y2 + z2

at the point P (2; 1; 3) in the direction of the origin.[Solution]First, the directional vector is v = h0; 0; 0i � h2; 1; 3i = h�2;�1;�3i. But, we need

an unit vector of v. The unit vector u = vjvj =

h�2;�1;�3ip(2)2+(1)2+(3)2

=D� 2p

14;� 1p

14;� 3p

14

E.

Second, the gradient of f is

rf (x; y; z) =�@f

@x;@f

@y;@f

@z

�= h2x; 2y; 2zi .

Thus, the directional derivative at the point (2; 1; 3) is

Duf (2; 1; 3) = rf (2; 1; 3)��� 2p

14;� 1p

14;� 3p

14

�= h4; 2; 6i�

�� 2p

14;� 1p

14;� 3p

14

�= � 28p

14.

�11.6 #20 Find the maximum rate of change of

f (p; q) = qe�p + pe�q

at the point (0; 0) in the direction in which it occurs.[Solution]The gradient is

rf (p; q) =�qe�p + e�q; e�p � pe�q

�which is the direction with the maximum rate of change. At (0; 0), the rate is

jrf (0; 0)j = jh1; 1ij =q(1)2 + (1)2 =

p2.

�

3

11.6 #24 Find the directions in which the direction derivative of f (x; y) = x2+ sin xy at the point(1; 0) has the value 1.[Solution]Assume that the direction is ha; bi where 1 = jha; bij =

pa2 + b2.

We know that the gradient rf = h2x+ y cosxy; x cosxyi.The directional derivative at the point (1; 0) is

Duf (1; 0) = rf (1; 0) � ha; bi = h2; 2i � ha; bi = 2a+ 2b.If the directional derivative has the value 1, we have 2a + 2b = 1, that is, b = 1�2a

2.

Note that we need a unit vector to calculate the directional derivative. So, we needa2 +

�1�2a2

�2= 1. This tells us that 8a2 � 4a � 3 = 0, or, a = 1�

p7

4. So, we have two

directionsD1+p7

4; 1�

p7

4

EorD1�p7

4; 1+

p7

4

E.

�11.6 #30 Suppose you are climbing a hill whose shape is given by the equation z = 1000�0:005x2�

0:01y2, where x, y, and z are measured in meters, and you are standing at a point whichcoordinates (60; 40; 966). The positive x-axis point points east and the positive y-axispoints north.(a) If you walk due south, will you start to ascend or descend? At what rate?(b) If you walk northwest, will you start to ascend or descend? At what rate?(c) In which direction is the slope largest? What is the rate of ascent in that direction?

At what angle above the horizontal does the path in that direction begin?[Solution]Consider f (x; y) = 1000� 0:005x2 � 0:01y2. We have fx = �0:01x and fy = �0:02y.(a) We can use a unit vector h0;�1i to represent the direction toward south. So, the

directional vector is

Duf (60; 40) = rf (60; 40) � h0;�1i = h�0:6;�0:8i � h0;�1i = 0:8.So, you start to ascend at the rate 0:8.

(b) We can use a vector h�1; 1i to represent the direction toward northwest. The unitvector is

D� 1p

2; 1p

2

E. So, the directional vector is

Duf (60; 40) = rf (60; 40) ��� 1p

2;1p2

�= h�0:6;�0:8i �

�� 1p

2;1p2

�= �0:8p

2.

So, you start to descend at the rate 0:8p2.

(c) The gradient isrf (60; 40) = h�0:6;�0:8i

which is the direction with the largest slope. The rate is jh�0:6;�0:8ij =q(�0:6)2 + (�0:8)2 =

1. Since the rate is 1 means that when you move 1 meter, you climb up 1 meter. So,the angle here is tan � = 1

1. This tells us that � = 45�.

�11.6 #36 Find the equation of the tangent plane and the normal line to the surface

x� z = 4arctan (yz)at the point (1 + �; 1; 1).[Solution]

4

Let F (x; y; z) = x� z� 4 arctan (yz). Then the surface becomes a level surface F = 0.To write these two equations, we need the gradient of F (x; y; z) which is

rF (x; y; z) =�@F

@x;@F

@y;@F

@z

�=

�1;� 4z

1 + (yz)2;�1� 4y

1 + (yz)2

�.

[Note that ddt(arctan t) = 1

1+t2.] So, at the point (1 + �; 1; 1), the gradient is

rF (1 + �; 1; 1) =�1;� 4 � 1

1 + (1 � 1)2;�1� 4 � 1

1 + (1 � 1)2�= h1;�2;�3i .

Thus, the tangent plane equation is

(1) (x� (1 + �)) + (�2) (y � 1) + (�3) (z � 1) = 0,or,

x� 2y � 3z + 4 = � � 4.The normal line equation is

x� (1 + �)1

=y � 1�2 =

z � 1�3 .

�11.6 #44 Find the points on the ellipsoid x2 + 2y2 + 3z2 = 1 where the tangent plane is parallel to

the plane 3x� y + 3z = 1.[Solution]We can treat the ellipsoid x2 + 2y2 + 3z2 = 1 as a level surface F (x; y; z) = 1 where

F (x; y; z) = x2 + 2y2 + 3z2. According to the tangent plane equation in textbook page796, the normal vector of the tangent plane to the level surface F (x; y; z) = 1 is rF =h2x; 4y; 6zi. If the tangent plane is parallel to the plane 3x � y + 3z = 1, their normalvector must be parallel. Thus, h2x; 4y; 6zi = k h3;�1; 3i. This implies that hx; y; zi =32k;�1

4k; 3

6k�. Since this point must be on the ellipsoid, we have�

3

2k

�2+ 2

��14k

�2+ 3

�3

6k

�2= 1.

By solving it, we have k2 = 825, or, k = �2

p25. Therefore, there are two points

3p2

5;�p2

10;

p2

5

!and

�3p2

5;

p2

10;�p2

5

!.

�11.6 #46 Show that the ellipsoid 3x2+2y2+z2 = 9 and the sphere x2+y2+z2�8x�6y�8z+24 = 0

are tangent to each other at the point (1; 1; 2). (This means that they have a commontangent plane at the point.)[Solution]We can treat the ellipsoid 3x2 + 2y2 + z2 = 9 as a level surface F (x; y; z) = 9 where

F (x; y; z) = 3x2 + 2y2 + z2. According to the tangent plane equation in textbook page796, the normal vector of the tangent plane to the level surface F (x; y; z) = 9 is rF =h6x; 4y; 2zi.Similarly, we can treat the sphere x2+y2+ z2�8x�6y�8z+24 = 0 as a level surface

G (x; y; z) = 0 where G (x; y; z) = x2 + y2 + z2 � 8x � 6y � 8z + 24. According to the

5

tangent plane equation in textbook page 796, the normal vector of the tangent plane tothe level surface G (x; y; z) = 0 is rG = h2x� 8; 2y � 6; 2z � 8i.At the point (1; 1; 2), rF (1; 1; 2) = h6; 4; 4i and rG (1; 1; 2) = h�6;�4;�4i. Since

these two normal vectors generate the same tangent plane at (1; 1; 2), we know that ourtwo graph share a common tangent plane at (1; 1; 2). Thus, the ellipsoid 3x2+2y2+z2 = 9and the sphere x2+ y2+ z2� 8x� 6y� 8z+24 = 0 are tangent to each other at the point(1; 1; 2).

�11.6 #48 Show that every normal line to the sphere x2 + y2 + z2 = r2 passes through the center of

the sphere.[Solution]We can treat the sphere x2 + y2 + z2 = r2 as a level surface F (x; y; z) = r2 where

F (x; y; z) = x2 + y2 + z2. According to the normal line equation in textbook page796, the directinoal vector of the normal line to the level surface F (x; y; z) = r2 isrF = h2x; 2y; 2zi. For a point (x0; y0; z0) on the sphere, the normal line equation at thatpoint is

x� x02x0

=y � y02y0

=z � z02z0

.

We notice that the center of the sphere, which is the origin (0; 0; 0), satis�es this normalline equation. Therefore, this normal line passes through the center of the sphere. Since(x0; y0; z0) is chosen arbitrarily, we prove that every normal line to the sphere x2+y2+z2 =r2 passes through the center of the sphere.

�11.7 #6 Find the local maximum and minimum values and saddle point(s) of the function

f (x; y) = x3y + 12x2 � 8y.[Solution]First, we have

rf (x; y) =�@f

@x;@f

@y

�=3x2y + 24x; x3 � 8

�.

By assuming rf (x; y) = 0, we have�3x2y + 24x = 0x3 � 8 = 0 .

x3 � 8 = 0 implies that x = 2. Thus, we have 3 (2)2 y + 24 (2) = 0, or, y = �4. So, thereis only one critical point (2;�4).To classify it, we need the determint

� = fxxfyy � f 2xy = (6xy) (0)��3x2�2= �9x4.

Since � < 0 at the critical point (2;�4), we know that it is a saddle point.�

11.7 #14 Find the local maximum and minimum values and saddle point(s) of the function

f (x; y) =�2x� x2

� �2y � y2

�.

[Solution]

6

First, we have

rf (x; y) =�@f

@x;@f

@y

�=(2� 2x)

�2y � y2

�;�2x� x2

�(2� 2y)

�.

By assuming rf (x; y) = 0, we have�(2� 2x) (2y � y2) = 0(2x� x2) (2� 2y) = 0 .

From the �rst equation, we have 2 � 2x = 0, or 2y � y2 = 0. This tells us that x = 1,or, y = 0 or 2. Similarly, From the second equation, we have 2x� x2 = 0, or 2� 2y = 0.This tells us that x = 0 or 2, or, y = 1. There are several combinations:(a) If x = 1, then y must be 1 from the second equation.(b) If y = 0, then x can be 0 or 2 from the second equation.(c) If y = 2, then x can be 0 or 2 from the second equation.Therefore, we have severl critical points (1; 1), (0; 0), (2; 0), (0; 2) and (2; 2).To classify them, we need the determint

�(x; y) = fxxfyy � f 2xy =�(�2)

�2y � y2

�� ��2x� x2

�(�2)

�� ((2� 2x) (2� 2y))2

= 4xy (2� y) (2� x)� 8 (1� x)2 (1� y)2

and fxx (x; y) = (�2) (2y � y2) = �2y (2� y).(a) For the critical point (1; 1), �(1; 1) = 4 > 0 and fxx (1; 1) = �2 < 0. Thus, it is a

local maximum.(b) For the critical point (0; 0), �(0; 0) = �8 < 0. Thus, it is a saddle point.(c) For the critical point (2; 0), �(2; 0) = �8 < 0. Thus, it is a saddle point.(d) For the critical point (0; 2), �(0; 2) = �8 < 0. Thus, it is a saddle point.(e) For the critical point (2; 2), �(2; 2) = �8 < 0. Thus, it is a saddle point.The function looks like

00

1

23

45

z1 x

23

54

10

5

y0

­5

­1­1

�11.7 #26 Find the absolute maximum and minimum values of

f (x; y) = 3 + xy � x� 2yon the set D whish is the closed triangular region with vertices (1; 0), (5; 0), and (1; 4).

7

[Solution]D can be seen as a closed region bounded by x = 1, y = 0 and x + y = 5 (the line

passes through (5; 0), and (1; 4).)For the interior part of D, we calculate the rf = hy � 1; x� 2i �rst. If rf = 0, then

we have (x; y) = (2; 1). It is a critical point with value f (2; 1) = 1.For the boundary part, there are three di¤erent edges, E1 = f(x; 0) j 1 � x � 5g, E2 =

f(1; y) j 0 � y � 4g, and E3 = f(x; 5� x) j 1 � x � 5g.For E1, our function becomes f (x; 0) = 3 � x. Thus, with 1 � x � 5, we have

�2 � f (x; 0) � 2. The maximum 2 happens at (1; 0) and the minimum �2 happens at(5; 0).For E2, our function becomes f (1; y) = 2 � y. Thus, with 0 � y � 4, we have

�2 � f (1; y) � 2. The maximum 2 happens at (1; 0) and the minimum �2 happens at(1; 4).For E3, our function becomes f (x; 5� x) = 3+x (5� x)�x�2 (5� x) = �x2+6x�7.

If the derivative�2x+6 is zero, we have x = 3. So, there are three candidates for extrema:x = 1; 3; 5 . When x = 1, y = 4 and f (1; 4) = �2. When x = 3, y = 2 and f (3; 2) = 2.When x = 5, y = 0 and f (5; 0) = �2. Thus, we have the maximum 2 happens at (3; 2)and the minimum �2 happens at (1; 4) and (5; 0).Hence, the absolute maximum is 2 which is happened at (1; 0) and (3; 2) and the

absolute minimum is �2 which is happened at (5; 0) and (1; 4).�

11.7 #30 Find the absolute maximum and minimum values of f (x; y) = xy2 on the set D =f(x; y) j x � 0; y � 0; x2 + y2 � 3g.[Solution]For the interior part f(x; y) j x > 0; y > 0; x2 + y2 < 3g, we calculate therf = hy2; 2xyi

�rst. If rf = 0, then we have (x; y) = (0; 0). It sits on the boundary, not in our interiorpart.For the boundary part, there are three di¤erent edges.The �rst one is B1 = f(x; 0) j 0 < x < 3g. In B1, we have 0 � f (x; y) � 0. Thus,

f (x; y) = 0.The second one is B2 = f(0; y) j 0 < y < 3g. In B2, we have 0 � f (x; y) � 0. Thus,

f (x; y) = 0.The third one is B3 = f(x; y) j x > 0; y > 0; x2 + y2 = 3g. On B3, our function becomes

f (x; y) = xy2 = x (3� x2). Let g (x) = x (3� x2) = 3x�x3. To �nd the extrema, we set0 = g0 (x) = 3� 3x2. This implies that x2 = 1, or, x = 1 (since x > 0.) When x = 1, onthe B3, we have y =

p2 (since y > 0.) Therefore, the extreme happens at

�1;p2�with

the value f�1;p2�= 2.

So, the maximum is 2 at�1;p2�and the minimum is 0 when x = 0 or y = 0.

[Another way to get the extreme on B3] We can use the polar coordinate to representB3. This gives us that B3 =

�(r; �) j r =

p3 and 0 < � < �

2

where x =

p3 cos � and

y =p3 sin �. Thus, we have f = xy2 =

p3 cos �

p3 sin2 � = 3 cos � sin2 �. Set g (�) =

3 cos � sin2 �. To �nd the maximum and the minimum values, we consider g0 (�) = 0, thatis, 3 (� sin �) sin2 �+3 cos � (2 sin � cos �) = (3 sin �)

�� sin2 � + 2 cos2 �

�. Since 0 < � < �

2,

we know that sin � 6= 0. Thus, � sin2 �+2 cos2 � = 0, or sin � =p2 cos � (since 0 < � < �

2.)

This implies that tan � =p2. Therefore, we have sin � =

p2p3and cos � = 1p

3. It means

that the extreme happens at�1;p2�with the value f

�1;p2�= 2.

8

�11.7 #36 Find the points on the surface y2 = 9 + xz that are closest to the origin.

[Solution]Let (x; y; z) be a point on the surface y2 = 9 + xz. The distance to the origin is

d =

q(x� 0)2 + (y � 0)2 + (z � 0)2 =

px2 + y2 + z2.

To minimize the distance, it is equivalent to consider d2 = x2+y2+z2. Since y2 = 9+xz,we have

d2 = x2 + (9 + xz) + z2 = x2 + xz + z2 + 9.

Let f (x; z) = x2 + xz + z2 + 9. We have

fx (x; z) = 2x+ z and fz (x; z) = x+ 2z.

If fx = fy = 0, we have 2x+ z = 0 and x+2z = 0. These two equations imply that thereis only one critical point (x; z) = (0; 0). Also, D = fxxfzz � f 2xz = 2 � 2� 12 = 3 > 0 andfxx = 2 > 0. Thus, (x; z) = (0; 0) is a local minimum. It gives us y2 = 9 + 0 � 0 = 9, or,y = �3. So, there are two points (0; 3; 0) and (0;�3; 0) on the surface y2 = 9 + xz thatare closest to the origin.

�11.7 #38 Find three positive numbers x, y, and z whose sum is 100 such that xaybzc is a maximum.

[Solution]Assume that all a, b, c are not zero and a+ b+ c 6= 0.Since the sum of x, y, and z is 100, we have x+ y+ z = 100. Assuma that f (x; y; z) =

xaybzc. To maximize f under the constraint g (x; y; z) = x+ y + z � 100 = 0, we use thelagrange multiplier. Assume that

rf = �rg.

This tells us thataxa�1ybzc; xabyb�1zc; xaybczc�1

�= � h1; 1; 1i. So, we have axa�1ybzc =

xabyb�1zc = xaybczc�1 and x + y + z = 100. Obviously, if f (x; y; z) is a maximum, thenall x, y, z are not zero. If all x, y, z are not one, then we have ayz = bxz = cxy. Thisimplies that x = a

by and z = c

by. By putting it back to the g = 0, we have y = 100b

a+b+c.

Thus, x = 100aa+b+c

and z = 100ca+b+c

.

�11.7 #42 Find the dimensions of the rectangular box with largest volume if the total surface area

is given as 64 cm2.[Solution]Assume that the dimensions of the box is (x; y; z) where x > 0, y > 0, and z > 0. The

volume is V = xyz. The surface area is 2xy + 2yz + 2xz = 64. It tells us that

z =32� xyx+ y

.

9

Consider f (x; y) = xy�32�xyx+y

�. We have

fx (x; y) = y

�32� xyx+ y

�+ xy

�(�y) (x+ y)� (32� xy) (1)

(x+ y)2

�= y

�32� xyx+ y

�+ xy

��y2 � 32(x+ y)2

�=

y2 (�x2 + 32� 2xy)(x+ y)2

and

fy (x; y) = x

�32� xyx+ y

�+ xy

�(�x) (x+ y)� (32� xy) (1)

(x+ y)2

�= x

�32� xyx+ y

�+ xy

��x2 � 32(x+ y)2

�=

x2 (�y2 + 32� 2xy)(x+ y)2

.

Note that x+y 6= 0 since x and y all positive. If fx = 0 = fy, we have �x2+32�2xy = 0and �y2+32� 2xy = 0 since x 6= 0 and y 6= 0. This tells us that x2 = y2, or x = y(againsince x and y are all positive.)Similarly, if we write x in terms of y and z, we will get y = z.So, we can conclude that x = y = z. Then, since 2xy + 2yz + 2xz = 64, we have

6x2 = 64, or x =q

323= y = z.

[Alternative Solution]Assume that the dimensions of the box is (x; y; z). The surface area is 2xy + 2yz +

2xz = 64. The volume is V = xyz. To maximize V under the constraint g (x; y; z) =2xy + 2yz + 2xz � 64 = 0, we use the lagrange multiplier. Assume that

rf = �rg.This tells us that hyz; xz; xyi = � h2 (y + z) ; 2 (x+ z) ; 2 (x+ y)i. So, we have

xyz = 2� (xy + xz) = 2� (xy + yz) = 2� (xz + yz) .

If � 6= 0, then xy + xz = xy + yz = xz + yz. This implies that x = y = z. By the

constraint, we have x = y = z =q

323.

If � = 0, then hyz; xz; xyi = 0. Therefore, the volume is 0 which is not what we want.�

11.7 #50 Find an equation of the plane that passes through the point (1; 2; 3) and cut o¤ thesmallest volume in the �rst octant.[Solution]An equation of the plane that passes through the point (1; 2; 3) looks like

a (x� 1) + b (y � 2) + c (z � 3) = 0.We need the intersection points of the plane and three axes to calculate the volume.Notice that if any of a; b; c equals zero, this plane contains one of the three axes and thevolume becomes in�nity. So, we can assume that a 6= 0, b 6= 0, and c 6= 0. By assumingy = 0 and z = 0, the intersection points of the plane and the x-axis is

�a+2b+3c

a; 0; 0

�.

10

Similarly, the intersection points of the plane and the y-axis is�0; a+2b+3c

b; 0�and the

intersection points of the plane and the z-axis is�0; 0; a+2b+3c

c

�. Therefore, the volume is

V =1

6

�a+ 2b+ 3c

a

��a+ 2b+ 3c

b

��a+ 2b+ 3c

c

�since it looks like a pyramid.

�11.8 #4 Use Lagrange multipliers to �nd the maximum and minimum values of the function

f (x; y) = 4x+ 6y

subject to the constraintx2 + y2 = 13.

[Solution]Let g (x; y) = x2 + y2 � 13. First, we calculate

rf (x; y) =�@f

@x;@f

@y

�= h4; 6i

and

rg (x; y) =�@g

@x;@g

@y

�= h2x; 2yi .

If rg (x; y) = 0, then (x; y) = (0; 0) which does not satisfy the constraint g = 0.Assume that rf = �rg. With the constraint, we have8<: 4 = 2�x

6 = 2�yx2 + y2 = 13

.

From the �rst and second equations, we know that x 6= 0, y 6= 0 and � 6= 0. We cansolve them to get x = 4

2�= 2

�and y = 6

2�= 3

�. By plugging into the constraint, we have�

2�

�2+�3�

�2= 13. This implies that �2 = 1, or, � = �1. Thus, we have x = �2 and

y = �3. Therefore, we have two critical points (2; 3) and (�2;�3).The values of the function f are f (2; 3) = 26 and f (�2;�3) = �26. Hence, the

maximum is 26 at the point (2; 3) and the minimum is �26 at the point (�2;�3).�

11.8 #12 Use Lagrange multipliers to �nd the maximum and minimum values of the function

f (x; y; z) = x4 + y4 + z4

subject to the constraintx2 + y2 + z2 = 1.

[Solution]Let g (x; y; z) = x2 + y2 + z2 � 1. First, we calculate

rf (x; y; z) =�@f

@x;@f

@y;@f

@z

�=4x3; 4y3; 4z3

�and

rg (x; y; z) =�@g

@x;@g

@y;@g

@z

�= h2x; 2y; 2zi .

If rg (x; y; z) = 0, then (x; y; z) = (0; 0; 0) which does not satisfy the constraint g = 0.

11

Assume that rf = �rg. With the constraint, we have8>><>>:4x3 = 2�x4y3 = 2�y4z3 = 2�z

x2 + y2 + z2 = 1

.

Notice that if � = 0, then x = 0, y = 0 and z = 0. It is not the case we need to considersince (x; y; z) = (0; 0; 0) which does not satisfy the constraint g = 0. Thus, we know that� 6= 0.Assume that x 6= 0, y 6= 0 and z 6= 0. Then we have8>><>>:

x2 = �2

y2 = �2

z2 = �2

x2 + y2 + z2 = 1

.

By plugging into the constraint, we have �2+ �

2+ �

2= 1. This implies that � = 2

3. Thus,

we have x = � 1p3, y = � 1p

3and z = � 1p

3. Therefore, we have 8 critical points in all the

combinations. The values of the function in this case are all f�� 1p

3;� 1p

3;� 1p

3

�= 1

3.

If x = 0, then we have 8<: 4y3 = 2�y4z3 = 2�z

y2 + z2 = 1.

Similarly, assume that y 6= 0 and z 6= 0. Then we have8<: y2 = �2

z2 = �2

y2 + z2 = 1.

By plugging into the constraint, we have �2+ �

2= 1. This implies that � = 1. Thus, we

have y = � 1p2and z = � 1p

2. Therefore, we have 4 critical points in all the combinations.

The values of the function in this case are all f�0;� 1p

2;� 1p

2

�= 1

2.

If y = 0, then we have z2 = 1, or, z = �1. We have 2 critical points in all thecombinations. The values of the function in this case are all f (0; 0;�1) = 1.Similarly, if y = 0, we have critical points

�� 1p

2; 0;� 1p

2

�with f

�� 1p

2; 0;� 1p

2

�= 1

2

and (0;�1; 0) with f (0;�1; 0) = 1. If z = 0, we have critical points�� 1p

2;� 1p

2; 0�with

f�� 1p

2;� 1p

2; 0�= 1

2and (0; 0;�1) with f (0; 0;�1) = 1.

By combining everything together, we have the maximum is 1 at the points (0; 0;�1),(0;�1; 0) and (�1; 0; 0) and the minimum is 1

3at the points in all 8 conbinations of�

� 1p3;� 1p

3;� 1p

3

�.

�11.8 #16 Use Lagrange multipliers to �nd the maximum and minimum values of the function

f (x; y; z) = 3x� y � 3zsubject to the constraints

x+ y � z = 0

12

andx2 + 2z2 = 1.

[Solution]Let g (x; y; z) = x2 + y2 + z2 � 1 and h (x; y; z) = x2 + 2z2 � 1. First, we calculate

rf (x; y; z) =�@f

@x;@f

@y;@f

@z

�= h3;�1; 3i ,

rg (x; y; z) =�@g

@x;@g

@y;@g

@z

�= h1; 1;�1i ,

and

rh (x; y; z) =�@h

@x;@h

@y;@h

@z

�= h2x; 0; 4zi .

Note that rg (x; y; z) 6= 0. If rh (x; y; z) = 0, then (x; y; z) = (0; 0; 0) which does notsatisfy the constraint h = 0.Assume that rf = �rg + �rh. With the constraints, we have8>>>><>>>>:

3 = �+ 2�x�1 = �3 = ��+ 4�z

x+ y � z = 0x2 + 2z2 = 1

.

It becomes 8>>>><>>>>:x = 2

�

� = �1z = 1

2�

y = z � xx2 + 2z2 = 1

.

By plugging into the last constraint, we have�2�

�2+ 2

�12�

�2= 1. This implies that

�2 = 92, or, � = � 3p

2. Thus, x = �2

p23, z = �

p26and y = z � x = �

p22. Therefore, there

are two critical points�2p23;�

p22;p26

�and

��2

p23;p22;�

p26

�.

The values of the function f are f�2p23;�

p22;p26

�= 2

p2and f

��2

p23;p22;�

p26

�=

�2p2. Hence, the maximum is 2

p2 at the point

�2p23;�

p22;p26

�and the minimum is

�2p2 at the point

��2

p23;p22;�

p26

�.

�11.8 #18 Find the extreme values of

f (x; y) = 2x2 + 3y2 � 4x� 5on the region described by

x2 + y2 � 16.[Solution]Let g (x; y) = x2 + y2 � 16. First, we calculate

rf (x; y) =�@f

@x;@f

@y

�= h4x� 4; 6yi

13

and

rg (x; y) =�@g

@x;@g

@y

�= h2x; 2yi .

First, we �nd the extreme values for the interior of x2 + y2 � 16. Consider the regionx2+ y2 < 16. Assume that rf = 0. This implies that (x; y) = (1; 0). Since (1; 0) satis�esx2 + y2 < 16, (1; 0) is a critical point. The value is f (1; 0) = �7.Next, we consider the boundary x2 + y2 = 16. Thus, g = 0 becomes a constraint.If rg (x; y) = 0, then (x; y) = (0; 0; 0) which does not satisfy the constraint g = 0.Assume that rf = �rg. With the constraint, we have8<: 4x� 4 = 2�x

6y = 2�yx2 + y2 = 16

.

If y 6= 0, then by the second equation, we have � = 3. This implies that x = �2 andy = �

p12. In this case, we have two critical points

��2; 2

p3�and

��2;�2

p3�. The

values are both f��2; 2

p3�= f

��2;�2

p3�= 47.

If y = 0, then x = �4. In this case, we have two critical points (4; 0) and (�4; 0). Thevalues of them are f (4; 0) = 11 and f (�4; 0) = 43.Hence, the maximum is 47 at the points

��2; 2

p3�and

��2;�2

p3�and the minimum

is �7 at the point (1; 0).�

11.8 #38 Find the maximum and minimum volumes of a rectangular box whose surface area is1500 cm2 and whose total edge length is 200 cm.[Solution]Assume that the box has edges x, y and z where x > 0, y > 0 and z > 0. Thus, the

volume can be written as V (x; y; z) = xyz. The surface area is 2xy + 2yz + 2xz = 1500.This gives us a constraint xy+yz+xz = 750. Also, the total edge length is 4x+4y+4z =200. This gives us another constraint x+ y + z = 50.Let us rewrite our problem as the following: to maximize or to minimize

F (x; y) = xy (50� x� y) = 50xy � x2y � xy2

subject to the constraint xy+ y (50� x� y)+x (50� x� y) = 750, or, 50x+50y�xy�x2 � y2 = 750.Let G (x; y) = 50x+ 50y � xy � x2 � y2 � 750. First, we calculate

rF (x; y) =�@F

@x;@F

@y

�=50y � 2xy � y2; 50x� 2xy � x2

�and

rG (x; y) =�@G

@x;@G

@y

�= h50� y � 2x; 50� x� 2yi .

If rG (x; y) = 0, then we have 50�y�2x = 0 = 50�x�2y. It implies that x = y = 503

which does not satisfy the constraint G = 0.Assume that rf = �rg. With the constraints, we have8<: 50y � 2xy � y2 = � (50� y � 2x)

50x� 2xy � x2 = � (50� x� 2y)50x+ 50y � xy � x2 � y2 = 750

.

14

If 50 � y � 2x 6= 0, then � = y. So, the second equation becomes 50x � 2xy � x2 =y (50� x� 2y), or, (50� 2y � x) (x� y) = 0. Thus, we have x = y or 50� 2y � x = 0.If x = y, then the constraint becomes 3x2� 100x+ 750 = 0, or, x = 100�

p1002�4�3�7506

=100�10

p10

6= y. The values are F

�100+10

p10

6; 100+10

p10

6

�� 2947:937259 and F

�100�10

p10

6; 100�10

p10

6

��

3533:544227.If 50 � 2y � x = 0, then the constraint becomes 50 (50� 2y) + 50y � (50� 2y) y �

(50� 2y)2 � y2 = 0, or, �y (3y � 100) = 0. Since y is an edge, y > 0. So, we havey = 100

3. Therefore, x = �140

3which is a contradiction since x is an edge and x > 0.

Since x; y; z are symmetric in our equations, by assuming y = z or x = z, we can haveanother two sets of critical points

�100�10

p10

6; 50+10

p10

3; 100�10

p10

6

�,�100+10

p10

6; 50�10

p10

3; 100+10

p10

6

�and

�50+10

p10

3; 100�10

p10

6; 100�10

p10

6

�,�50�10

p10

3; 100+10

p10

6; 100+10

p10

6

�.

Hence, the maximum is 3533:544227 with edges�100�10

p10

6; 100�10

p10

6; 50+10

p10

3

�and

the minimum is 2947:937259 at the point�100+10

p10

6; 100+10

p10

6; 50�10

p10

3

�.

[Alternative Solution]Let g (x; y; z) = xy+ yz+ xz� 750 and h (x; y; z) = x+ y+ z� 50. First, we calculate

rV (x; y; z) =�@V

@x;@V

@y;@V

@z

�= hyz; xz; xyi ,

rg (x; y; z) =�@g

@x;@g

@y;@g

@z

�= hy + z; x+ z; x+ yi .

and

rh (x; y; z) =�@h

@x;@h

@y;@h

@z

�= h1; 1; 1i .

Note that rh (x; y; z) 6= 0. If rg (x; y; z) = 0, then (x; y; z) = (0; 0; 0) which does notsatisfy the constraint g = 0.Assume that rf = �rg + �rh. With the constraints, we have8>>>><>>>>:

yz = � (y + z) + �xz = � (x+ z) + �xy = � (x+ y) + �

xy + yz + xz = 750x+ y + z = 50

.

It becomes 8>>>><>>>>:xyz = �xy + �xz + �xxyz = �xy + �yz + �yxyz = �xz + �yz + �z

xy + yz + xz = 750x+ y + z = 50

.

First three equations tell us that8<: �xy + �xz + �x = �xy + �yz + �y�xy + �yz + �y = �xz + �yz + �z�xy + �xz + �x = �xz + �yz + �z

.

15

It becomes that 8<: (�z + �) (x� y) = 0(�x+ �) (y � z) = 0(�y + �) (x� z) = 0

.

We notice that if two of the three equations x� y = 0, y� z = 0 and x� z = 0 are hold,then we have x = y = z. But, by two constraints, it implies that 3x2 = 750 and 3x = 50.There is no such kind of x. Thus, we can assume that we can only have at most one ofthe three equations x� y = 0, y � z = 0 and x� z = 0 hold.Also, if x, y and z are three distinct numbers and � 6= 0, then we have z = ��

�= x = y.

It cannot happen.So, we can assune that x = y, z 6= y, z 6= x. Therefore, the constraints becomes�

x2 + 2xz = 7502x+ z = 50

.

By replancing z in the �rst equation by z = 50 � 2x, we have x2 + 2x (50� 2x) = 750,or, 3x2 � 100x + 750 = 0. It tells us that x = 100�

p1002�4�3�7506

= 100�10p10

6= y and the

results are the same as the �rst way.�

11.8 #40(b) The plane 4x� 3y + 8z = 5 intersects the cone z2 = x2 + y2 in an ellipse. Use Lagrangemultipliers to �nd the highest and lowest points on the ellipse.[Solution]Assume that (x; y; z) is a point on the ellipse. To �nd the highest and lowest points is

equivalent to �nd the largest z and the smallest z.Let f (x; y; z) = z, g (x; y; z) = 4x � 3y + 8z � 5 and h (x; y; z) = x2 + y2 � z2. First,

we calculate

rf (x; y; z) =�@f

@x;@f

@y;@f

@z

�= h0; 0; 1i ,

rg (x; y; z) =�@g

@x;@g

@y;@g

@z

�= h4;�3; 8i ,

and

rh (x; y; z) =�@h

@x;@h

@y;@h

@z

�= h2x; 2y;�2zi .

Note that rg (x; y; z) 6= 0. If rh (x; y; z) = 0, then (x; y; z) = (0; 0; 0) which does notsatisfy the constraint g = 0.Assume that rf = �rg + �rh. With the constraints, we have8>>>><>>>>:

0 = 4�+ 2�x0 = �4�+ 2�y1 = 8�� 2�z

4x� 3y + 8z = 5x2 + y2 � z2 = 0

.

Note that if � = 0, then the �rst equations tells us that � = 0. But, it will turn the thirdequation into 1 = 0 which is a contradiction. So, we know that � 6= 0. Thus, we have8<:

x = �2��

y = 2��

z = 8��12�

.

16

This tells us that x = �y. Thus, the constraints become�7x+ 8z = 52x2 � z2 = 0

.

Here, 2x2 � z2 = 0 implies that z = �p2x. If z =

p2x, then the �rst equation tells us

that x = 57+8

p2. In this case, z =

p2x = 5

p2

7+8p2. If z = �

p2x, then the �rst equation

tells us that x = 57�8

p2. In this case, z =

p2x = 5

p2

7�8p2.

Hence, the highest point on the ellipse is�

57+8

p2; 57+�8

p2; 5

p2

7+8p2

�and the lowest points

on the ellipse�

57�8

p2; 57�8

p2; 5

p2

7�8p2

�because 5

p2

7+8p2= 0:38611 > �1:6392 = 5

p2

7�8p2.

�