Section 14.8 Lagrange Multipliers

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Section 14.8 Lagrange Multipliers

7. Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint.

f (x, y) = 2x²+ 6y², x⁴+ 3y⁴= 1

Solution:

SECTION 14.8 LAGRANGE MULTIPLIERS ¤ 1451

2²= 8² ⇒ ²= 4², and substitution into the third equation gives 4²+ 4²= 8 ⇒  = ±1. If  = ±1 then

²= 4 ⇒  = ±2, so  has possible extreme values at (1 ±2) and (−1 ±2). Evaluating  at these points, we see that the maximum value is (1 2) = (−1 −2) = 2 and the minimum is (1 −2) = (−1 2) = −2.

6.  ( ) = ^, ( ) = ²+ ²= 2, and ∇ = ∇ ⇒ h^ ^i = h2 2i, so ^= 2, ^= 2, and

²+ ²= 2. First note that from the first equation  6= 0. If  = 0, the second equation implies  = 0, so  6= 0. Then from the first two equations we have ^

2=  = ^

2 ⇒ 2^= 2²^ ⇒  = ², and substituting into the third equation gives ²+ (²)²= 2 ⇒ ⁴+ ²− 2 = 0 ⇒ (²+ 2)(²− 1) = 0 ⇒  = ±1. From  = ²we have  = 1, so  has possible extreme values at (±1 1). Evaluating  at these points, we see that the maximum value is

 (1 1) = and the minimum is (−1 1) = −.

7.  ( ) = 2²+ 6², ( ) = ⁴+ 3⁴= 1, and ∇ = ∇ ⇒ h4 12i =

4³ 12³, so we get the three equations 4 = 4³, 12 = 12³, and ⁴+ 3⁴ = 1. The first equation implies that  = 0 or ²= 1

. The second equation implies that  = 0 or ²= 1

. Note that  and  cannot both be zero as this contradicts the third equation. If  = 0, the third equation implies  = ± 1

√4

3. If  = 0, the third equation implies that  = ±1. Thus,  has possible extreme values at

 0 ± 1

√4

3



and (±1 0). Next, suppose ²= ²= 1

. Then the third equation gives

1



2

+ 3

1



2

= 1 ⇒  = ±2.

 = −2 results in a nonreal solution, so consider  = 2 ⇒  =  = ± 1

√2. Therefore,  also has possible extreme values

at



± 1

√2 ± 1

√2



(all 4 combinations). Substituting all 8 points into , we find the maximum value is





± 1

√2 ± 1

√2



= 4and the minimum value is (±1 0) = 2.

8.  ( ) = ^−²^−², ( ) = 2 −  = 0, and ∇ = ∇ ⇒



^−²^−²− 2²^−²^−² ^−²^−²− 2²^−²^−²

= h2 −i, so we get the three equations

^−²^−²− 2²^−²^−²= 2, ^−²^−²− 2²^−²^−² = −, and 2 −  = 0. Multiplying the second equation by 2 and adding it to the first gives 2^−²^−²− 4²^−²^−²+ ^−²^−²− 2²^−²^−²= −2 + 2 = 0 ⇒

2 − 4²+  − 2² = 0(as ^−²^−²6= 0). From the third equation, 2 = , and substituting into the new equation, we have 2 − 4(2)²+ 2 − 2²(2) = 0 ⇒ 4 − 20³= 0 ⇒ 4(1 − 5²) = 0 ⇒  = 0 or  = ± 1

√5, so 

has possible extreme values at (0 0),

 1

√5 2

√5

 , and



− 1

√5 − 2

√5



. Substituting these into , we see that the minimum

value is (0 0) = 0 and the maximum value is 

 1

√5 2

√5



= 



− 1

√5 − 2

√5



= 2 5.

10. Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint.

f (x, y, z) = e^xyz; 2x²+ y²+ z²= 24

Solution:

²= 1 ⇒  = ±1. If  = 1 then substitution into the second equation gives −2 = 2 ⇒  = 0, and from the constraint we must have  = ±1. Thus the possible points for the extreme values of  are (0 ±1) and (±1 0). Evaluating  at these points, we see that the maximum value of  is (±1 0) = 1 and the minimum is (0 ±1) = −1.

4.  ( ) = ^, ( ) = ³+ ³ = 16, and ∇ = ∇ ⇒ h^ ^i =

3² 3²

, so ^= 3²and

^= 3². Note that  = 0 ⇔  = 0 which contradicts ³+ ³= 16, so we may assume  6= 0,  6= 0, and then

 = ^(3²) = ^(3²) ⇒ ³= ³ ⇒  = . But ³+ ³= 16, so 2³= 16 ⇒  = 2 = .

Here there is no minimum value, since we can choose points satisfying the constraint ³+ ³= 16that make ( ) = ^

arbitrarily close to 0 (but never equal to 0). The maximum value is (2 2) = ⁴.

5.  ( ) = , ( ) = 4²+ ² = 8, and ∇ = ∇ ⇒ h i = h8 2i, so  = 8,  = 2, and

4²+ ²= 8. First note that if  = 0 then  = 0 by the ﬁrst equation, and if  = 0 then  = 0 by the second equation. But this contradicts the third equation, so  6= 0 and  6= 0. Then from the ﬁrst two equations we have



8=  = 

2 ⇒ 2² = 8² ⇒ ² = 4², and substitution into the third equation gives

4²+ 4² = 8 ⇒  = ±1. If  = ±1 then ²= 4 ⇒  = ±2, so  has possible extreme values at (1 ±2) and (−1 ±2). Evaluating  at these points, we see that the maximum value is (1 2) = (−1 −2) = 2 and the minimum is

 (1 −2) = (−1 2) = −2.

6.  ( ) = ^, ( ) = ²+ ²= 2, and ∇ = ∇ ⇒ h^ ^i = h2 2i, so ^= 2, ^= 2, and

²+ ²= 2. First note that from the ﬁrst equation  6= 0. If  = 0, the second equation implies  = 0, so  6= 0. Then from the ﬁrst two equations we have ^

2=  = ^

2 ⇒ 2^= 2²^ ⇒  = ², and substituting into the third equation gives ²+ (²)²= 2 ⇒ ⁴+ ²− 2 = 0 ⇒ (²+ 2)(²− 1) = 0 ⇒  = ±1. From  = ²we have  = 1, so  has possible extreme values at (±1 1). Evaluating  at these points, we see that the maximum value is

 (1 1) = and the minimum is (−1 1) = −.

7.  (  ) = ²+ ²+ ², (  ) =  +  +  = 12, and ∇ = ∇ ⇒ h2 2 2i = h  i. Then 2 =  = 2 = 2 ⇒  =  = , and substituting into  +  +  = 12 we have

 +  +  = 12 ⇒  = 4 =  = . Here there is no maximum value, since we can choose points satisfying the constraint  +  +  = 12 that make (  ) = ²+ ²+ ²arbitrarily large. The minimum value is (4 4 4) = 48.

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

486 ¤ CHAPTER 14 PARTIAL DERIVATIVES

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 ⇒

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.

[continued]

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28. Find the extreme values of f on the region described by the inequality.

f (x, y) = 2x²+ 3y²− 4x − 5, x²+ y²≤ 16

Solution:

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^√¹₂. 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

57. The plane x + y + 2z = 2 intersects the paraboloid z = x²+ y²in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.

Solution:

494 ¤ CHAPTER 14 PARTIAL DERIVATIVES

 = ⁵₂[again  6= 2 or else (3) implies  = 0]. Hence⁵2³=  and the dimensions which minimize cost are

 =  = ³

2

5 units,  = ^135 2

23

units.

43.If the dimensions of the box are given by , , and , then we need to ﬁnd the maximum value of (  ) = 

[    0] subject to the constraint  =

²+ ²+ ²or (  ) = ²+ ²+ ² = ². ∇ = ∇ ⇒ h  i = h2 2 2i, so  = 2 ⇒  = 

2,  = 2 ⇒  = 

2, and  = 2 ⇒  = 

2. Thus  = 

2= 

2 ⇒ ²= ² [since  6= 0] ⇒  =  and  = 

2 =

2 ⇒  =  [since  6= 0].

Substituting into the constraint equation gives ²+ ²+ ²= ² ⇒ ²= ²3 ⇒  = √

3 =  = and the maximum volume is

√ 3³

= ³ 3√

3.

44.Let the dimensions of the box be , , and , so its volume is (  ) = , its surface area is 2 + 2 + 2 = 1500 and its total edge length is 4 + 4 + 4 = 200. We ﬁnd the extreme values of (  ) subject to the

constraints (  ) =  +  +  = 750 and (  ) =  +  +  = 50. Then

∇ = h  i = ∇ + ∇ = h( + ) ( + ) ( + )i + h  i. So  = ( + ) +  (1),

 = ( + ) +  (2), and  = ( + ) +  (3). Notice that the box can’t be a cube or else  =  =  = ⁵⁰₃ but then  +  +  = ²⁵⁰⁰₃ 6= 750. Assume  is the distinct side, that is,  6= ,  6= . Then (1) minus (2) implies

 ( − ) = ( − ) or  = , and (1) minus (3) implies ( − ) = ( − ) or  = . So  =  =  and  +  +  = 50 implies  = 50 − 2; also  +  +  = 750 implies (2) + ²= 750. Hence 50 − 2 =750 − ²

2 or

3²− 100 + 750 = 0 and  =50 ± 5√ 10

3 , giving the points1 3

50 ∓ 10√ 10, ¹₃

50 ± 5√ 10,¹₃

50 ± 5√ 10. Thus the minimum of  is 1

3

50 − 10√ 3, ¹₃

50 + 5√ 10, ¹₃

50 + 5√ 10

=₂₇¹

87,500 − 2500√

10, and its maximum is 1

3

50 + 10√ 10,¹₃

50 − 5√ 10, ¹₃

50 − 5√ 10

= ₂₇¹87,500 + 2500√ 10. Note: If either  or  is the distinct side, then symmetry gives the same result.

45.We need to ﬁnd the extreme values of (  ) = ²+ ²+ ²subject to the two constraints (  ) =  +  + 2 = 2 and (  ) = ²+ ²−  = 0. ∇ = h2 2 2i, ∇ = h  2i and ∇ = h2 2 −i. Thus we need 2 =  + 2 (1), 2 =  + 2 (2), 2 = 2 −  (3),  +  + 2 = 2 (4), and ²+ ²−  = 0 (5).

From (1) and (2), 2( − ) = 2( − ), so if  6= ,  = 1. Putting this in (3) gives 2 = 2 − 1 or  =  +¹2, but putting

 = 1into (1) says  = 0. Hence  + ¹₂ = 0or  = −¹2. Then (4) and (5) become  +  − 3 = 0 and ²+ ²+¹₂ = 0. The last equation cannot be true, so this case gives no solution. So we must have  = . Then (4) and (5) become 2 + 2 = 2 and 2²−  = 0 which imply  = 1 −  and  = 2². Thus 2²= 1 −  or 2²+  − 1 = (2 − 1)( + 1) = 0 so  =¹2 or

 = −1. The two points to check are1 2¹₂¹₂

and (−1 −1 2): 1 2¹₂¹₂

= ³₄and (−1 −1 2) = 6. Thus1

2¹₂¹₂is the point on the ellipse nearest the origin and (−1 −1 2) is the one farthest from the origin.

46. (a) After plotting  =

²+ ², the top half of the cone, and the plane

 = (5 − 4 + 3)8 we see the ellipse formed by the intersection of the surfaces. The ellipse can be plotted explicitly using cylindrical coordinates (see Section 15.7): The cone is given by  = , and the plane is

4 cos  − 3 sin  + 8 = 5. Substituting  =  into the plane equation gives 4 cos  − 3 sin  + 8 = 5 ⇒  = 5

4 cos  − 3 sin  + 8. Since  =  on the ellipse, parametric equations (in cylindrical coordinates) are  = ,  =  = 5

4 cos  − 3 sin  + 8, 0 ≤  ≤ 2.

(b) We need to ﬁnd the extreme values of (  ) =  subject to the two

constraints (  ) = 4 − 3 + 8 = 5 and (  ) = ²+ ²− ²= 0.

∇ = ∇ + ∇ ⇒ h0 0 1i = h4 −3 8i + h2 2 −2i, so we need 4 + 2 = 0 ⇒  = −^2 (1),

−3 + 2 = 0 ⇒  =^32 (2), 8 − 2 = 1 ⇒  = ^82⁻¹ (3), 4 − 3 + 8 = 5 (4), and

²+ ²= ² (5). [Note that  6= 0, else  = 0 from (1), but substitution into (3) gives a contradiction.]

Substituting (1), (2), and (3) into (4) gives 4

−^2



− 3

3

2

 + 8

8−1 2



= 5 ⇒  = ^3910⁻⁸ and into (5) gives



−^2

2

+

3

2

2

=

8−1 2

2

⇒ 16²+ 9²= (8 − 1)² ⇒ 39²− 16 + 1 = 0 ⇒  = 13¹ or  = ¹₃. If  =₁₃¹ then  = −¹2and  =₁₃⁴,  = −13³,  =₁₃⁵. If  = ¹₃then  = ¹₂and  = −⁴3,  = 1,  = ⁵₃. Thus the highest point on the ellipse is

−⁴3 1⁵₃

and the lowest point is₄

13 −13³₁₃⁵ .

47. (  ) = ^^−, (  ) = 9²+ 4²+ 36²= 36, (  ) =  +  = 1. ∇ = ∇ + ∇ ⇒

^^− ^^− −^^−

= h18 8 72i + h  +  i, so ^^−= 18 + , ^^−= 8 + ( + ),

−^^−= 72 + , 9²+ 4²+ 36²= 36,  +  = 1. Using a CAS to solve these 5 equations simultaneously for ,

, , , and  (in Maple, use the allvalues command), we get 4 real-valued solutions:

 ≈ 0222444,  ≈ −2157012,  ≈ −0686049,  ≈ −0200401,  ≈ 2108584

 ≈ −1951921,  ≈ −0545867,  ≈ 0119973,  ≈ 0003141,  ≈ −0076238

 ≈ 0155142,  ≈ 0904622,  ≈ 0950293,  ≈ −0012447,  ≈ 0489938

 ≈ 1138731,  ≈ 1768057,  ≈ −0573138,  ≈ 0317141,  ≈ 1862675 Substituting these values into  gives (0222444 −2157012 −0686049) ≈ −53506,

2