Minimize the square of the distance,

2 2 2

( , , ) ,

f x y z =x +y +z subject to

2 2

( , , ) – 9 0.

g x y z =x y z + = 2 , 2 , 2x y z =λ 2 ,xy x2, – 2z 2x = 2λxy, 2y=λx², 2z = –2λz,

2 – 2 9 0

x y z + =

Critical points are (0, 0, ±3) [case x = 0];

(

^{± 2, –1,± 7}

)

^[casex≠0, λ = –1]; and

(

^{±3 2 / 9,6 −39 / 2, 0}

)

^{[case x≠0,λ≠ −1].}

Evaluating f at each of these eight points yields 9 (case x = 0), 10 (case x≠0, λ = –1), and

( )

²

3 3

3 2 9

2 (case x≠0,λ≠ −1). The latter is the smallest, so the least distance between the origin and the surface is ⁶3

3 2.8596.

4 ≈

788 Section 12.9 Instructor’s Resource Manual 11. Maximize f(x, y, z) = xyz, subject to

2 2 2 2 2 2 2 2 2 2 2 2

( , , ) –

g x y z =b c x +a c y +a b z a b c = 0

2 2 2 2 2 2

, , 2 , 2 , 2

yz xz xy =λ b c x a c y a b z

2 2 2 2 2 2

2 , 2 , 2 ,

yz= b c x xz= a c y xy= a b z

2 2 2 2 2 2 2 2 2 2 2 2

b c x +a c y +a b z =a b c Critical point is , , .

3 3 3

a b c

⎛ ⎞

⎜ ⎟

⎝ ⎠

, , 8 ,

3 3 3 3 3

a b c abc

V⎛⎜ ⎞⎟=

⎝ ⎠ which is the

maximum.

12. Maximize V(x, y, z) = xyz, subject to ( , , ) x y z–1 0.

g x y z

a b c

= + + = Let ( , , ) ( , , ), V x y z λ g x y z

∇ = ∇ so

1 1 1

, , , , .

yz xz xy

a b c λ

= Then

x y z

a b c

λ λ λ

= = (each equals xyz).

λ ≠ 0 since λ = 0 would imply x = y = z = 0 which would not satisfy the constraint.

Thus, x y z.

a= =b c These along with the

constraints yield , , .

3 3 3

a b c

x= y= z= The maximum value of .

27 V =abc

13. Maximize f(x,y,z) = x + y + z with the constraint

2 2 2

( , , ) 81 0.

g x y z =x +y +z − = Let ( , , ) ( , , )

f x y z λ g x y z

∇ = ∇ , so

1,1,1 =λ 2 , 2 , 2 ;x y z Thus, x = y = z and 3x² = 81 or x= = = ±y z 3 3.

The maximum value of f is 9 3 when , , 3 3,3 3,3 3

x y z =

14. Minimize d²= f x y z( , , )=x²+y²+z² with the constraint ( , , ) 2g x y z = x+4y+3z−12 0=

( , , ) ( , , ) f x y z λ g x y z

∇ = ∇

2 , 2 , 2x y z =λ 2,3, 4 ;

2x=2 ; 2λ y=4 ; 2λ z=3λ leads to a critical point of

(

²⁴29 29 29,^{48 36},

)

The nature of the problem indicates this will give a minimum rather than a maximum value. The minimum distance is

2 2

2 48 36

24

29 +29 +29 ≈2.2283

15. Minimize d²= f x y z( , , )

2 2 2

(x 1) (y 2) z

= − + − + with the constraint

2 2

( , , ) 0;

g x y z =x +y − =z

2x−2, 2y−4, 2z =λ 2 , 2 , 1x y −

Setting up, solving each equation for λ, and substituting into equation x²+y²− =z 0 produces λ ≈ −1.5445; The resulting critical point is approximately (0.393, 0.786, 0.772). The nature of this problem indicates this will give a minimum value rather than a maximum. The minimum distance is approximately 1.5616.

16. Minimize d²= f x y z( , , )

2 2 2

(x 1) (y 2) z

= − + − + with the constraint

2 2 2

( , , ) 0

g x y z =x +y −z =

2x−2, 2y−4, 2z =λ 2 , 2 , 2x y− z 1 5

2 2

1,x ,y 1,z

λ = − = = = ± ; The critical points are

(

^{12,1, 25}

)

^and

(

^{12,1,− 25}

)

which both lead to a minimum distance of ¹⁰.

2

17. (See problem 37, section 12.8). Let the dimensions of the box be l, w, and h . Then the cost of the box is

.25(2hl+2hw lw+ ) .4( )+ lw or ( , , ) .5 .5 .65 C l w h = hl+ hw+ lw.

We want to minimize C subject to the constraint 2

lhw= ; set V l h w( , , )=lwh−2. Now:

( , , ) (.5 .65 ) (.5 .65 ) .5( )

C l w h h w h l l w

∇ = + i + + j + + k

and ( , , )

V l w h wh lh lw

∇ = i + j + k

Thus the Lagrange equations are .5h+.65w=λwh (1) .5h+.65l=λlh (2) .5(l+w)=λlw (3)

2

lwh= (4)

Solving (4) for h and putting the result in (1) and (2), we get

1 .65w 2

lw l

+ = λ (5)

1 2

.65l

lw w

+ = λ (6)

Multiply (5) by l and (6) by w to get 1 .65lw 2

w+ = λ (7)

1 .65lw 2

l+ = λ (8)

Instructor’s Resource Manual Section 12.9 789 from which we conclude that l=w. Putting this

result into (3) we have

l=λl2 (9)

Since V ≠0,l≠0 and (9) tells us that l ¹

=λ ;

thus 1 1 2 ²

, , 2

l w l h

lw λ

λ λ

= = = = = .

Putting these results into equation (1), we conclude

2 1 1 2

.5(2λ ) .65 λ (2λ )

λ λ

⎛ ⎞ ⎛ ⎞ + ⎜ ⎟= ⎜ ⎟

⎝ ⎠ ⎝ ⎠ or

1 2

.65 λ

⎛ ⎞ =λ

⎜ ⎟⎝ ⎠ .

Hence: λ =³.65 .866≈ , so the minimum cost is obtained when:

1 1.154 and 2 2 1.5

l w h λ

= =λ ≈ = ≈

18. (See problem 38, section 12.8). Let the dimensions of the box be l, w, and h . Then the cost of the box is

1(2hl+2hw) 4( ) 3(2+ lw + l+2 ) 4w + h or

( , , ) 2 2 4 6 6 4

C l w h = hl+ hw+ lw+ +l w+ h. We want to minimize C subject to the constraint

60

lhw= ; set ( , , )V l h w =lwh−60. Now:

( ) ( ) ( )

( )

, , 2 4 6 2 4 6

2 2 4

C l w h h w h l

l w

∇ = + + + + +

+ + +

i j

k

and ( , , )

V l w h wh lh lw

∇ = i + j + k

Thus the Lagrange equations are

2h+4w+ =6 λwh (1)

2h+ + =4l 6 λlh (2)

2l+2w+ =4 λlw (3)

60

lwh= (4)

Solving (4) for 60

h=lw and putting the result in (1) and (2), we get

120 4w 6 60

lw l

+ + = λ (5)

120 60

4l 6

lw w

+ + = λ (6)

Multiply (5) by l and (6) by w to get 120 4lw 6l 60

w + + = λ (7)

120 4lw 6w 60

l + + = λ (8)

from which we conclude that

120 120

6w 6 or (l l w lw)( 20) 0

l + = w + − + = .

Since lw cannot be negative (= 20− ), we conclude that l=w; putting this result into

equation (3), we have

2w+2w+ =4 λw2 or ₂1 4 w

w λ ⎛ + ⎞

= ⎜ ⎟

⎝ ⎠. Therefore, from equation (1), we have

2 2 2

120 4 6 4 w 1 60

w w

w w w

+

⎛ ⎞ ⎛ ⎞

+ + = ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ or (multiplying through by w³ and simplifying)

4 3

2w +3w −60w−120 0=

Using one of several techniques available to solve, we conclude that w= =l 3.213 and

2

60 5.812

(3.213)

h= ≈ .

19. (See problem 40, section 12.8) Let

c = circumference of circle p = perimeter of first square q = perimeter of second square Then the sum of the areas is

2 2 2 1 2 2 2

( , , )

4 16 16 4 4 4

c p q c p q

A c p q

π π

⎡ ⎤

= + + = ⎢ + + ⎥

⎢ ⎥

⎣ ⎦

so we wish to maximize and minimize

2 2 2

( , , )

4 4

c p q

A c p q

= π + + subject to the constraint L c p q( , , )= + + − =c p q k 0. Now

( , , ) 2

2 2

( , , )

c p q

A c p q L c p q

∇ = π + +

∇ = + +

i j k

i j k

so the Lagrange equations are 2c λ

π ⁼ (1)

2

p=λ (2)

2

q=λ (3) c+ + =p q k (4) Putting (1), (2) and (3) into (4) we get

(4 ) or = 2k

2 k 8+

π λ λ

+ = π Therefore:

0

0

0

0.282 8

4 0.359

8

4 0.359

8

c k k

p k k

q k k

π π

π π

= ≈

+

= ≈

+

= ≈

+

Now A c( ,₀ p q₀, ₀) 0.0224≈ k² while ( ,0,0) .079 2

A k ≈ k , so we conclude that

790 Section 12.9 Instructor’s Resource Manual

0 0 0

( , , )

A c p q is a minimum value. There is also a maximum value (see problem 40, section 12.8) but our Lagrange approach does not capture this.

The reason is that the maximum exists because c, p, and q must all be 0≥ . Our constraint,

however, does not require this and allows negative values for any or all of the variables.

Under these conditions, there is no global maximum.

20. (See problem 42, section 12.8). Let P be the plane x y z 1

A+ +B C = . This plane will cross the first octant, forming a triangle, T, in P; the vertices of this triangle occur where P intersects the coordinate axes. They are:

( ,0,0), (0, ,0), (0,0, )

x y z

V = A V = B V = C .

a. Define the vectors g= −A B, ,0 and A,0,C

= −

h . From example 3 in 11.4, we know the area of T is

2 2 2

1 1 ( ) ( ) ( )

2g h× =2 BC + AC + AB . b. The height of the tetrahedron in question is

the distance is the distance between (0,0,0) and P . By example 10 in 11.3, this distance is

1 1 1

2 2 2

2

2 2 2

1 ( )

( ) ( ) ( )

A B C

h ABC

BC AC AB

+ +

= =

+ +

c. Finally, the volume of the tetrahedron is 1 (area of )

3h T , or

2 2 2

2

2 2 2

( , , ) 1 ( ) ( ) ( )

6

( )

( ) ( ) ( )

V A B C BC AC AB

ABC

BC AC AB

= + +

⎡ ⎤

⎢ ⎥

⋅⎢⎣ + + ⎥⎦

That is, 1 ² 1

( , , ) ( )

6 6

V A B C = ABC = ABC . Hence we want to minimize

( , , )

V A B C = ABC subject to the constraint a b c 1

A+ +B C = ; define

( , , ) a b c 1

g A B C

A B C

= + + − . Now

( , , ) A B C

V A B C

ABC ABC ABC

∇ = i+ j+ k

and ( , , ) a₂ b₂ c₂ g A B C

A B C

− − −

∇ = i+ j+ k. Thus

the Lagrange equations are

2

A a

ABC A

λ

= − (1)

2

B b

ABC B

λ

= − (2)

2

C c

ABC C

λ

= − (3)

a b c 1

A+ +B C = (4) From (1) – (3) we have

3 3 3

A B C

ABC a b c

λ =⁻ =⁻ =⁻ (5)

Solving in pairs we get

3b , 3c

B A C A

a a

⎛ ⎞ ⎛ ⎞

=⎜⎜ ⎟⎟ =⎜⎜ ⎟⎟

⎝ ⎠ ⎝ ⎠ (6)

and putting these results into (4) we obtain

3 2 3 2

3 2 3 2 3 2

3

A a ab ac

a a b c

= + +

⎛ ⎞

= ⎜ + + ⎟

⎝ ⎠

Similarly, we have

3 2 3 2

3 2 3 2 3 2

3

3 2 3 2

3 2 3 2 3 2

3

B a b b bc

b a b c

C a c b c c

c a b c

= + +

⎛ ⎞

= ⎜ + + ⎟

⎝ ⎠

= + +

⎛ ⎞

= ⎜ + + ⎟

⎝ ⎠

Finally, the volume of the tetrahedron is

3 2 3 2 3 2 3 3

6 6

abc a b c

ABC

⎛ + + ⎞

⎜ ⎟

⎝ ⎠

= .

21. Finding critical points on the interior first:

f 1 0 x

∂ = ≠

∂ f 1 0;

y

∂ = ≠

∂ There are no critical points on the interior. Finding critical points on the boundary: ∇f x y( , )= ∇λ g x y( , );

1,1 =λ 2 , 2 ;x y The solution to the system

2 2

1= ⋅λ 2 , 1x = ⋅λ 2 ,y x +y =1is ¹

2, λ= ±

1 1

2, 2

x= ± y= ± The four critical points are

(

¹2,± ¹2

)

and

(

− ¹2,± ¹2

)

.

(

¹2, ¹2

)

10 2

f = + is the maximum value.

(

¹2, ¹2

)

10 2

f − − = − is the minimum value.

Instructor’s Resource Manual Section 12.9 791 22. Finding critical points on the interior:

1 0 1;

f y y

x

∂ = − = ⇒ =

∂

1 0 1;

f x x

y

∂ = − = ⇒ =

∂ The only critical point on

the interior is c₁= (1,1). Finding critical points on the boundary: Solve the system of equations 1− = ⋅y λ 2 ;x 1− = ⋅x λ 2 ;y x²+y²=9 Using substitution, it can be found that the critical points on the boundary are

2 3 3

2, 2

c ⎛ ⎞

= ⎜ ⎟

⎝ ⎠ , ₃ ³ , ³

2 2

c = −⎛⎜ − ⎞⎟

⎝ ⎠,

c4 = (2.56155, -1,56155), c5= (-1.56155, 2.56155)

The maximum value of 5 is obtained substituting either c₄ or c₅ into f. The minimum value of about -8.7426 is obtained by substituting c₃ into f.

23. Finding critical points on the interior:

2 3 0;

f x y

x

∂ = + − =

∂ ^{2 0}

f y x

y

∂ = − =

∂

The solution to this system is the only critical point on the interior, c₁=(-2,-1).

Critical points on the boundary will come from the solutions to the following system of equations:

2x+ − = ⋅3 y λ 2 ,x 2y− = ⋅x λ 2 ,y

2 2 9

x +y = . From the solutions to this system, the critical points are c₂=(0,3),

3 3 3 3

, ,

2 2

c ⎛ ⎞

=⎜⎜ − ⎟⎟

⎝ ⎠ ⁴

3 3 3 2 , 2

c ⎛ ⎞

= −⎜⎜ − ⎟⎟

⎝ ⎠

( )1 3,

f c = − f c( ) 9,₂ = f c( ) 20.6913,₃ ≈ ( )4 2.6913

f c ≈ − The max value of f is

≈20.6913 and the min value is -3.

24. ^{( , ) 2}on the set

{

( , ): ^{42 92} 1

}

1

x y

f x y x S x y

y

= = + ≤

+

We first find the max and min for f on the set

{

^{( , ):x42 y92 1}

}

S = x y + < using the methods of section 12.8:

2 2 2

1 2

( , )

1 (1 )

f x y xy

y y

∇ = + −

+ i + j so setting ( , )

f x y

∇ =0 we have ¹ ₂ 0 1 y

+ = (impossible).

Thus f has no max or min on S .

We now look for the max and min of f on the

boundary ^{S =}

{

^{( , ):x y x42+y92 =1}

}

; this is done using Lagrange multipliers. Let

2 2

( , ) 1

4 9

x y

g x y = + − ; then

2 2 2

1 2

( , )

1 (1 )

f x y xy

y y

∇ = + −

+ i + j and

( , ) 2

2 9

x y

g x y

∇ = i+ j

The Lagrange equations are

2

1 1 2

x y

=λ

+ (1)

2 2

2 2

(1 ) 9

xy y

y

λ

− =

+ (2)

2 2

9x +4y =36 (3)

Putting (1) into (2) yields

2 3 2

2 9

x y y

λ λ

− = (4)

One solution to (4) is y=0 which yields, from (3), x= ±2. Thus (2,0) and ( 2,0)− are candidates for optimization points.

If y≠0, (4) can be reduced to

2 3 2

2 9

λ x λ

− = (5)

so that ⁴₃

λ=9x⁻ . Putting this result into (1) yields ¹ ₂ ²₂

1 y 9x

+ = − , which has no solutions (left side always +, right side always -).

Therefore the only two candidates for max/min are (2,0) and ( 2,0)− . Since f(2,0) 2= and

( 2,0) 2

f − = − we conclude that the max value of on S

f is 2 and the min value is −2.

25. f f 2(1 ) 0 1

x y x y

x y

∂ ∂

= = + + = ⇒ + = −

∂ ∂

There is no minimum or maximum value on the interior since there are an infinite number of critical points. The critical points on the boundary will come from the solutions to the following system of equations:

2(1 ) 1 2 2(1 ) 1

8

x y x

x y y

λ λ + + = ⋅ + + = ⋅

Solving these two equations for λ leads to 1

y= − −x or y=4x. Together with the constraint

2 2

4 16 1 0

x +y − = leads to the critical

792 Section 12.9 Instructor’s Resource Manual points on the boundary: ^{1 2 19}, ^{4 2 19}

5 5

− − − +

⎛ ⎞

⎜ ⎟

⎜ ⎟

⎝ ⎠,

1 2 19, 4 2 19

5 5

⎛− + − − ⎞

⎜ ⎟

⎜ ⎟

⎝ ⎠, 2 , 8

5 5

⎛ ⎞

− −

⎜ ⎟

⎝ ⎠ and

2 8

5, 5

⎛ ⎞

⎜ ⎟

⎝ ⎠. Respectively, the maximum value is

≈29.9443 and the minimum value is 0.

26. It is clear that the maximum will occur for a