Answers and Solutions to Section 11.7 HomeworkProblems 1—13 (odd) and 23—45 (odd)

S. F. Ellermeyer

1. (a)

D = fxx (1, 1) fyy (1, 1)− (fxy (1, 1))2

= (4) (2)− (1)2= 7 > 0

andfxx (1, 1) = 4 > 0

so the point (1, 1) corresponds to a local minimum value of f (by theSecond Derivative Test).

(b)

D = fxx (1, 1) fyy (1, 1)− (fxy (1, 1))2

= (4) (2)− (3)2= −1 < 0

so the point (1, 1) corresponds to a saddle point of f (by the SecondDerivative Test).

3. Looking at the level curves pictured in the book, it looks like the graphof f has a “pit” at the point (1, 1) (meaning that f has a local minimumthere). It also looks like f has a saddle point at (0, 0), because if we startin the third quadrant and travel through the origin into the first quadrant,the values of f go up (reaching a maximum value of 4) and then go backdown again; whereas, if we start in the second quadrant and travel throughthe origin to the fourth quadrant, then the values of f go down (reachinga minimum value of 4) and then go back up again. We will now confirmthis using Calculus. The function under consideration is

f (x, y) = 4 + x3 + y3 − 3xy.Its partial derivatives are

fx = 3x2 − 3y

fy = 3y2 − 3x.

Both of these partial derivatives are equal to zero when

x2 − y = 0y2 − x = 0.

1

The first of these equations tells us that y = x2 at any critical point.Substituting this into the second equation gives

x4 − x = 0which gives x = 0 or x = 1. For x = 0, we must have y = 0, and for x = 1,we must have y = 1. The critical points of f are thus (0, 0) and (1, 1).

We now use the Second Derivative Test to classify these critical points.The second derivatives of f are

fxx = 6x

fyy = 6y

fxy = −3.For the critical point (0, 0), we have

D = fxx (0, 0) fyy (0, 0)− (fxy (0, 0))2

= (0) (0)− (−3)2= −9 < 0

which tells us that (0, 0) is a saddle point.

For the critical point (1, 1), we have

D = fxx (1, 1) fyy (1, 1)− (fxy (1, 1))2

= (6) (6)− (−3)2= 27 > 0

andfxx (1, 1) = 6 > 0

which tells us that (1, 1) corresponds to a local minimum.

5. For the function

f (x, y) = 9− 2x+ 4y − x2 − 4y2,we have

fx = −2− 2xfy = 4− 8y

from which it can be seen that the only critical point of f is¡−1, 12¢.

The second derivatives of f are

fxx = −2fyy = −8fxy = 0

2

At the point¡−1, 12¢, we have

D = fxx

µ−1, 1

2

¶fyy

µ−1, 1

2

¶−µfxy

µ−1, 1

2

¶¶2= (−2) (−8)− (0)2= 16 > 0

and

fxx

µ−1, 1

2

¶= −2 < 0

which tells us that¡−1, 12¢ corresponds to a local maximum of f . Note

that the local maximum value of f at this point is

f

µ−1, 1

2

¶= 9− 2 (−1) + 4

µ1

2

¶− (−1)2 − 4

µ1

2

¶2= 11.

In fact, we can verify this by rewriting the formula of f using “completingthe square” to see that the graph of f is a paraboloid:

f (x, y) = 9− 2x+ 4y − x2 − 4y2= 9− ¡x2 + 2x¢− 4 ¡y2 − y¢= 9− ¡x2 + 2x+ 1¢+ 1− 4µy2 − y + 1

4

¶+ 1

= − (x+ 1)2 − 4µy − 1

2

¶2+ 11.

A graph of f is shown below.

Graph of f (x, y) = 9− 2x+ 4y − x2 − 4y2

3

7. For the functionf (x, y) = x2 + y2 + x2y + 4,

we have

fx = 2x+ 2xy

fy = 2y + x2.

Setting

2x (1 + y) = 0

2y + x2 = 0

we see that the critical points of f are (0, 0),¡√2,−1¢, and ¡−√2,−1¢.

The second derivatives of f are

fxx = 2 + 2y

fyy = 2

fxy = 2x

At the point (0, 0), we have

D = fxx (0, 0) fyy (0, 0)− (fxy (0, 0))2

= (2) (2)− (0)2= 4 > 0

andfxx (0, 0) = 2 > 0

which tells us that¡−1, 12¢ corresponds to a local minimum of f .

At the point¡√2,−1¢, we have

D = fxx

³√2,−1

´fyy

³√2,−1

´−³fxy

³√2,−1

´´2= (0) (2)−

³2√2´2

= −8 < 0which tells us that

¡√2,−1¢ corresponds to a saddle point of f .

At the point¡−√2,−1¢, we have

D = fxx

³−√2,−1

´fyy

³−√2,−1

´−³fxy

³−√2,−1

´´2= (0) (2)−

³−2√2´2

= −8 < 0which tells us that

¡−√2,−1¢ corresponds to a saddle point of f .4

9. For the functionf (x, y) = xy − 2x− y,

we have

fx = y − 2fy = x− 1.

Setting

y − 2 = 0x− 1 = 0

we see that the only critical point of f is (1, 2).

The second derivatives of f are

fxx = 0

fyy = 0

fxy = 1

At the point (1, 2), we have

D = fxx (1, 2) fyy (1, 2)− (fxy (1, 2))2

= (0) (0)− (1)2= −1 < 0

which tells us that (1, 2) corresponds to a saddle point of f . A graph of fis shown below.

Graph of f (x, y) = xy − 2x− y

5

11. For the functionf (x, y) = ex cos (y) ,

we have

fx = ex cos (y)

fy = −ex sin (y) .The critical points occur at values (x, y) such that

ex cos (y) = 0

−ex sin (y) = 0.Since ex is never zero and cos (y) and sin (y) are never zero for the samevalue of y, there are no critical points for this function.

13. For the functionf (x, y) = x sin (y) ,

we have

fx = sin (y)

fy = x cos (y) .

The critical points occur at values (x, y) such that

sin (y) = 0

x cos (y) = 0.

In the above system of equations, if x = 0, then any value of y such thatsin (y) = 0 will work. This gives us the critical points (0, nπ) (where n canbe any integer). On the other hand, if x 6= 0 in the above equations, thenwe must have both cos (y) = 0 and sin (y) = 0 (which is not possible).Therefore, the only critical points of f are (0, nπ) (where n can be anyinteger).

The second derivatives of f are

fxx = 0

fyy = −x sin (y)fxy = cos (y) .

At any point (0, nπ), we have

D = fxx (0, nπ) fyy (0, nπ)− (fxy (0, nπ))2= (0) (0)− cos2 (nπ)= −1 < 0

so all of the critical points are saddle points.

A graph of f is shown below.

6

Graph of f (x, y) = x sin (y)

23. For the functionf (x, y) = 1 + 4x− 5y,

since fx = 4 and fy = −5, we see that there are no critical points. Hencethe absolute maximum and absolute minimum values of f on the set Dmust occur somewhere along the boundary of D. The set D is picturedbelow. The boundary of this set consists of three lines segments:

x = 0, 0 ≤ y ≤ 3y = 0, 0 ≤ x ≤ 2y = −3

2x+ 3, 0 ≤ x ≤ 2

7

On the line segment x = 0, we have

f (0, y) = 1− 5y, 0 ≤ y ≤ 3

and this function clearly has maximum value

f (0, 0) = 1

and minimum valuef (0, 3) = −14.

On the line segment y = 0, we have

g (x, 0) = 1 + 4x, 0 ≤ x ≤ 2

and this function clearly has maximum value

f (2, 0) = 9

and minimum valuef (0, 0) = 1.

On the line segment y = −32x+ 3, we have

f

µx,−3

2x+ 3

¶= 1 + 4x− 5

µ−32x+ 3

¶=23

2x− 14

and this function clearly has maximum value

f (2, 0) = 9

and minimum valuef (0, 3) = −14.

From these considerations, we can now see that the absolute maximumvalue of f on the region D is 9 and that this absolute maximum occurs atthe point (2, 0). Also, the absolute minimum value of f on the region Dis −14 and this absolute minimum occurs at the point (0, 3).

25. The set D = {(x, y) | |x| ≤ 1 and |y| ≤ 1} is pictured below.

8

The boundary of this set consists of the curves

x = −1, − 1 ≤ y ≤ 1y = 1, − 1 ≤ x ≤ 1x = 1, − 1 ≤ y ≤ 1y = −1, − 1 ≤ x ≤ 1.

Here is a picture of D along with some of the level curves of f (x, y) =x2 + y2 + x2y + 4.

9

It looks like the minimum value of f occurs somewhere in the middle ofD (perhaps at (0, 0)) and it looks like it has a maximum value of 7 thatoccurs at both of the points (−1, 1) and (1, 1). Let us check this out withCalculus:

fx = 2x+ 2xy

fy = 2y + x2

and

2x (1 + y) = 0

2y + x2 = 0

implies that (x, y) = (0, 0) or¡√2, 1¢or¡−√2, 1¢. All three of these are

critical points of f , but only (0, 0) lies in the set D, so it is the only onewe are interested in.

Since

fxx = 2 + 2y

fyy = 2

fxy = 2x

we have

D = fxx (0, 0) fyy (0, 0)− (fxy (0, 0))2= (2) (2)− ¡02¢= 4 > 0

andfxx (0, 0) = 2 > 0

which tells us that (0, 0) corresponds to a local minimum of f . Note that

f (0, 0) = 4.

Now we check the boundary curves of D: On the curve x = −1, we havef (−1, y) = y2 + y + 5.

Sinced

dy

¡y2 + y + 5

¢= 2y + 1,

we see that this function has a local minimum (as a function of one vari-able) at y = −1/2. Observe that

f

µ−1,−1

2

¶= 4.75

f (−1,−1) = 5f (−1, 1) = 7.

10

On the curve x = 1, we have

f (1, y) = y2 + y + 5.

Sinced

dy

¡y2 + y + 5

¢= 2y + 1,

we see that this function has a local minimum (as a function of one vari-able) at y = −1/2. Observe that

f

µ1,−1

2

¶= 4.75

f (1,−1) = 5f (−1, 1) = 7.

On the curve y = −1, we have

f (x,−1) = 5

so f is constantly equal to 5 along this curve (as one can see by lookingat that level curve drawn in the figure above).

On the curve y = 1, we have

f (x, 1) = 2x2 + 5.

Sinced

dx

¡2x2 + 5

¢= 4x,

we see that this function has a local minimum (as a function of one vari-able) at x = 0. Observe that

f (0, 1) = 5

f (−1, 1) = 7f (1, 1) = 7.

Our conclusion is that, on the set D, f has an absolute minimum value of4 that occurs at the point (0, 0) and has an absolute maximum value of 7that occurs at each of the points (−1, 1) and (1, 1).

27. The region, D, bounded by the parabola y = x2 and the line y = 4 ispictured below.

11

For the function f (x, y) = 1 + xy − x− y, we have

fx = y − 1fy = x− 1

so the only critical point of f is (1, 1). This point lies on the boundary ofD, so we just need to check f along the boundary of D.

Along the curve y = x2, −2 ≤ x ≤ 2, we have

f¡x, x2

¢= 1 + x3 − x− x2.

Sinced

dx

¡1 + x3 − x− x2¢ = 3x2 − 2x− 1

and the solutions of3x2 − 2x− 1 = 0

are x = −1/3 and x = 1, we observe that

f

µ−13,1

9

¶= 1 +

µ−13

¶3−µ−13

¶−µ−13

¶2=32

27

f (1, 1) = 1 + (1)3 − (1)− (1)2 = 0f (−2, 4) = 1 + (−2)3 − (−2)− (−2)2 = −9f (2, 4) = 1 + (2)3 − (2)− (2)2 = 3

so the smallest value of f that occurs along the curve y = x2 is −9 andthe largest value of f that occurs along this curve is 3.

Along the curve y = 4, −2 ≤ x ≤ 2, we have

f (x, 4) = 1 + 4x− x− 4 = 3x− 3.

12

This is an increasing linear function of x with minimum value

f (−2, 4) = −9and maximum value

f (2, 4) = 3.

Our conclusion is that the absolute maximum value of f on D is 3. Thismaximum occurs at the point (2, 4). The absolute minimum value of f ,which occurs at the point (−2, 4), is −9.

29. For the function

f (x, y) = − ¡x2 − 1¢2 − ¡x2y − x− 1¢2 ,we have

fx = −2¡x2 − 1¢ (2x)− 2 ¡x2y − x− 1¢ (2xy − 1)

= −4x3 + 2x− 4x3y2 + 6x2y + 4xy − 2and

fy = −2¡x2y − x− 1¢ ¡x2¢

= −2x2 ¡x2y − x− 1¢ .Observe that there is no solution of fx (x, y) = 0 for which x = 0 (forthis would make fx (x, y) = −2). Since no critical point can have x = 0,we observe see that the solutions of fy = 0 that we are looking for mustsatisfy

yx2 − x− 1 = 0or

y =x+ 1

x2.

Placing this result into the equation fx = 0, we obtain

−4x3 + 2x− 4x3µx+ 1

x2

¶2+ 6x2

µx+ 1

x2

¶+ 4x

µx+ 1

x2

¶− 2 = 0

which simplifies to

−4x3 + 2x− 4(x+ 1)2

x+ 6 (x+ 1) + 4

µx+ 1

x

¶− 2 = 0,

which upon multiplication of both sides by x gives

−4x4 + 2x2 − 4 ¡x2 + 2x+ 1¢+ 6x (x+ 1) + 4 (x+ 1)− 2x = 0or simply

−4x4 + 4x2 = 0.

13

The non—zero solutions of this equation are x = 1 and x = −1. Thus, theonly critical points of f are (x, y) = (1, 2) and (x, y) = (−1, 0). (It can bechecked that both of these satisfy fx (x, y) = 0 and fy (x, y) = 0.)

We now compute the second partial derivatives of f :

fxx = −12x2 + 2− 12x2y2 + 12xy + 4yfyy = −2x4fxy = −8x3y + 6x2 + 4x.

At the critical point (x, y) = (1, 2), we have

D = fxx (1, 2) fyy (1, 2)− (fxy (1, 2))2

= (−26) (−2)− (−6)2= 16 > 0

andfxx (1, 2) = −26 < 0

so the point (1, 2) corresponds to a local maximum of f .

At the critical point (x, y) = (−1, 0), we have

D = fxx (−1, 0) fyy (−1, 0)− (fxy (−1, 0))2

= (−10) (−2)− (2)2= 26 > 0

andfxx (1, 2) = −10 < 0

so the point (−1, 0) corresponds to a local maximum of f . A graph of fis shown below.

Graph of f (x, y) = − ¡x2 − 1¢2 − ¡x2y − x− 1¢214

31. The square of the distance from the point (2, 1,−1) to any point in theplane x+ y − z = 1 is

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

= (x− 2)2 + (y − 1)2 + (x+ y)2 .

We want to find the absolute minimum value of this function. (The squareroot of the answer we get will be the actual distance from the point to theplane.)

First, we compute

fx = 2 (x− 2) + 2 (x+ y) = 4x+ 2y − 4fy = 2 (y − 1) + 2 (x+ y) = 2x+ 4y − 2.

The only solution of

4x+ 2y − 4 = 02x+ 4y − 2 = 0

is (x, y) = (1, 0).

Now we check to see that this gives a minimum:

fxx = 4 > 0

fyy = 4

fxy = 2

D = (4) (4)− (2)2 = 12 > 0.

Evaluating f at the point (1, 0), we obtain

f (1, 0) = 3.

Our conclusion is that the point in the plane x+ y − z = 1 that is closestto the point (2, 1,−1) is (1, 0, 0). The distance from the point (2, 1,−1)to the plane x+ y − z = 1 is √3.

33. The square of the distance from the point (0, 0, 0) to any point on thesurface z2 = xy + 1 is

f (x, y) = (x− 0)2 + (y − 0)2 + (z − 0)2= x2 + y2 + xy + 1.

Since

fx = 2x+ y

fy = 2y + x,

15

the only critical point of f is (0, 0). Also,

fxx = 2 > 0

fyy = 2

fxy = 1

D = (2) (2)− (1)2 = 3 > 0so this critical point corresponds to the absolute minimum value of f . Atthis point,we have

f (0, 0) = 1.

Note that there are two points on the surface z2 = xy+1 that correspondto (x, y) = (0, 0). They are (0, 0, 1) and (0, 0,−1). These two points arethe two points on the surface that are closest to the origin. Both have adistance of 1 from the origin.

35. We want to find three numbers x > 0, y > 0, and z > 0 such thatx+ y + z = 100 and such that the value of the function

f (x, y) = xy (100− x− y) = 100xy − x2y − xy2

is as large as possible.

Since

fx = 100y − 2xy − y2 = y (100− 2x− y)fy = 100x− x2 − 2xy = x (100− x− 2y) ,

we see that the only critical points of f are (0, 0), (0, 100), (100, x), and(100/3, 100/3). These four critical points correspond to the sets of num-bers (0, 0, 100) , (0, 100, 0), (100, 0, 0) , and (100/3, 100/3, 100/3). The firstthree sets of numbers each satisfy xyz = 0, so this is obviously not themaximum possible product. The last set satisfies

xyz =

µ100

3

¶3=1, 000, 000

27≈ 37, 037

and this is obviously the maximum possible product. To check to makesure that this is a maximum, note that

fxx = −2yfyy = −2xfxy = 100− 2x− 2y

and at the point (x, y) = (100/3, 100/3), we have

fxx = −2003< 0

D =

µ−2003

¶µ−2003

¶−µ−1003

¶2=10, 000

3> 0.

16

37. We want to find the volume of the largest rectangular box with edgesparallel to the axes that can be inscribed in the ellipsoid

9x2 + 36y2 + 4z2 = 36.

If we let x, y, and z be half of each dimension of the box, then the squareof the volume of the box is

((2x) (2y) (2z))2= 64x2y2z2

which we can write in terms of x and y only as

V (x, y) = 64x2y2z2

= 16x2y2¡4z2¢

= 16x2y2¡36− 9x2 − 36y2¢

= 576x2y2 − 144x4y2 − 576x2y4

with domain D = {(x, y) | 0 ≤ x ≤ 2 and 0 ≤ y ≤ 1}.Observe that

Vx = 1152xy2 − 576x3y2 − 1152xy4

Vy = 1152x2y − 288x4y − 2304x2y3.

Setting Vx = 0 and Vy = 0, we obtain

−576xy2 ¡−2 + x2 + 2y2¢ = 0−288x2y ¡−4 + x2 + 8y2¢ = 0.

Any point of the form (0, y) or of the form (x, 0) satisfies these equations,but these are clearly not the critical points that give the largest possiblerectangular box. (In fact, they give “boxes” with volume 0). The onlyother critical point is

¡2√3/3,√3/3¢. Since

Vxx = 1152y2 − 1728x2y2 − 1152y4

Vyy = 1152x2 − 288x4 − 6912x2y2

Vxy = 2304xy − 1152x3y − 4608xy3,

at the critical point¡2√3/3,√3/3¢we have

D = Vxx

³2√3/3,√3/3´Vyy

³2√3/3,√3/3´−³Vxy

³2√3/3,√3/3´´2

= (−512) (−2, 048)− (262, 144)= 786, 432 > 0

andVxx

³2√3/3,√3/3´= −512 < 0.

17

This tells us that the rectangular box of maximum volume that can beinscribed in the ellipsoid 9x2 + y2 + 4z2 = 36 is a box with side lengths4√3/3, 2

√3/3, and

2z =

r36− 9

³2√3/3´2−³√3/3´2=√213/3.

The volume of this box is³4√3/3´³2√3/3´³√

213/3´= 8√213/9.

39. The volume of the box in question is

V = xyz =1

3xy (3z) =

1

3xy (6− x− 2y) .

Note that

Vx =1

3xy (−1) + 1

3y (6− x− 2y) = −2

3xy + 2y − 2

3y2

Vy =1

3xy (−2) + 1

3x (6− x− 2y) = −4

3xy + 2x− 1

3x2.

Setting

−23y (x+ y − 3) = 0

−13x (4y − 6 + x) = 0,

we see that the critical points are (0, 0), (6, 0), (0, 3), and (2, 1). Fromthe geometric nature of the problem, it is obvious that none of the firstthree critical points listed gives the rectangular box of maximum possiblevolume. Let us check the point (2, 1). Since

Vxx = −23y

Vyy = −43x

Vxy = −23x+ 2− 4

3y

we have

D =

µ−23

¶µ−83

¶−µ−23

¶2=4

3

Vxx = −23

18

which shows that (2, 1) corresponds to a local maximum of V . The corre-sponding value of z is

z =1

3(6− (2)− 2 (1)) = 2

3.

The volume of the box is

xyz = (2) (1)

µ2

3

¶=4

3.

41. With the constraint that 4x + 4y + 4z = c, we want to find the absolutemaximum value of xyz. To do this, we find the absolute maximum valueof the function

f (x, y) =1

4xy (c− 4x− 4y)

with domain 0 ≤ x ≤ c/4, 0 ≤ y ≤ c/4.Since

fx =1

4xy (−4) + 1

4y (c− 4x− 4y) = −2xy + 1

4cy − y2

fy =1

4xy (−4) + 1

4x (c− 4x− 4y) = −2xy + 1

4cx− x2,

we find the critical points of f by solving

y (−8x+ c− 4y) = 0x (−8y + c− 4x) = 0.

It can be seen that the critical points are (0, 0), (c/4, 0), (0, c/4), and(c/12, c/12). The box of maximum volume obviously does not come fromany of the first three critical points. Let us check the third one:

fxx = −2yfyy = −2xfxy = −2x+ 1

4c− 2y

so for the point (c/12, c/12), we have

D =³− c6

´³− c6

´−³− c12

´2=c2

48> 0

fxx = − c6< 0.

We conclude that the box has maximum volume when all of its edges havethe same length (that is, the box is a cube).

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43. Suppose that the dimensions of the box are x, y, and z. We are given that

xyz = 32, 000.

The surface area of the box (which is what we are supposed to minimize)is

xy + 2xz + 2yz = xy + (2x+ 2y) z.

Hence, we must find the minimum value of the function

f (x, y) = xy + (2x+ 2y)

µ32000

xy

¶= xy +

64000

y+64000

x.

Since

fx = y − 64000x2

fy = x− 64000y2

,

we see that the critical points, (x, y), of f must satisfy

x2y = 64000

xy2 = 64000.

In particular, it must be true that x2y = xy2 and hence that x = y. Theonly critical point of f is thus (40, 40). To see that this gives a minimum,note that

fxx =128000

x3

fyy =128000

y3

fxy = 1

and hence (at (40, 40)),

D = (2) (2)− 1 = 3 > 0fxx = 2 > 0.

The dimensions of the box (with open top) and minimum surface area arethus 40× 40× 20.

45. Given the points (x1, y1), (x2, y2),. . . ,(xn, yn) and defining (for each i =1 . . . n)

di = yi − (mxi + b) ,

20

we want to find the value of m and b that minimize the function

f (m, b) =nXi=1

d2i

=nXi=1

(yi − (mxi + b))2 .

Since

fm =nXi=1

2 (yi − (mxi + b)) (−xi)

= −2nXi=1

¡xiyi −mx2i − bxi

¢= −2

ÃnXi=1

xiyi −mnXi=1

x2i − bnXi=1

xi

!

fb =nXi=1

2 (yi − (mxi + b)) (−1)

= −2nXi=1

(yi − (mxi + b))

= −2Ã

nXi=1

yi −mnXi=1

xi − nb!.

The critical points of f must thus satisfy

nXi=1

xiyi −mnXi=1

x2i − bnXi=1

xi = 0

nXi=1

yi −mnXi=1

xi − nb = 0

or ÃnXi=1

x2i

!m+

ÃnXi=1

xi

!b =

nXi=1

xiyiÃnXi=1

xi

!m+ nb =

nXi=1

yi

In general, there is a unique m and b that satisfy both of these equations.It is a good exercise to try to use the second derivative test to show thatthis m and b will give a minimum.

21