Section 14.8 Lagrange Multipliers

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Section 14.8 Lagrange Multipliers

486 ¤ CHAPTER 14 PARTIAL DERIVATIVES

8.  (  ) = ^, (  ) = 2²+ ²+ ²= 24, and ∇ = ∇ ⇒ h^ ^ ^i = h4 2 2i.

Then ^= 4, ^= 2, ^= 2, and 2²+ ²+ ²= 24. If any of , , , or  is zero, then the ﬁrst three equations imply that two of the variables , ,  must be zero. If  =  =  = 0 it contradicts the fourth equation, so exactly two are zero, and from the fourth equation the possibilities are

±2√

3 0 0, 0 ±2√

6 0,

0 0 ±2√ 6, all with an -value of ⁰= 1. If none of , , ,  is zero then from the ﬁrst three equations we have

4

 = ^= 2

 =2

 ⇒ 2

 = 

 = 

. This gives 2² = ² ⇒ 2² = ² and ² = ² ⇒

²= ². Substituting into the fourth equation, we have ²+ ²+ ²= 24 ⇒ ²= 8 ⇒  = ±2√ 2, so

²= 4 ⇒  = ±2 and ²= ² ⇒  = ±2√

2, giving possible points

±2 ±2√ 2 ±2√

2(all combinations).

The value of  is ¹⁶when all coordinates are positive or exactly two are negative, and the value is ⁻¹⁶when all are negative or exactly one of the coordinates is negative. Thus the maximum of  subject to the constraint is ¹⁶and the minimum is ⁻¹⁶.

9.  (  ) = ², (  ) = ²+ ²+ ²= 4, and ∇ = ∇ ⇒ 

² 2 ²

=  h2 2 2i. Then

² = 2, 2 = 2, ² = 2, and ²+ ²+ ²= 4.

Case 1: If  = 0, then the ﬁrst equation implies that  = 0 or  = 0. If  = 0, then any values of  and  satisfy the ﬁrst three equations, so from the fourth equation all points ( 0 ) such that ²+ ²= 4are possible points. If  = 0 then from the third equation  = 0 or  = 0, and from the fourth equation, the possible points are (0 ±2 0), (±2 0 0). The -value in all these cases is 0.

Case 2: If  6= 0 but any one of , ,  is zero, the ﬁrst three equations imply that all three coordinates must be zero, contradicting the fourth equation. Thus if  6= 0, none of , ,  is zero and from the ﬁrst three equations we have

²

2 =  =  =²

2 . This gives ² = 2² ⇒ ²= 2² and 2²²= 2²² ⇒ ²= ². Substituting into the fourth equation, we have ²+ 2²+ ²= 4 ⇒ ²= 1 ⇒  = ±1, so  = ±√

2and  = ±1, giving possible points

±1 ±√ 2 ±1

(all combinations). The value of  is 2 when  and  are the same sign and −2 when they are opposite.

Thus the maximum of  subject to the constraint is (1 ±√

2 1) =  (−1 ±√

2 −1) = 2 and the minimum is

 (1 ±√

2 −1) = (−1 ±√

2 1) = −2.

10.  (  ) = ln(²+ 1) + ln(²+ 1) + ln(²+ 1), (  ) = ²+ ²+ ²= 12. Then ∇ = ∇ ⇒

 2

²+ 1 2

²+ 1 2

²+ 1



=  h2 2 2i, so 2

²+ 1= 2, 2

²+ 1= 2, 2

²+ 1= 2, and ²+ ²+ ²= 12.

First, if  = 0 then  =  =  = 0 which contradicts the last equation, so we may assume that  6= 0.

Case 1: If  6= 0,  6= 0, and  6= 0, then from the ﬁrst three equations we have 1

²+ 1=  = 1

²+ 1= 1

²+ 1 ⇒

8.  (  ) = ^, (  ) = 2²+ ²+ ²= 24, and ∇ = ∇ ⇒ h^ ^ ^i = h4 2 2i.

Then ^= 4, ^= 2, ^= 2, and 2²+ ²+ ²= 24. If any of , , , or  is zero, then the ﬁrst three equations imply that two of the variables , ,  must be zero. If  =  =  = 0 it contradicts the fourth equation, so exactly two are zero, and from the fourth equation the possibilities are

±2√

3 0 0, 0 ±2√

6 0,

0 0 ±2√ 6, all with an -value of ⁰= 1. If none of , , ,  is zero then from the ﬁrst three equations we have

4

 = ^= 2

 =2

 ⇒ 2

 = 

 = 

. This gives 2² = ² ⇒ 2² = ² and ² = ² ⇒

²= ². Substituting into the fourth equation, we have ²+ ²+ ²= 24 ⇒ ²= 8 ⇒  = ±2√2, so

²= 4 ⇒  = ±2 and ²= ² ⇒  = ±2√

2, giving possible points

±2 ±2√ 2 ±2√

2

(all combinations).

The value of  is ¹⁶when all coordinates are positive or exactly two are negative, and the value is ⁻¹⁶when all are negative or exactly one of the coordinates is negative. Thus the maximum of  subject to the constraint is ¹⁶and the minimum is ⁻¹⁶.

9.  (  ) = ², (  ) = ²+ ²+ ²= 4, and ∇ = ∇ ⇒ 

² 2 ²

=  h2 2 2i. Then

² = 2, 2 = 2, ² = 2, and ²+ ²+ ²= 4.

Case 1: If  = 0, then the ﬁrst equation implies that  = 0 or  = 0. If  = 0, then any values of  and  satisfy the ﬁrst three equations, so from the fourth equation all points ( 0 ) such that ²+ ²= 4are possible points. If  = 0 then from the third equation  = 0 or  = 0, and from the fourth equation, the possible points are (0 ±2 0), (±2 0 0). The -value in all these cases is 0.

Case 2: If  6= 0 but any one of , ,  is zero, the ﬁrst three equations imply that all three coordinates must be zero, contradicting the fourth equation. Thus if  6= 0, none of , ,  is zero and from the ﬁrst three equations we have

²

2 =  =  =²

2 . This gives ² = 2² ⇒ ²= 2² and 2²²= 2²² ⇒ ²= ². Substituting into the fourth equation, we have ²+ 2²+ ²= 4 ⇒ ²= 1 ⇒  = ±1, so  = ±√

2and  = ±1, giving possible points

±1 ±√ 2 ±1

(all combinations). The value of  is 2 when  and  are the same sign and −2 when they are opposite.

Thus the maximum of  subject to the constraint is (1 ±√

2 1) =  (−1 ±√

2 −1) = 2 and the minimum is

 (1 ±√

2 −1) = (−1 ±√

2 1) = −2.

10.  (  ) = ln(²+ 1) + ln(²+ 1) + ln(²+ 1), (  ) = ²+ ²+ ²= 12. Then ∇ = ∇ ⇒

 2

²+ 1 2

²+ 1 2

²+ 1



=  h2 2 2i, so 2

²+ 1= 2, 2

²+ 1= 2, 2

²+ 1= 2, and ²+ ²+ ²= 12.

First, if  = 0 then  =  =  = 0 which contradicts the last equation, so we may assume that  6= 0.

Case 1: If  6= 0,  6= 0, and  6= 0, then from the ﬁrst three equations we have 1

²+ 1=  = 1

²+ 1= 1

²+ 1 ⇒

SECTION 14.8 LAGRANGE MULTIPLIERS ¤ 487

² = ² = ², and substitution into the last equation gives 3²= 12 ⇒  = ±2. Thus possible points are (±2 ±2 ±2) (all combinations), all of which have an -value of 3 ln 5.

Case 2: If exactly one of , ,  is zero, say  = 0, then from the second and third equations we have ²= ². Substitution into the last equation gives 2²= 12 ⇒  = ±√

6. The situation is similar for  = 0 or  = 0, giving possible points

0 ±√ 6 ±√

6,

±√ 6 0 ±√

6,

±√ 6 ±√

6 0(all combinations), all with an -value of 2 ln 7.

Case 3: If exactly two of , ,  are zero, then the square of the nonzero variable is 12, giving possible points

0 0 ±2√ 3,

0 ±2√ 3 0,

±2√

3 0 0, all with an -value of ln 13.

Thus the maximum of  subject to the constraint is 3 ln 5 ≈ 483 and the minimum is ln 13 ≈ 256.

11.  (  ) = ²+ ²+ ², (  ) = ⁴+ ⁴+ ⁴= 1 ⇒ ∇ = h2 2 2i, ∇ =

4³ 4³ 4³. Case 1: If  6= 0,  6= 0, and  6= 0, then ∇ = ∇ implies  = 1(2²) = 1(2²) = 1(2²)or ²= ²= ²and 3⁴= 1or  = ±√4¹

3giving the points

±√4¹ 3_√₄¹

3 _√₄¹

3

,

±√4¹ 3 −√4¹

3_√₄¹

3

,

±√4¹ 3_√₄¹

3 −√4¹ 3

,

±√4¹ 3 −√4¹

3 −√4¹ 3



all with an -value of√ 3.

Case 2: If one of the variables equals zero and the other two are not zero, then the squares of the two nonzero coordinates are equal with common value√¹

2and corresponding -value of√2.

Case 3: If exactly two of the variables are zero, then the third variable has value ±1 with the corresponding -value of 1.

Thus on ⁴+ ⁴+ ⁴= 1, the maximum value of  is√

3and the minimum value is 1.

12.  (  ) = ⁴+ ⁴+ ⁴, (  ) = ²+ ²+ ²= 1 ⇒ ∇ =

4³ 4³ 4³

, ∇ = h2 2 2i.

Case 1: If  6= 0,  6= 0, and  6= 0 then ∇ = ∇ implies  = 2² = 2²= 2²or ²= ²= ²=¹₃ giving 8 points each with an -value of¹₃.

Case 2: If one of the variables is 0 and the other two are not, then the squares of the two nonzero coordinates are equal with

common value¹₂ and the corresponding -value is¹₂.

Case 3: If exactly two of the variables are 0, then the third variable has value ±1 with corresponding -value of 1.

Thus on ²+ ²+ ²= 1, the maximum value of  is 1 and the minimum value is¹₃.

13.  (   ) =  +  +  + , (   ) = ²+ ²+ ²+ ²= 1 ⇒ h1 1 1 1i = h2 2 2 2i, so

 = 1(2) = 1(2) = 1(2) = 1(2)and  =  =  = . But ²+ ²+ ²+ ²= 1, so the possible points are

±¹2 ±¹2 ±¹2 ±¹2

. Thus the maximum value of  is 1

2¹₂¹₂¹₂

= 2and the minimum value is



−¹2 −¹2 −¹2 −¹2

= −2.

14.  (1 2     ) = 1+ 2+ · · · + ^, (1 2     ) = ²1+ ²2+ · · · + ²^= 1 ⇒

h1 1     1i = h2¹ 22     2i, so  = 1(2¹) = 1(22) = · · · = 1(2^)and 1= 2= · · · = ^.

1

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² = ² = ², and substitution into the last equation gives 3²= 12 ⇒  = ±2. Thus possible points are (±2 ±2 ±2) (all combinations), all of which have an -value of 3 ln 5.

Case 2: If exactly one of , ,  is zero, say  = 0, then from the second and third equations we have ²= ². Substitution into the last equation gives 2²= 12 ⇒  = ±√

6. The situation is similar for  = 0 or  = 0, giving possible points

0 ±√ 6 ±√

6 ,

±√ 6 0 ±√

6 ,

±√ 6 ±√

6 0

(all combinations), all with an -value of 2 ln 7.

Case 3: If exactly two of , ,  are zero, then the square of the nonzero variable is 12, giving possible points

0 0 ±2√ 3,

0 ±2√ 3 0,

±2√

3 0 0, all with an -value of ln 13.

Thus the maximum of  subject to the constraint is 3 ln 5 ≈ 483 and the minimum is ln 13 ≈ 256.

11.  (  ) = ²+ ²+ ², (  ) = ⁴+ ⁴+ ⁴= 1 ⇒ ∇ = h2 2 2i, ∇ =

4³ 4³ 4³. Case 1: If  6= 0,  6= 0, and  6= 0, then ∇ = ∇ implies  = 1(2²) = 1(2²) = 1(2²)or ²= ²= ²and 3⁴= 1or  = ±√4¹

3giving the points

±√4¹ 3_√₄¹

3 _√₄¹

3

,

±√4¹ 3 −√4¹

3_√₄¹

3

,

±√4¹ 3_√₄¹

3 −√4¹ 3

,

±√4¹ 3 −√4¹

3 −√4¹ 3



all with an -value of√3.

Case 2: If one of the variables equals zero and the other two are not zero, then the squares of the two nonzero coordinates are equal with common value√¹

2and corresponding -value of√2.

Case 3: If exactly two of the variables are zero, then the third variable has value ±1 with the corresponding -value of 1.

Thus on ⁴+ ⁴+ ⁴= 1, the maximum value of  is√

3and the minimum value is 1.

12.  (  ) = ⁴+ ⁴+ ⁴, (  ) = ²+ ²+ ²= 1 ⇒ ∇ =

4³ 4³ 4³

, ∇ = h2 2 2i.

Case 1: If  6= 0,  6= 0, and  6= 0 then ∇ = ∇ implies  = 2² = 2²= 2²or ²= ²= ²=¹₃ giving 8 points each with an -value of¹₃.

Case 2: If one of the variables is 0 and the other two are not, then the squares of the two nonzero coordinates are equal with

common value¹₂ and the corresponding -value is¹₂.

Case 3: If exactly two of the variables are 0, then the third variable has value ±1 with corresponding -value of 1.

Thus on ²+ ²+ ²= 1, the maximum value of  is 1 and the minimum value is¹₃.

13.  (   ) =  +  +  + , (   ) = ²+ ²+ ²+ ²= 1 ⇒ h1 1 1 1i = h2 2 2 2i, so

 = 1(2) = 1(2) = 1(2) = 1(2)and  =  =  = . But ²+ ²+ ²+ ²= 1, so the possible points are

±¹2 ±¹2 ±¹2 ±¹2

. Thus the maximum value of  is 1

2¹₂¹₂¹₂

= 2and the minimum value is



−¹2 −¹2 −¹2 −¹2

= −2.

14.  (1 2     ) = 1+ 2+ · · · + ^, (1 2     ) = ²1+ ²2+ · · · + ²^= 1 ⇒

h1 1     1i = h2¹ 22     2i, so  = 1(2¹) = 1(22) = · · · = 1(2^)and 1= 2= · · · = ^.

488 ¤ CHAPTER 14 PARTIAL DERIVATIVES But ²1+ ²2+ · · · + ²^= 1, so = ±1√

for  = 1,   , . Thus the maximum value of  is

 (1√

 1√

,    , 1√ ) = √ and the minimum value is (−1√

 −1√

,    , − 1√

 ) = −√.

15.  ( ) = ²+ ², ( ) =  = 1, and ∇ = ∇ ⇒ h2 2i = h i, so 2 = , 2 = , and  = 1.

From the last equation,  6= 0 and  6= 0, so 2 =  ⇒  = 2. Substituting, we have 2 = (2)  ⇒

²= ² ⇒  = ±. But  = 1, so  =  = ±1 and the possible points for the extreme values of  are (1 1) and (−1 −1). Here there is no maximum value, since the constraint  = 1 ⇔  = 1 allows  or  to become arbitrarily large, and hence ( ) = ²+ ²can be made arbitrarily large. The minimum value is (1 1) = (−1 −1) = 2.

16.  (  ) = ²+ 2²+ 3², ( ) =  + 2 + 3 = 10, and ∇ = ∇ ⇒ h2 4 6i = h 2 3i, so 2 = , 4 = 2, 6 = 3, and  + 2 + 3 = 10. From the ﬁrst three equations we have 2 =  = 2 = 2 ⇒  =  = , and substituting into the fourth equation gives  + 2 + 3 = 10 ⇒  =⁵3 =  = . Thus the only possible point for an extreme value of  is5

3⁵₃⁵₃

. Notice here that the constraint  + 2 + 3 = 10 allows any of ||, ||, or || to be arbitrarily large, and hence (  ) = ²+ 2²+ 3²can be made arbitrarily large. So  has no maximum value subject to the constraint. The minimum value is 5

3⁵₃⁵₃

= 65 3

2

=⁵⁰₃.

17.  (  ) =  +  + , (  ) = ²+ ² = 2, (  ) =  +  = 1, and ∇ = ∇ + ∇ ⇒

h1 1 1i = h2 0 2i + h  0i. Then 1 = 2 + , 1 = , 1 = 2, ²+ ²= 2, and  +  = 1. Substituting

 = 1into the ﬁrst equation gives  = 0 or  = 0. But  = 0 contradicts 1 = 2, so  = 0. Then  +  = 1 ⇒  = 1 and ²+ ²= 2 ⇒  = ±√2, so the possible points are

0 1 ±√

2. The maximum value of  subject to the constraints is (0 1√

2) = 1 +√

2 ≈ 241 and the minimum is (0 1 −√

2) = 1 −√

2 ≈ −041.

Note: Since  +  = 1 is one of the constraints, we could have solved the problem by solving ( ) = 1 +  subject to

²+ ²= 2.

18.  (  ) = , (  ) = ²+ ²− ² = 0, (  ) =  +  +  = 24, and ∇ = ∇ + ∇ ⇒

h0 0 1i = h2 2 −2i + h  i. Then 0 = 2 + , 0 = 2 + , 1 = −2 + , ²+ ²− ²= 0, and

 +  +  = 24. From the ﬁrst two equations we have −2 =  = −2 ⇒  = 0 or  = . But  = 0 ⇒  = 0 which contradicts the third equation, so  =  and substitution into the last equation gives  = 24 − 2. From the fourth equation we have ²+ ²− (24 − 2)²= 0 ⇒ −2²+ 96 − 576 = 0 ⇒ ²− 48 + 288 = 0 ⇒

 =48 ±√ 1152

2 = 24 ± 12√

2 = . Now  = 24 − 2, so the possible points are

24 + 12√

2 24 + 12√

2 −24 − 24√ 2 and

24 − 12√

2 24 − 12√

2 −24 + 24√

2. The maximum of  subject to the constraints is



24 − 12√

2 24 − 12√

2 −24 + 24√ 2

= −24 + 24√

2 ≈ 994 and the minimum is



24 + 12√

2 24 + 12√

2 −24 − 24√ 2

= −24 − 24√

2 ≈ −5794.

19. (  ) =  + 2, (  ) =  +  +  = 1, (  ) = ²+ ²= 4 ⇒ ∇ = h1 2 0i, ∇ = h  i and ∇ = h0 2 2i. Then 1 = , 2 =  + 2 and 0 =  + 2 so  = ¹2 = − or  = 1(2),  = −1(2).

Thus  +  +  = 1 implies  = 1 and ²+ ²= 4implies  = ±2√¹

2. Then the possible points are 1 ±√

2 ∓√ 2 and the maximum value is 

1√ 2 −√

2

= 1 + 2√

2and the minimum value is  1 −√

2√ 2

= 1 − 2√2.

20. (  ) = 3 −  − 3, (  ) =  +  −  = 0, (  ) = ²+ 2²= 1 ⇒ ∇ = h3 −1 −3i,

∇ = h  −i, ∇ = (2 0 4). Then 3 =  + 2, −1 =  and −3 = − + 4, so  = −1,  = −1,

 = 2. Thus (  ) = 1 implies 4

² + 2

 1

²



= 1or  = ±√

6, so  = ∓^√¹₆;  = ±^√²₆; and (  ) = 0

implies  = ∓^√³₆. Hence the maximum of  subject to the constraints is √ 6

3  −^√2⁶ −^√6⁶



= 2√

6and the minimum is 

−^√3⁶^√₂⁶^√₆⁶

= −2√6.

21. ( ) = ²+ ²+ 4 − 4. For the interior of the region, we ﬁnd the critical points: ^= 2 + 4, = 2 − 4, so the only critical point is (−2 2) (which is inside the region) and (−2 2) = −8. For the boundary, we use Lagrange multipliers.

( ) = ²+ ²= 9, so ∇ = ∇ ⇒ h2 + 4 2 − 4i = h2 2i. Thus 2 + 4 = 2 and 2 − 4 = 2.

Adding the two equations gives 2 + 2 = 2 + 2 ⇒  +  = ( + ) ⇒ ( + )( − 1) = 0, so

 +  = 0 ⇒  = − or  − 1 = 0 ⇒  = 1. But  = 1 leads to a contradition in 2 + 4 = 2, so  = − and

²+ ²= 9implies 2² = 9 ⇒  = ±^√³₂. We have 

√3 2 −^√³₂

= 9 + 12√

2 ≈ 2597 and



−^√³₂√³ 2



= 9 − 12√

2 ≈ −797, so the maximum value of  on the disk ²+ ²≤ 9 is 

√3 2 −^√³₂

= 9 + 12√ 2and the minimum is (−2 2) = −8.

22. ( ) = 2²+ 3²− 4 − 5 ⇒ ∇ = h4 − 4 6i = h0 0i ⇒  = 1,  = 0. Thus (1 0) is the only critical point of , and it lies in the region ²+ ²  16. On the boundary, ( ) = ²+ ²= 16 ⇒ ∇ = h2 2i, so 6 = 2 ⇒ either  = 0 or  = 3. If  = 0, then  = ±4; if  = 3, then 4 − 4 = 2 ⇒  = −2 and

 = ±2√

3. Now (1 0) = −7, (4 0) = 11, (−4 0) = 43, and 

−2 ±2√ 3

= 47. Thus the maximum value of

 ( )on the disk ²+ ²≤ 16 is 

−2 ±2√ 3

= 47, and the minimum value is (1 0) = −7.

23. ( ) = ^−. For the interior of the region, we ﬁnd the critical points: = −^−, = −^−, so the only critical point is (0 0), and (0 0) = 1. For the boundary, we use Lagrange multipliers. ( ) = ²+ 4²= 1 ⇒

∇ = h2 8i, so setting ∇ = ∇ we get −^−= 2and −^−= 8. The ﬁrst of these gives

^−= −2, and then the second gives −(−2) = 8 ⇒ ²= 4². Solving this last equation with the

19.  (  ) =  + , (  ) =  = 1, (  ) = ²+ ²= 1 ⇒ ∇ = h  +  i, ∇ = h  0i,

∇ = h0 2 2i. Then  =  implies  = 1 [ 6= 0 since (  ) = 1],  +  =  + 2 and  = 2. Thus

 = (2) = (2)or ²= ², and so ²+ ²= 1implies  = ±^√¹₂,  = ±^√¹₂. Then  = 1 implies  = ±√ 2and the possible points are

±√

2 ±^√¹₂√¹ 2

,

±√

2 ±^√¹₂ −^√¹₂

. Hence the maximum of  subject to the constraints is



±√

2 ±^√¹₂ ±^√¹₂

= ³₂and the minimum is 

±√

2 ±^√¹₂ ∓^√¹₂

=¹₂.

Note: Since  = 1 is one of the constraints we could have solved the problem by solving ( ) =  + 1 subject to

²+ ²= 1.

20.  (  ) = ²+ ²+ ², (  ) =  −  = 1, (  ) = ²− ²= 1 ⇒ ∇ = h2 2 2i,

∇ = h − 0i, and ∇ = h0 2 −2i. Then 2 = , 2 = − + 2, and 2 = −2 ⇒  = 0or  = −1.

If  = 0 then ²− ²= 1implies ²= 1 ⇒  = ±1. If  = 1,  −  = 1 implies  = 2, and if  = −1 we have

 = 0, so possible points are (2 1 0) and (0 −1 0). If  = −1 then 2 = − + 2 implies 4 = −, but  = 2 so 4 = −2 ⇒  = −2 and  −  = 1 implies −3 = 1 ⇒  = −¹3. But then ²− ²= 1implies ²= −⁸9, an impossibility. Thus the maximum value of  subject to the constraints is (2 1 0) = 5 and the minimum is (0 −1 0) = 1.

Note: Since  −  = 1 ⇒  =  + 1 is one of the constraints we could have solved the problem by solving

 ( ) = ( + 1)²+ ²+ ²subject to ²− ²= 1.

21.  ( ) = ²+ ²+ 4 − 4. For the interior of the region, we ﬁnd the critical points: ^= 2 + 4, = 2 − 4, so the only critical point is (−2 2) (which is inside the region) and (−2 2) = −8. For the boundary, we use Lagrange multipliers.

( ) = ²+ ²= 9, so ∇ = ∇ ⇒ h2 + 4 2 − 4i = h2 2i. Thus 2 + 4 = 2 and 2 − 4 = 2.

Adding the two equations gives 2 + 2 = 2 + 2 ⇒  +  = ( + ) ⇒ ( + )( − 1) = 0, so

 +  = 0 ⇒  = − or  − 1 = 0 ⇒  = 1. But  = 1 leads to a contradition in 2 + 4 = 2, so  = − and

²+ ² = 9implies 2²= 9 ⇒  = ±^√³₂. We have 

√3

2 −^√³₂

= 9 + 12√

2 ≈ 2597 and



−^√³₂√³ 2

= 9 − 12√

2 ≈ −797, so the maximum value of  on the disk ²+ ²≤ 9 is 

√3 2 −^√³₂

= 9 + 12√ 2and the minimum is (−2 2) = −8.

22.  ( ) = 2²+ 3²− 4 − 5 ⇒ ∇ = h4 − 4 6i = h0 0i ⇒  = 1,  = 0. Thus (1 0) is the only critical point of , and it lies in the region ²+ ² 16. On the boundary, ( ) = ²+ ²= 16 ⇒ ∇ = h2 2i, so 6 = 2 ⇒ either  = 0 or  = 3. If  = 0, then  = ±4; if  = 3, then 4 − 4 = 2 ⇒  = −2 and

 = ±2√

3. Now (1 0) = −7, (4 0) = 11, (−4 0) = 43, and 

−2 ±2√

3= 47. Thus the maximum value of

 ( )on the disk ²+ ²≤ 16 is 

−2 ±2√ 3

= 47, and the minimum value is (1 0) = −7.

(b) The constraint is ²+ ⁴− ³= 0 ⇔ ²= ³− ⁴. The left side is non-negative, so we must have ³− ⁴≥ 0 which is true only for 0 ≤  ≤ 1. Therefore the minimum possible value for ( ) =  is 0 which occurs for  =  = 0.

However, ∇(0 0) =  h0 − 0 0i = h0 0i and ∇(0 0) = h1 0i, so ∇(0 0) 6= ∇(0 0) for all values of .

(c) Here ∇(0 0) = 0 but the method of Lagrange multipliers requires that ∇ 6= 0 everywhere on the constraint curve.

26. (a) The graphs of ( ) = 37 and ( ) = 350 seem to be tangent to the circle, and so 37 and 350 are the approximate minimum and maximum values of the function ( ) subject to the constraint ( − 3)²+ ( − 3)²= 9.

(b) Let ( ) = ( − 3)²+ ( − 3)². We calculate ( ) = 3²+ 3,

( ) = 3²+ 3, ( ) = 2 − 6, and ^( ) = 2 − 6, and use a CAS to search for solutions to the equations ( ) = ( − 3)²+ ( − 3)²= 9,

= , and = . The solutions are ( ) = 3 −³2

√2 3 −³2

√2

≈ (0879 0879) and ( ) =

3 +³₂√

2 3 +³₂√ 2

≈ (5121 5121). These give  3 −³2

√2 3 −³2

√2

=³⁵¹₂ −²⁴³2

√2 ≈ 3673 and

 3 +³₂√

2 3 +³₂√ 2

=³⁵¹₂ +²⁴³₂ √

2 ≈ 34733, in accordance with part (a).

27.  ( ) = ^¹^−, ( ) =  +  =  ⇒ ∇ =

^⁻¹¹^− (1 − )^^−

, ∇ = h i.

Then ()¹^−= and (1 − )()^= and  +  = , so ()¹^− = (1 − )()^or

[(1 − )] = ()^()¹^−or  = [(1 − )]. Substituting into  +  =  gives  = (1 − )

and  =  for the maximum production.

28. ( ) =  + , ( ) = ^¹^−=  ⇒ ∇ = h i, ∇ =

^⁻¹¹^− (1 − )^^−.

Then 







1−

= 

(1 − )







and ^¹^−=  ⇒ 

(1 − )=





1−







⇒

 = 

(1 − )and so 

 

(1 − )



¹^−= . Hence  = 

 ([(1 − )])^ =^(1 − )^

^^ and  =^⁻¹(1 − )^⁻¹

^⁻¹^⁻¹ = ¹^−¹^−

¹^−(1 − )¹^− minimizes cost.

29. Let the sides of the rectangle be  and . Then ( ) = , ( ) = 2 + 2 =  ⇒ ∇( ) = h i,

∇ = h2 2i. Then  =¹2 =¹₂implies  =  and the rectangle with maximum area is a square with side length¹₄.

30. Let (  ) = ( − )( − )( − ), (  ) =  +  + . Then

∇ = h−( − )( − ) −( − )( − ) −( − )( − )i, ∇ = h  i. Thus

( − )( − ) = ( − )( − ) (1), and ( − )( − ) = ( − )( − ) (2). (1) implies  =  while (2) implies  = , so  =  =  = 3 and the triangle with maximum area is equilateral.

2

(3)

 (−1951921 −0545867 0119973) ≈ −00688, (0155142 0904622 0950293) ≈ 04084,

 (1138731 1768057 −0573138) ≈ 97938. Thus the maximum is approximately 97938, and the minimum is approximately −53506.

48.  (  ) =  +  + , (  ) = ²− ²−  = 0, (  ) = ²+ ²= 4.

∇ = ∇ + ∇ ⇒ h1 1 1i = h2 −2 −1i + h2 0 2i, so 1 = 2 + 2, 1 = −2, 1 = − + 2,

²− ²= , ²+ ²= 4. Using a CAS to solve these 5 equations simultaneously for , , , , and , we get 4 real-valued solutions:

 ≈ −1652878,  ≈ −1964194,  ≈ −1126052,  ≈ 0254557,  ≈ −0557060

 ≈ −1502800,  ≈ 0968872,  ≈ 1319694,  ≈ −0516064,  ≈ 0183352

 ≈ −0992513,  ≈ 1649677,  ≈ −1736352,  ≈ −0303090,  ≈ −0200682

 ≈ 1895178,  ≈ 1718347,  ≈ 0638984,  ≈ −0290977,  ≈ 0554805 Substituting these values into  gives (−1652878 −1964194 −1126052) ≈ −47431,

 (−1502800 0968872 1319694) ≈ 07858, (−0992513 1649677 −1736352) ≈ −10792,

 (1895178 1718347 0638984) ≈ 42525. Thus the maximum is approximately 42525, and the minimum is approximately −47431.

49. (a) We wish to maximize (1 2,    , ) = √^12· · · ^subject to

(1 2,    , ) = 1+ 2+ · · · + ^= and  0.

∇ =

1

(12· · · ^)^¹⁻¹(2· · · ^),_¹(12· · · ^)¹^⁻¹(13· · · ^),    ,_¹(12· · · ^)^¹⁻¹(1· · · ^−1) and ∇ = h ,    , i, so we need to solve the system of equations

1

(12· · · ^)¹^⁻¹(2· · · ^) =  ⇒ ^1₁ ^1₂ · · · ^1^ = 1 1

(12· · · ^)¹^⁻¹(13· · · ^) =  ⇒ ^1₁ ^1₂ · · · ^1^ = 2

...

1

(12· · · ^)^¹⁻¹(1· · · ^−1) =  ⇒ ^1₁ ^1₂ · · · ^1^ = 

This implies 1= 2 = · · · = ^. Note  6= 0, otherwise we can’t have all ^ 0. Thus 1= 2= · · · = ^. But 1+ 2+ · · · + ^=  ⇒ ¹=  ⇒ 1= 

 = 2= 3= · · · = ^. Then the only point where  can have an extreme value is 



,    , 



. Since we can choose values for (1 2     )that make  as close to

zero (but not equal) as we like,  has no minimum value. Thus the maximum value is

 

 

,    , 





= ^



· 

· · · 

 = 

.

APPLIED PROJECT ROCKET SCIENCE ¤ 497 (b) From part (a), 

is the maximum value of . Thus (1 2,    , ) = √^12· · · ^≤ 

. But

1+ 2+ · · · + ^= , so √^12· · · ^≤1+ 2+ · · · + ^

 . These two means are equal when  attains its maximum value 

, but this can occur only at the point 



,    , 



we found in part (a). So the means are equal only

when 1= 2= 3= · · · = ^= 

.

50. (a) Let (1      1     ) =



 = 1

, (1     ) =



 = 1

²_, and (1     ) =



 = 1

_². Then

∇ = ∇



 = 1

= h¹ 2      1 2     i, ∇ = ∇



 = 1

²_ = h2¹ 22     2 0 0     0i and

∇ = ∇



 = 1

²_ = h0 0     0 2¹ 22     2i. So ∇ = ∇ + ∇ ⇔ ^= 2and = 2,

1 ≤  ≤ . Then 1 =



 = 1

_²=



 = 1

4²²_ = 4²



 = 1

²_ = 4² ⇒  = ±¹2. If  =¹₂ then = 2₁

2

= ,

1 ≤  ≤ . Thus



 = 1

=



 = 1

²_ = 1. Similarly if  = −¹2we get = −^and ^

 = 1

= −1. Similarly we get

 = ±¹2 giving = ±^, 1 ≤  ≤ , and ^

 = 1

= ±1. Thus the maximum value of



 = 1

is 1.

(b) Here we assume ^

 = 1

²_ 6= 0 and



 = 1

²_ 6= 0. (If



 = 1

²_ = 0, then each = 0and so the inequality is trivially true.)

= 

²_ ⇒ 

²_ =

²_

²_ = 1, and = 

²_ ⇒ 

²_ =

²_

²_ = 1. Therefore, from part (a),

= 

²_

²_ ≤ 1 ⇔ 

≤²_²_.

APPLIED PROJECT Rocket Science

1. Initially the rocket engine has mass  = 1and payload mass  = 2+ 3+ . Then the change in velocity resulting from the ﬁrst stage is ∆1 = − ln



1 − (1 − )¹

2+ 3+  + 1



. After the ﬁrst stage is jettisoned we can consider the rocket engine to have mass  = 2and the payload to have mass  = 3+ . The resulting change in velocity from the second stage is ∆2= − ln



1 − (1 − )²

3+  + 2



. When only the third stage remains, we have = 3and  = , so

the resulting change in velocity is ∆3= − ln



1 −(1 − )³

 + 3



. Since the rocket started from rest, the ﬁnal velocity

APPLIED PROJECT ROCKET SCIENCE ¤ 497 (b) From part (a), 

is the maximum value of . Thus (1 2,    , ) = √^12· · · ^≤ 

. But

1+ 2+ · · · + ^= , so √^12· · · ^≤1+ 2+ · · · + ^

 . These two means are equal when  attains its maximum value 

, but this can occur only at the point 



,    , 



we found in part (a). So the means are equal only

when 1= 2= 3= · · · = ^= 

.

50. (a) Let (1      1     ) =



 = 1

, (1     ) =



 = 1

²_, and (1     ) =



 = 1

_². Then

∇ = ∇



 = 1

= h¹ 2      1 2     i, ∇ = ∇



 = 1

²_ = h2¹ 22     2 0 0     0i and

∇ = ∇



 = 1

²_ = h0 0     0 2¹ 22     2i. So ∇ = ∇ + ∇ ⇔ ^= 2and = 2,

1 ≤  ≤ . Then 1 =



 = 1

_²=



 = 1

4²²_ = 4²



 = 1

²_ = 4² ⇒  = ±¹2. If  =¹₂ then = 2₁

2

= ,

1 ≤  ≤ . Thus



 = 1

=



 = 1

²_ = 1. Similarly if  = −¹2we get = −^and ^

 = 1

= −1. Similarly we get

 = ±¹2 giving = ±^, 1 ≤  ≤ , and ^

 = 1

= ±1. Thus the maximum value of



 = 1

is 1.

(b) Here we assume ^

 = 1

²_ 6= 0 and



 = 1

²_ 6= 0. (If



 = 1

²_ = 0, then each = 0and so the inequality is trivially true.)

= 

²_ ⇒ 

²_ =

²_

²_ = 1, and = 

²_ ⇒ 

²_ =

²_

²_ = 1. Therefore, from part (a),

= 

²_

²_ ≤ 1 ⇔ 

≤²_²_.

APPLIED PROJECT Rocket Science

1. Initially the rocket engine has mass  = 1and payload mass  = 2+ 3+ . Then the change in velocity resulting from the ﬁrst stage is ∆1 = − ln



1 − (1 − )¹

2+ 3+  + 1



. After the ﬁrst stage is jettisoned we can consider the rocket engine to have mass  = 2and the payload to have mass  = 3+ . The resulting change in velocity from the second stage is ∆2= − ln



1 − (1 − )²

3+  + 2



. When only the third stage remains, we have = 3and  = , so

the resulting change in velocity is ∆3= − ln



1 −(1 − )³

 + 3



. Since the rocket started from rest, the ﬁnal velocity