Section 14.7 Maximum and Minimum Values

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Section 14.7 Maximum and Minimum Values

15. Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.

f (x, y) = x⁴− 2x²+ y³− 3y

Solution:

462 ¤ CHAPTER 14 PARTIAL DERIVATIVES

13. ( ) = ⁴− 2²+ ³− 3 ⇒ ^= 4³− 4, ^= 3²− 3, ^= 12²− 4, ^= 0, = 6.

Then = 0implies 4(²− 1) = 0 ⇒  = 0 or  = ±1, and ^= 0implies 3(²− 1) = 0 ⇒  = ±1.

Thus there are six critical points: (0 ±1), (±1 1), and (±1 −1).

(0 1) = (−4)(6) − (0)²= −24  0 and

(±1 −1) = (8)(−6) = −48  0, so (0 1) and (±1 −1) are saddle points. (0 −1) = (−4)(−6) = 24  0 and ^(0 −1) = −4  0, so

 (0 −1) = 2 is a local maximum. (±1 1) = (8)(6) = 48  0 and

(±1 1) = 8  0, so (±1 1) = −3 are local minima.

14. ( ) =  +1

+1

 ⇒ ^=  − 1

², =  − 1

², = 2

³,

= 1, = 2

³. Then = 0implies  = 1

² and = 0implies

 = 1

². Substituting the ﬁrst equation into the second gives

 = 1

(1²)² ⇒  = ⁴ ⇒ (³− 1) = 0 ⇒  = 0 or  = 1.

is not deﬁned when  = 0, and when  = 1 we have  = 1, so the only critical point is (1 1).

(1 1) = (2)(2) − 1²= 3  0and (1 1) = 2  0, so (1 1) = 3 is a local minimum.

15. ( ) = ^cos  ⇒ ^= ^cos , = −^sin .

Now = 0implies cos  = 0 or  = ^₂ + for  an integer.

But sin 2 + 

6= 0, so there are no critical points.

16. ( ) = ^−(²^+²^)2 ⇒ ^=  · ^−(²^+²^)2(−) + ^−(²^+²^)2·  = (1 − ²)^−(²^+²^)2,

=  · ^−(²^+²^)2(−) + ^−(²^+²^)2·  = (1 − ²)^−(²^+²^)2,

= 

(1 − ²) · ^−(²^+²^)2(−) + ^−(²^+²^)2(−2)

= (²− 3)^−(²^+²^)2,

= (1 − ²)

 · ^−(²^+²^)2(−) + ^−(²^+²^)2(1)

= (1 − ²)(1 − ²)^−(²^+²^)2,

= 

(1 − ²) · ^−(²^+²^)2(−) + ^−(²^+²^)2(−2)

= (²− 3)^−(²^+²^)2.

Then = 0implies (1 − ²) = 0 ⇒  = 0 or  = ±1. Substituting  = 0 into ^= 0gives ^−²^2= 0 ⇒

 = 0, and substituting  = ±1 into ^= 0gives ±(1 − ²)^−(1+²^)2= 0 ⇒  = ±1, so the critical points are (0 0), (1 ±1), and (−1 ±1). (0 0) = (0)(0) − (1)²= −1  0, so (0 0) is a saddle point.

[continued]

21. Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.

f (x, y) = y²− 2y cos x, −1 ≤ x ≤ 7 Solution:

SECTION 14.7 MAXIMUM AND MINIMUM VALUES ¤ 463

(1 1) = (−1 −1) = (−2⁻¹)(−2⁻¹) − (0)²= 4⁻² 0and

(1 1) = (−1 −1) = −2⁻¹ 0, so (1 1) = (−1 −1) = ⁻¹ are local maxima.

(1 −1) = (−1 1) = (2⁻¹)(2⁻¹) − (0)²= 4⁻² 0and

(1 −1) = ^(−1 1) = 2⁻¹ 0, so (1 −1) = (−1 1) = −⁻¹ are local minima.

17. ( ) =  + ^− ⇒ ^=  − ^−,  =  − ^−, = ²^−,

= 1 −

(−^−) + ^−(1)

= 1 + ( − 1)^−, = ²^−. Then = 0implies (1 − ^−) = 0 ⇒

 = 0 or ^−= 1 ⇒  = 0 or  = 0. If  = 0 then ^= 0for any -value, so all points of the form (0 0)are critical points. If  = 0, then =  − ⁰= 0for any -value, so all points of the form (0 0)are critical points. We have

(0 0) = (0)(²0) − (0)²= 0 and (0 0) = (²0)(0) − (0)²= 0, so the Second Derivatives Test gives no information.

Notice that if we let  = , then ( ) = () =  + ^− ⇒

⁰() = 1 − ^−. Now ⁰() = 0only for  = 0, and ⁰()  0for   0,

⁰()  0for   0. Thus (0) = 1 is a local and absolute minimum, so

 ( ) =  + ^−≥ 1 for all ( ) with equality if and only if  = 0 or  = 0. Hence all points on the - and -axes are local (and absolute) minima, where ( ) = 1.

18. ( ) = (²+ ²)^− ⇒ ^= (²+ ²)(−^−) + ^−(2) = (2 − ²− ²)^−, = 2^−,

= (2 − ²− ²)(−^−) + ^−(2 − 2) = (²+ ²− 4 + 2)^−, = −2^−, = 2^−. Then = 0 implies  = 0 and substituting into = 0gives (2 − ²)^−= 0 ⇒

(2 − ) = 0 ⇒  = 0 or  = 2, so the critical points are (0 0) and (2 0). (0 0) = (2)(2) − (0)²= 4  0and (0 0) = 2  0, so

 (0 0) = 0is a local minimum.

(2 0) = (−2⁻²)(2⁻²) − (0)² = −4⁻⁴ 0so (2 0) is a saddle point.

19. ( ) = ²− 2 cos  ⇒ ^= 2 sin , = 2 − 2 cos ,

= 2 cos , = 2 sin , = 2. Then = 0implies  = 0 or sin  = 0 ⇒  = 0, , or 2 for −1 ≤  ≤ 7. Substituting  = 0 into

= 0gives cos  = 0 ⇒  =^2 or ^3₂ , substituting  = 0 or  = 2

into = 0gives  = 1, and substituting  =  into = 0gives  = −1.

Thus the critical points are (0 1),_

2 0

, ( −1),_3

2 0, and (2 1).

464 ¤ CHAPTER 14 PARTIAL DERIVATIVES

 2 0

= 3

2  0

= −4  0 so

2 0and3

2  0

are saddle points. (0 1) = ( −1) = (2 1) = 4  0 and

(0 1) = ( −1) = ^(2 1) = 2  0, so (0 1) = ( −1) = (2 1) = −1 are local minima.

20.  ( ) = sin  sin  ⇒ ^= cos  sin , = sin  cos , = − sin  sin , ^= cos  cos ,

= − sin  sin . Here we have −     and −    , so ^= 0implies cos  = 0 or sin  = 0. If cos  = 0 then  = −^2 or^₂, and if sin  = 0 then  = 0. Substituting  = ±^2 into = 0gives cos  = 0 ⇒  = −^2 or ^₂, and substituting  = 0 into = 0gives sin  = 0 ⇒  = 0. Thus the critical points are

−^2 ±^2

, 2 ±^2

, and (0 0).

(0 0) = −1  0 so (0 0) is a saddle point.



−^2 ±^2

=  2 ±^2

= 1  0and



−^2 −^2

=  2^₂

= −1  0 while



−^2^₂

=  2 −^2

= 1  0, so 

−^2 −^2

=  2^₂

= 1 are local maxima and 

−^2^₂

=  2 −^2

= 1are local minima.

21.  ( ) = ²+ 4²− 4 + 2 ⇒ ^= 2 − 4, ^= 8 − 4, ^= 2, = −4, ^= 8. Then = 0 and = 0each implies  = ¹₂, so all points of the form

0¹₂0

are critical points and for each of these we have



0¹₂0

= (2)(8) − (−4)² = 0. The Second Derivatives Test gives no information, but

 ( ) = ²+ 4²− 4 + 2 = ( − 2)²+ 2 ≥ 2 with equality if and only if  =¹2. Thus 

0¹₂0

= 2are all local (and absolute) minima.

22.  ( ) = ²^−²^−² ⇒

= ²^−²^−²(−2) + 2^−²^−² = 2(1 − ²)^−²^−²,

= ²^−²^−²(−2) + ²^−²^−²= ²(1 − 2²)^−²^−²,

= 2(2⁴− 5²+ 1)^−²^−²,

= 2(1 − ²)(1 − 2²)^−²^−², = 2²(2²− 3)^−²^−².

 = 0implies  = 0,  = 0, or  = ±1. If  = 0 then ^= 0for any -value, so all points of the form (0 ) are critical points. If  = 0 then = 0 ⇒ ²^−²= 0 ⇒  = 0, so (0 0) (already included above) is a critical point. If  = ±1 then (1 − 2²)^−1−² = 0 ⇒  = ±^√¹₂, so

±1^√¹₂ and

±1 −^√¹₂

are critical points. Now



±1^√¹₂

= 8⁻³ 0, 

±1^√¹₂

= −2√

2 ^−32 0and 

±1 −^√¹₂

= 8⁻³ 0,



±1 −^√¹₂

= 2√

2 ^−32  0, so 

±1^√¹₂

= √¹

2^−32are local maximum points while



±1 −^√¹₂

= −^√¹₂^−32are local minimum points. At all critical points (0 ) we have (0 ) = 0, so the Second Derivatives Test gives no information. However, if   0 then ²^−²^−² ≥ 0 with equality only when  = 0, so we have local minimum values (0 ) = 0,   0. Similarly, if   0 then ²^−²^−² ≤ 0 with equality when  = 0 so

 (0 ) = 0,   0 are local maximum values, and (0 0) is a saddle point.

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39. Find the absolute maximum and minimum values of f on the set D.

f (x, y) = 2x³+ y⁴, D = {(x, y)|x²+ y²≤ 1}

Solution:

 = 3where (2 3) = 7 and a minimum value at  = 1 where (2 1) = −1. Along ³:  = 3, so

 ( 3) = ²− 2 + 7 = ( − 1)²+ 6, 0 ≤  ≤ 2, which has a maximum value both at  = 0 and  = 2 where

 (0 3) =  (2 3) = 7and a minimum value at  = 1, where (1 3) = 6. Along 4:  = 0, so

 (0 ) = 2²− 4 + 1 = 2( − 1)²− 1, 0 ≤  ≤ 3, which has a maximum value at  = 3 where (0 3) = 7 and a minimum value at  = 1 where (0 1) = −1. Thus the absolute maximum is attained at both (0 3) and (2 3), where (0 3) = (2 3) = 7, and the absolute minimum is

 (1 1) = −2.

38. ( ) = ² ⇒ ^= ²and = 2, and since = 0 ⇔

 = 0, there are no critical points in the interior of . Along 1:  = 0 and

 ( 0) = 0. Along 2:  = 0 and (0 ) = 0. Along 3:  =√ 3 − ², so let () = 

√ 3 − ²

= 3 − ³for 0 ≤  ≤√3. Then

⁰() = 3 − 3²= 0 ⇔  = 1. The maximum value is  1√

2

= 2 and the minimum occurs both at  = 0 and  =√

3where  0√

3

= √

3 0

= 0. Thus the absolute maximum of  on

is  1√

2= 2, and the absolute minimum is 0 which occurs at all points along 1and 2.

39. ( ) = 2³+ ⁴ ⇒ ^( ) = 6²and ( ) = 4³. And so = 0and = 0only occur when  =  = 0.

Hence, the only critical point inside the disk is at  =  = 0 where (0 0) = 0. Now on the circle ²+ ²= 1, ² = 1 − ² so let () = ( ) = 2³+ (1 − ²)²= ⁴+ 2³− 2²+ 1, −1 ≤  ≤ 1. Then ⁰() = 4³+ 6²− 4 = 0 ⇒

 = 0, −2, or ¹2. (0 ±1) =  (0) = 1, 

1 2 ±^√2³



= 1 2

=¹³₁₆, and (−2 −3) is not in . Checking the endpoints, we

get (−1 0) = (−1) = −2 and (1 0) = (1) = 2. Thus the absolute maximum and minimum of  on  are (1 0) = 2 and (−1 0) = −2.

Another method: On the boundary ²+ ²= 1we can write  = cos ,  = sin , so (cos  sin ) = 2 cos³ + sin⁴, 0 ≤  ≤ 2.

40. ( ) = ³− 3 − ³+ 12 ⇒ ^( ) = 3²− 3 and ^( ) = −3²+ 12and the critical points are (1 2), (1 −2), (−1 2), and (−1 −2). But only (1 2) and (−1 2) are in  and (1 2) = 14, (−1 2) = 18. Along ¹:  = −2 and (−2 ) = −2 − ³+ 12, −2 ≤  ≤ 3, which has a maximum at  = 2 where (−2 2) = 14 and a minimum at

 = −2 where (−2 −2) = −18. Along ²:  = 2 and (2 ) = 2 − ³+ 12, 2 ≤  ≤ 3, which has a maximum at

 = 2where (2 2) = 18 and a minimum at  = 3 where (2 3) = 11. Along 3:  = 3 and ( 3) = ³− 3 + 9,

−2 ≤  ≤ 2, which has a maximum at  = −1 and  = 2 where (−1 3) =  (2 3) = 11 and a minimum at  = 1

55. A cardboard box without a lid is to have a volume of 32,000 cm³. Find the dimensions that minimize the amount of cardboard used.

Solution:

= 0implies  = 32 − ²

2 and substituting into = 0gives 32(4²) − (32 − ²)(4²) − (32 − ²)²= 0or 3⁴+ 64²− (32)²= 0. Thus ²=⁶⁴₆ or  = √⁸

6,  = ^643

16√

6 = ^√⁸₆and  = √⁸

6. Thus the box is a cube with edge length√⁸

6 cm.

51.Let the dimensions be , , and ; then 4 + 4 + 4 =  and the volume is

 =  = ₁

4 −  − 

= ¹₄ − ² − ²,   0,   0. Then =¹₄ − 2 − ²and =¹₄ − ²− 2, so = 0 = when 2 +  =¹₄and  + 2 = ¹₄. Solving, we get  =₁₂¹,  =₁₂¹and  =¹₄ −  −  =12¹. From the geometrical nature of the problem, this critical point must give an absolute maximum. Thus the box is a cube with edge length₁₂¹.

52.The cost equals 5 + 2( + ) and  =  , so ( ) = 5 + 2 ( + )() = 5 + 2 (⁻¹+ ⁻¹). Then

= 5 − 2 ⁻², = 5 − 2  ⁻², = 0implies  = 2 (5²), = 0implies  =³

2

5 = . Thus the dimensions of the aquarium which minimize the cost are  =  = ³

2

5 units,  = ^135 2

23

.

53.Let the dimensions be ,  and , then minimize  + 2( + ) if  = 32,000 cm³. Then

 ( ) =  + [64,000( + )] =  + 64,000(⁻¹+ ⁻¹), =  − 64,000⁻², =  − 64,000⁻². And = 0implies  = 64,000²; substituting into = 0implies ³= 64,000 or  = 40 and then  = 40. Now

( ) = [(2)(64,000)]²⁻³⁻³− 1  0 for (40 40) and ^(40 40)  0so this is indeed a minimum. Thus the dimensions of the box are  =  = 40 cm,  = 20 cm.

54.Let  be the length of the north and south walls,  the length of the east and west walls, and  the height of the building. The heat loss is given by  = 10(2) + 8(2) + 1() + 5() = 6 + 16 + 20 The volume is 4000 m³, so

 = 4000, and we substitute  =⁴⁰⁰⁰_ to obtain the heat loss function ( ) = 6 + 80,000 + 64,000.

(a) Since  = ⁴⁰⁰⁰_ ≥ 4,  ≤ 1000 ⇒  ≤ 1000. Also  ≥ 30 and

 ≥ 30, so the domain of  is  = {( ) |  ≥ 30 30 ≤  ≤ 1000}.

(b) ( ) = 6 + 80,000⁻¹+ 64,000⁻¹ ⇒

= 6 − 80,000⁻², = 6 − 64,000⁻².

= 0implies 6² = 80,000 ⇒  =80,000

6² and substituting into

= 0gives 6 = 64,000

 6² 80,000

2

⇒ ³= 80,000²

6 · 64,000 = 50,000 3 , so

 = ³

50,000 3 = 10³

50

3 ⇒  = 80

√3

60, and the only critical point of  is

 10³

50 3 80

√3

60



≈ (2554 2043)

which is not in . Next we check the boundary of .

2