Solutions for Chapter 15 Review

Transcription

SOLUTIONS to Review Problems for Chapter Fifteen
so
λ=
Thus
x1 =
ab
p1
1225
1
.
b
x2 =
(1 − a)b
p2
so the maximum satisfaction is given by
S = u(x1 , x2 ) = u
ab (1 − a)b
,
p1
p2
= a ln
ab
p1
+ (1 − a) ln
(1 − a)b
p2
= a ln a + a ln b − a ln p1 + (1 − a) ln(1 − a) + (1 − a) ln b − (1 − a) ln p2
= a ln a + (1 − a) ln(1 − a) + ln b − a ln p1 − (1 − a) ln p2 .
(b) We want to calculate the value of b needed to achieve u(x1 , x2 ) = c. Thus, we solve for b in the equation
c = a ln a + (1 − a) ln(1 − a) + ln b − a ln p1 − (1 − a) ln p2 .
Since
we have
ln b = c − a ln a − (1 − a) ln(1 − a) + a ln p1 + (1 − a) ln p2 ,
b=
ec pa1 p1−a
ec · ea ln p1 · e(1−a) ln p2
2
= a
.
a
ln
a
(1−a)
ln(1−a)
e
·e
a (1 − a)(1−a)
Solutions for Chapter 15 Review
Exercises
1. The critical points of f are obtained by solving fx = fy = 0, that is
so
fx (x, y) = 2y 2 − 2x = 0
and
fy (x, y) = 4xy − 4y = 0,
2(y 2 − x) = 0
and
4y(x − 1) = 0
The second equation gives either y = 0 or x = 1. If y = 0 then x = 0 by the first equation, so (0, 0) is a critical point. If
x = 1 then y 2 = 1 from which y = 1 or y = −1, so two further critical points are (1, −1), and (1, 1).
Since
D = fxx fyy − (fxy )2 = (−2)(4x − 4) − (4y)2 = 8 − 8x − 16y 2 ,
we have
D(0, 0) = 8 > 0,
D(1, 1) = D(1, −1) = −16 < 0,
and fxx = −2 < 0. Thus, (0, 0) is a local maximum; (1, 1) and (1, −1) are saddle points.
2. At a critical point
fx (x, y) = 6x2 − 6xy + 12x = 0
fy (x, y) = −3x2 − 12y = 0
From the second equation, we conclude that −3(x2 + 4y) = 0, so y = − 14 x2 . Substituting for y in the first equation
gives
1
6x2 − 6x − x2 + 12x = 0
4
or
1
x
x2 + x3 + 2x = (4x + x2 + 8) = 0.
4
4
Thus x = 0 or x2 + 4x + 8 = 0. The quadratic has no real solutions, so the only one critical point is (0, 0).
At (0, 0), we have
D(0, 0) = fxx fyy − (fxy )2 = (12)(−12) − 02 = −144 < 0,
so (0, 0) is a saddle point.
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Chapter Fifteen /SOLUTIONS
3. At a critical point
fx (x, y) = 2xy − 2y = 0
fy (x, y) = x2 + 4y − 2x = 0.
From the first equation, 2y(x − 1) = 0, so either y = 0 or x = 1. If y = 0, then x 2 − 2x = 0, so x = 0 or x = 2. Thus
(0, 0) and (2, 0) are critical points. If x = 1, then 12 + 4y − 2 = 0, so y = 1/4. Thus (1, 1/4) is a critical point. Now
D = fxx fyy − (fxy )2 = 2y · 4 − (2x − 2)2 = 8y − 4(x − 1)2 ,
so
D(0, 0) = −4, D(2, 0) = −4, D(1, 41 ) = 2
so (0, 0) and (2, 0) are saddle points. Since fyy = 4 > 0, we see that (1, 1/4) is a local minimum.
4. Critical points occur where fx = fy = 0:
fx (x, y) =
−80
+ 10 + 10y.
x2 y
fy (x, y) =
−80
+ 10x + 20.
xy 2
Substituting x = 2, y = 1 gives
−80
+ 10 + 10.1 = 0
22 · 1
−80
fy (2, 1) =
+ 10.2 + 20 = 0.
2 · 12
fx (2, 1) =
So (2, 1) is a critical point.
To determine if this critical point is a minimum we use the second derivative test.
160
, fxx (2, 1) = 20,
x3 y
160
=
, fyy (2, 1) = 80,
xy 3
80
= 2 2 + 10, fxy (2, 1) = 30.
x y
fxx =
fyy
fxy
So D = 20 · 80 − 302 = 700 > 0 and fxx (2, 1) > 0, therefore the point (2, 1) is a local minimum.
5. The partial derivatives are
fx = cos x + cos (x + y).
fy = cos y + cos (x + y).
Setting fx = 0 and fy = 0 gives
cos x = cos y
For 0 < x < π and 0 < y < π, cos x = cos y only if x = y. Then, setting f x = fy = 0:
cos x + cos 2x = 0,
cos x + 2 cos2 x − 1 = 0,
(2 cos x − 1)(cos x + 1) = 0.
So cos x = 1/2 or cos x = −1, that is x = π/3 or x = π. For the given domain 0 < x < π, 0 < y < π, we only
consider the solution when x = π/3 then y = x = π/3. Therefore, the critical point is ( π3 , π3 ).
Since
√
fxx (x, y) = − sin x − sin (x + y) fxx ( π3 , π3 ) = − sin π3 − sin 2π
=− 3
3
fxy (x, y) = − sin (x + y)
the discriminant is
Since
fxy ( π3 , π3 ) = − sin
fyy (x, y) = − sin y − sin (x + y) fyy ( π3 , π3 ) = − sin
fxx ( π3 , π3 )
√
= − 3 < 0,
2
D(x, y) = fxx fyy − fxy
√
√
√
= (− 3)(− 3) − (− 23 )2 =
( π3 , π3 )
is a local maximum.
9
4
2π
3
π
−
3
sin
> 0.
2π
3
=−
√
3
√2
=− 3
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6. We find critical points:
fx (x, y) = 12 − 6x = 0
fy (x, y) = 6 − 2y = 0
so (2, 3) is the only critical point. At this point
D = fxx fyy − (fxy )2 = (−6)(−2) = 12 > 0,
and fxx < 0, so (2, 3) is a local maximum. Since this is a quadratic, the local maximum is a global maximum.
Alternatively, we complete the square, giving
f (x, y) = 10 − 3(x2 − 4x) − (y 2 − 6y) = 31 − 3(x − 2)2 − (y − 3)2 .
This expression for f shows that its maximum value (which is 31) occurs where x = 2, y = 3.
7. The partial derivatives are fx = 2x − 3y, fy = 3y 2 − 3x. For critical points, solve fx = 0 and fy = 0 simultaneously.
From 2x − 3y = 0 we get x = 32 y. Substituting it into 3y 2 − 3x = 0, we have that
3
9
9
3y 2 − 3( y) = 3y 2 − y = y(3y − ) = 0.
2
2
2
So y = 0 or 3y − 29 = 0, that is, y = 0 or y = 3/2. Therefore the critical points are (0, 0) and ( 94 , 32 ).
The contour diagram for f in Figure 15.26 (drawn by a computer), shows that (0, 0) is a saddle point and that ( 94 , 23 ) is a
local minimum.
y
110
90
4
70
3
50
30
2
0
( 94 , 23 )
10
1
x
−4
−3
0
−2
−2
−30
−70
1
2
−1
−10
−90
−1
−50
3
4
10
0
30
−3
−4
Figure 15.26: Contour map of f (x, y) = x2 + y 3 − 3xy
We can also see that (0, 0) is a saddle point and ( 94 , 32 ) is a local minimum analytically. Since fxx = 2, fyy =
6y, fxy = −3, the discriminant is
2
= 12y − (−3)2 = 12y − 9.
D(0, 0) = −9 < 0, so (0, 0) is a saddle point.
D( 94 , 32 ) = 9 > 0 and fxx = 2 > 0, we know that ( 94 , 23 ) is a local minimum. The point ( 49 , 23 ) is not a global minimum
since f ( 49 , 32 ) = −1.6875, whereas f (0, −2) = −8.
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8. Note that the x-axis and the y-axis are not in the domain of f . Since x 6= 0 and y 6= 0, by setting f x = 0 and fy = 0 we
get
1
= 0 when x = ±1
x2
4
fy = 1 − 2 = 0 when y = ±2
y
fx = 1 −
So the critical points are (1, 2), (−1, 2), (1, −2), (−1, −2). Since fxx = 2/x3 and fyy = 8/y 3 and fxy = 0, the
discriminant is
16
8
2
2
− 02 =
=
.
x3
y3
(xy)3
Since D < 0 at the points (−1, 2) and (1, −2), these points are saddle points. Since D > 0 at (1, 2) and (−1, −2) and
fxx (1, 2) > 0 and fxx (−1, −2) < 0, the point (1, 2) is a local minimum and the point (−1, −2) is a local maximum.
No global maximum or minimum, since f (x, y) increases without bound if x and y increase in the first quadrant; f (x, y)
decreases without bound if x and y decrease in the third quadrant.
9. The partial derivatives are
1
, fy = x + 2y.
x
For critical points, solve fx = 0 and fy = 0 simultaneously. From fy = x + 2y = 0 we get that x = −2y. Substituting
into fx = 0, we have
1
1
1
1
y+ =y−
= (y 2 − ) = 0
x
2y
y
2
fx = y +
Since
1
y
6= 0, y 2 −
1
2
= 0, therefore
√
1
2
y = ±√ = ±
,
2
2
√
√
√ √
√
√
2, − 22 . But x must be greater than 0, so − 2,
and x = ∓ 2. So the critical points are − 2, 22 and
√
2
2
not in the domain.
√
√ 2, − 22 is a saddle point of f (x, y).
The contour diagram for f in Figure 15.27 (drawn by computer), shows that
y
4
10
0
−10
2
1
√
√
√
+ ln 2
f ( 2, − 22 ) = − 21
2
−1
30
20
3
?
-
−4
5
6
7
6
−2
−3
x
4
−10
−20
0
Figure 15.27: Contour map of f (x, y) = xy + ln x + y 2 − 10
We can also see that
Since fxx =
D
√
2, −
√
− x12 , fyy
2
2
√
2, −
√
2
2
is a saddle point analytically.
= 2, fxy = 1, the discriminant is:
= −2 < 0, so
√
2, −
√
2
2
2
2
= − 2 − 1.
x
is a saddle point.
8
is
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10. The objective function is f (x, y) = 3x − 4y and the constraint equation is g(x, y) = x 2 + y 2 = 5, so grad f = 3~i − 4~j
and grad g = (2x)~i + (2y)~j . Setting grad f = λ grad g gives
3 = λ(2x),
−4 = λ(2y).
From the first equation we have λ = 3/(2x), and from the second equation we have λ = −2/y. Setting these equal gives
3
x = − y.
4
√
√
Substituting this into the constraint equation x2 + y 2 = 5 gives y 2 = 16/5, so y = 4/ 5 and y = −4/ 5. Since
3
x = − 4 y, there are two points where a maximum or a minimum might occur:
√
√
√
√
(−3/ 5, 4/ 5) and (3/ 5, −4/ 5).
Since the
is closed
maximum
and
√ constraint
√
√ and bounded,
√
√
√ minimum values of f subject to the constraint
√ exist.
√ Since
f (−3/ 5, 4/ 5) = −5√ 5 and f√(3/ 5, −4/ 5) = 5 5, we see that f has a minimum value at (−3/ 5, 4/ 5) and
a maximum value at (3/ 5, −4/ 5).
11. The objective function is f (x, y) = x2 + y 2 and the equation of constraint is g(x, y) = x4 + y 4 = 2. Their gradients are
∇f (x, y) = 2x~i + 2y~j ,
∇g(x, y) = 4x3~i + 4y 3~j .
So the equation ∇f = λ∇g becomes 2x~i + 2y~j = λ(4x3~i + 4y 3~j ). This tells us that
2x = 4λx3 ,
2y = 4λy 3 .
Now if x = 0, the first equation is true for any value of λ. In particular, we can choose λ which satisfies the second
equation. Similarly, y = 0 is solution.
Assuming both x 6= 0 and y 6= 0, we can divide to solve for λ and find
λ=
2y
2x
=
4x3
4y 3
1
1
=
2x2
2y 2
y 2 = x2
y = ±x.
Going back to our equation of constraint, we find
g(0, y) = 04 + y 4 = 2,
g(x, 0) = x4 + 04 = 2,
√
4
so y = ± 2
√
4
so x = ± 2
g(x, ±x) = x4 + (±x)4 = 2, so x = ±1.
√
√
Thus, the critical points are (0, ± 4 2), (± 4 2, 0), (1, ±1) and (−1, ±1). Since the constraint is closed and bounded,
maximum and minimum values of f subject to the constraint exist. Evaluating f at the critical points, we find
f (1, 1) = f (1, −1) = f (−1, 1) = f (−1, −1) = 2,
√
√
√
√
√
4
4
4
4
2) = f (0, − 2) = f ( 2, 0) = f (− 2, 0) = 2.
√
Thus, the minimum value of f (x, y) on g(x, y) = 2 is 2 and the maximum value is 2.
f (0,
12. The objective function is f (x, y) = x2 + y 2 and the constraint equation is g(x, y) = 4x − 2y = 15, so grad f =
(2x)~i + (2y)~j and grad g = 4~i − 2~j . Setting grad f = λ grad g gives
2x = 4λ,
2y = −2λ.
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From the first equation we have λ = x/2, and from the second equation we have λ = −y. Setting these equal gives
y = −0.5x.
Substituting this into the constraint equation 4x − 2y = 15 gives x = 3. The only critical point is (3, −1.5).
We have f (3, −1.5) = (3)2 + (1.5)2 = 11.25. One way to determine if this point gives a maximum or minimum
value or neither for the given constraint is to examine the contour diagram of f with the constraint sketched in, Figure 15.28. It appears that moving away from the point P = (3, −1.5) in either direction along the constraint increases the
value of f , so (3, −1.5) is a point of minimum value.
y
y
x2 − y 2 = 1
Constraint: 4x − 2y = 15
1
5
10 15
20
x
1
2
3
−1
−2
4
−2
0.2
0.866
(−1.038, −0.278)
P = (3, −1.5)
(1.038, 0.278)
x
2
2
−3
Figure 15.28
Figure 15.29
13. The objective function is f (x, y) = x2 − xy + y 2 and the equation of constraint is g(x, y) = x2 − y 2 = 1. The gradients
of f and g are
∇f (x, y) = (2x − y)~i + (−x + 2y)~j ,
∇g(x, y) = 2x~i − 2y~j .
Therefore the equation ∇f (x, y) = λ∇g(x, y) gives
2x − y = 2λx
−x + 2y = −2λy
x2 − y 2 = 1.
Let us suppose that λ = 0. Then 2x = y and 2y = x give x = y = 0. But (0, 0) is not a solution of the third equation,
so we conclude that λ 6= 0. Now let’s multiply the first two equations
−2λy(2x − y) = 2λx(−x + 2y).
As λ 6= 0, we can cancel it in the equation above and after doing the algebra we get
√
√
which gives x = (2
√ + 3)y or x = (2 − 3)y.
If x = (2 + 3)y, the third equation gives
x2 − 4xy + y 2 = 0
(2 +
√
3)2 y 2 − y 2 = 1
so y ≈ ±0.278 and
√ x ≈ ±1.038. These give the critical points (1.038, 0.278), (−1.038, −0.278).
If x = (2 − 3)y, from the third equation we get
√
(2 − 3)2 y 2 − y 2 = 1.
√
But (2 − 3)2 − 1 ≈ −0.928 < 0 so the equation has no solution. Evaluating f gives
f (1.038, 0.278) = f (−1.038, −0.278) ≈ 0.866
Since y → ∞ on the constraint, rewriting f as
y 2 3 2
+ y
2
4
shows that f has no maximum on the constraint. The minimum value of f is 0.866. See Figure 15.29.
f (x, y) = x −
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14. The objective function is f (x, y) = x2 + 2y 2 and the constraint equation is g(x, y) = 3x + 5y = 200, so grad f =
(2x)~i + (4y)~j and grad g = 3~i + 5~j . Setting grad f = λ grad g gives
2x = 3λ,
4y = 5λ.
From the first equation, we have λ = 2x/3, and from the second equation we have λ = 4y/5. Setting these equal gives
x = 1.2y.
Substituting this into the constraint equation 3x + 5y = 200 gives y = 23.256. Since x = 1.2y, we have x = 27.907. A
maximum or minimum value of f can occur only at (27.907, 23.256).
We have f (27.907, 23.256) = 1860.484. From Figure 15.30, we see that the point (27.907, 23.256) is a minimum
value of f subject to the given constraint.
y
50
4000
40
3000
30
1860
(27.9, 23.6)
1000
20
3x + 5y = 200
10
x
10
20
30
40
50
Figure 15.30
15. The objective function is f (x, y) = 2xy and the constraint equation is g(x, y) = 5x + 4y = 100, so grad f =
(2y)~i + (2x)~j and grad g = 5~i + 4~j . Setting grad f = λ grad g gives
2y = 5λ,
2x = 4λ.
From the first equation we have λ = 2y/5, and from the second equation we have λ = x/2. Setting these equal gives
y = 1.25x.
Substituting this into the constraint equation 5x + 4y = 100 gives x = 10 and y = 12.5. A maximum or minimum value
for f subject to the constraint can occur only at (10, 12.5).
We have f (10, 12.5) = 250. From Figure 15.31, we see that the point (10, 12.5) gives a maximum.
y
30
400
250
20
100
(10, 12.5)
10
5x + 4y = 100
x
10
Figure 15.31
20
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16. We will use the Lagrange multipliers with:
Objective function: f (x, y) = −3x2 − 2y 2 + 20xy
Constraint: g(x, y) = x + y − 100
We first find
∇f = (−6x + 20y)~i + (−4y + 20x)~j
∇g = ~i + ~j .
To optimize f , we must solve the equations
∇f = λ∇g
(−6x + 20y)~i + (−4y + 20x)~j = λ(~i + ~j ) = λ~i + λ~j
We have a vector equation, so we equate the coordinates:
−6x + 20y = λ
So
20x − 4y = λ.
− 6x + 20y = 20x − 4y
24y = 26x
13
y=
x
12
Substituting into the constraint equation x + y = 100, we obtain:
x+
13
x = 100
12
25
x = 100
12
x = 48.
Consequently, y = 52, and f (48, 52) = 37,600. The point (48, 52) leads to the extreme value of f (x, y), given that
x + y = 100. Note that f has no minimum on the line x + y = 100 since f (x, 100 − x) = −3x 2 − 2(100 − x)2 +
20x(100 − x) = −25x2 + 2400x − 20000 which goes to −∞ as x goes to ±∞. Therefore, the point (48, 52) gives the
maximum value for f on the line x + y = 100.
17. Our objective function is f (x, y, z) = x2 − 2y + 2z 2 and our equation of constraint is g(x, y, z) = x2 + y 2 + z 2 − 1 = 0.
To optimize f (x, y, z) with Lagrange multipliers, we solve ∇f (x, y, z) = λ∇g(x, y, z) subject to g(x, y, z) = 0. The
gradients of f and g are
We get,
∇f (x, y, z) = 2x~i − 2~j + 4z~k ,
∇g(x, y) = 2x~i + 2y~j + 2z~k .
x = λx
−1 = λy
2z = λz
2
2
x + y + z 2 = 1.
From the first equation we get x = 0 or λ = 1.
If x = 0 we have
−1 = λy
2z = λz
2
y + z 2 = 1.
From the second equation z = 0 or λ = 2. So if z = 0, we have y = ±1 and √
we get the solutions
(0, 1, 0),(0, −1, 0). If
√
z 6= 0 then λ = 2 and y = − 12 . So z 2 = 34 which gives the solutions (0, − 21 , 23 ), (0, − 12 , − 23 ).
If x 6= 0, then λ = 1, so y = −1, which implies, from the equation x 2 + y 2 + z 2 = 1, that x = 0, which contradicts
the assumption.
Since the constraint is closed and bounded, maximum and minimum values of f subject to the
constraint exist. There√
√
fore, evaluating f at the critical points, we get f (0, 1, 0) = −2, f (0, −1, 0) = 2 and f (0, − 12 , 23 ) = f (0, − 12 , − 23 ) =
4. So the maximum value of f is 4 and the minimum is −2.
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18. Our objective function is f (x, y, z) = x2 − y 2 − 2z and our equation of constraint is g(x, y, z) = x2 + y 2 − z = 0.
To optimize f (x, y, z) with Lagrange multipliers, we solve ∇f (x, y, z) = λ∇g(x, y, z) subject to g(x, y, z) = 0. The
gradients of f and g are
∇f (x, y, z) = 2x~i − 2y~j − 2~k ,
∇g(x, y, z) = 2x~i + 2y~j − ~k .
We get
2x = 2λx
−2y = 2λy
−2 = −λ
x2 + y 2 = z.
The third equation gives λ = 2 and from the first x = 0, from the second y = 0 and from the fourth z = 0. So the only
solution is (0, 0, 0), and f (0, 0, 0) = 0.
To see what kind of extreme point is (0, 0, 0), let (a, b, c) be a point which satisfies the constraint, i.e. a 2 + b2 = c.
Then f (a, b, c) = a2 − b2 − 2c = −a2 − 3b2 ≤ 0. The conclusion is that 0 is the maximum value of f and that there is
no minimum.
19. We first find the critical points in the disk
∇z = (8x − y)~i + (8y − x)~j
Setting ∇z = 0 gives 8x − y = 0 and 8y − x = 0. The only solution is x = y = 0. So (0, 0) is the only critical point in
the disk.
Next we find the extremal values on the boundary using Lagrange multipliers. We have objective function z =
4x2 − xy + 4y 2 and constraint G = x2 + y 2 − 2 = 0.
∇z = (8x − y)~i + (8y − x)~j
∇G = 2x~i + 2y~j
∇z = λ∇G gives
8x − y = 2λx
8y − x = 2λy
If λ = 0 we get
8x − y = 0
8y − x = 0
with only solutions x = y = 0, which does not satisfy the constraint: x 2 + y 2 − 2 = 0. Therefore λ 6= 0 and we get:
2λy(8x − y) = 2λx(8y − x)
and
2
y(8x − y) = x(8y − x).
2
So x = y , x = ±y.
Substitute into G = 0, we get 2x2 − 2 = 0 so x = ±1. The extremal points on the boundary are therefore
(1, 1), (1, −1), (−1, 1), (−1, −1). The region x2 + y 2 ≤ 2 is closed and bounded, so minimum values of f in the region
exist. We check the values of z at these points :
z(1, 1) = 7,
z(−1, −1) = 7,
z(1, −1) = 9,
z(−1, 1) = 9,
z(0, 0) = 0
Thus (−1, 1) and (1, −1) give the maxima over the closed disk and (0, 0) gives the minimum.
20. The region x2 ≥ y is the shaded region in Figure 15.32 which includes the parabola y = x 2 .
1234
y
−70
70
−50
−30
30
−10
50
10
x
Figure 15.32
We first want to find the local maxima and minima of f in the interior of our region. So we need to find the extrema
of
f (x, y) = x2 − y 2 ,
x2 > y.
in the region
For this we compute the critical points:
fx = 2x = 0
fy = −2y = 0.
As (0, 0) does not belong to the region x2 > y, we have no critical points. Now let’s find the local extrema of f on
the boundary of our region, hence this time we have to solve a constraint problem. We want to find the extrema of
f (x, y) = x2 − y 2 subject to g(x, y) = x2 − y = 0. We use Lagrange multipliers:
and
grad f = λ grad g
x2 = y.
This gives
2x = 2λx
2y = λ
x2 = y.
From the first equation we get x = 0 or λ = 1.
If x = 0, from the third equation we get y = 0, so one solution is (0, 0). If x 6= 0, then λ = 1 and from the second
equation we get y = 21 . This gives x2 = 12 so the solutions ( √12 , 12 ) and (− √12 , 12 ).
So f (0, 0) = 0 and f ( √12 , 21 ) = f (− √12 , 21 ) = 14 . From Figure 15.32 showing the level curves of f and the region
2
x ≥ y, we see that (0, 0) is a local minimum of f on x2 = y, but not a global minimum and that ( √12 , 21 ) and (− √12 , 21 )
are global maxima of f on x2 = y but not global maxima of f on the whole region x2 ≥ y.
So there are no global extrema of f in the region x2 ≥ y.
21. The region x + y ≥ 1 is the shaded half plane (including the line x + y = 1) shown in Figure 15.33.
y
3
2
1
3
5
1
x
−1
−3
−5
1
Figure 15.33
2
3
1235
Let’s look for the critical points of f in the interior of the region. As
fx = 3x2
fy = 1
there are no critical points inside the shaded region. Now let’s find the extrema of f on the boundary of our region. We
want the extrema of f (x, y) = x3 + y subject to the constraint g(x, y) = x + y − 1 = 0. We use Lagrange multipliers
and x + y = 1,
grad f = λ grad g
which give
3x2 = λ
1=λ
x + y = 1.
From the first two equations we get 3x2 = 1, so the solutions are
1
1
(√ ,1 − √ )
3
3
1
1
(− √ , 1 + √ ).
3
3
and
Evaluating f at these points we get
1
f(√ , 1 −
3
1
f (− √ , 1 +
3
1
√ ) = 1−
3
1
√ ) = 1+
3
2
√
3 3
2
√ .
3 3
From the contour diagram in Figure 15.33, we see that ( √13 , 1 − √13 ) is a local minimum and (− √13 , 1 + √13 ) is a local
maximum of f on x + y = 1. Are they global extrema as well?
If we take x very big and y = 1 − x then f (x, y) = x3 + y = x3 − x + 1 which can be made as big as we want (if
we choose x big enough). So there will be no global maximum.
Similarly, taking x negative with big absolute value and y = 1 − x, f (x, y) = x 3 + y = x3 − x + 1 can be made as
small as we want (if we choose x small enough). So there is no global minimum. This can also be seen from Figure 15.33.
22. If x = 10 then f (x, y) = 100 − y 2 is a parabola opening downward, so it has a maximum but no minimum.
23. If y = 10 then f (x, y) = x2 − 100 is a parabola opening upward, so it has a minimum but no maximum.
√
√
24. If x2 + y 2 = 10, then f (x, y) = x2 − y 2 = 2x2 √
− 10 and x has values in the interval − 10
√≤ x ≤ 10. Hence f (x, y)
has a maximum on the constraint at (x, y) = (± 10, 0) and a minimum at (x, y) = (0, ± 10).
25. If xy = 10, then f (x, y) = x2 − y 2 = x2 − 100/x2 and x can take any nonzero value. Since
lim
x→∞
we see f has no maximum on the constraint. Since
x2 −
lim x2 −
x→0
we see f has no minimum on the constraint.
100
x2
100
x2
= ∞,
= −∞,
Problems
26. The maximum and minimum values change by approximately λ∆c. The Lagrange conditions give:
2x = λ4x3 ,
2y = λ4y 3 .
If x = 0, then y = ±21/4 from the constraint. If y = 0, then x = ±21/4 . If x 6= 0 and y 6= 0, we can solve for λ and set
the expressions equal to get x2 = y 2 , so y = ±x.
Thus, there are eight points satisfying the Lagrange conditions: four of the form (0, ±2 1/4 ) or (±21/4 , 0), and four
of the form (±1, ±1). Since f (x,√
y) = x2 + y 2 , we get a maximum value of 2 at the four points of the form (±1, ±1)
and a minimum value of 22/4 = 2 at the other four points. For the maximum value, we use λ = 1/(2x 2 ) = 1/2, so
2
the change is approximately
∆c/2. For the minimum value at (0, ±2 1/4 ), we
√
√ use λ = 1/(2y ), so there the change is
1/4
approximately ∆c/(2 2). Similarly, the change at ±2 , 0) is also ∆c/(2 2).
1236
27. Let the line be in the form y = b + mx. Then, when x equals 0, 1, and 2, y equals b, b + m, and b + 2m respectively.
The sum of the squares of the vertical distances, which is what we want to minimize, is
f (m, b) = (4 − b)2 + (3 − (b + m))2 + (1 − (b + 2m))2
To find critical points, set each partial derivative equal to zero.
fm = 0 + 2(3 − (b + m))(−1) + 2(1 − (b + 2m))(−2)
= 6b + 10m − 10
fb = 2(4 − b)(−1) + 2(3 − (b + m))(−1) + 2(1 − (b + 2m))(−1)
= 6b + 6m − 16
Setting both partial derivatives equal to zero and dividing by 2, we get a system of equations:
3b + 5m = 5
3b + 3m = 8
with solutions m = − 23 and b =
25
.
6
28. Since fxx < 0 and D = fxx fyy −
Thus, the line is y =
2
fxy
25
6
− 32 x.
> 0, the point (1, 3) is a maximum. See Figure 15.34.
y
−
32
−
16
−
− 4
1
−
64
−
12
0
0
3
x
1
Figure 15.34
29. (a)
(i) Suppose N = kAp . Then the rule of thumb tells us that if A is multiplied by 10, the value of N doubles. Thus
2N = k(10A)p = k10p Ap .
Thus, dividing by N = kAp , we have
2 = 10p
so taking logs to base 10 we have
p = log 2 = 0.3010.
(where log 2 means log10 2). Thus,
N = kA0.3010 .
(ii) Taking natural logs gives
ln N = ln(kAp )
ln N = ln k + p ln A
ln N ≈ ln k + 0.301 ln A
Thus, ln N is a linear function of ln A.
1237
(b) Table 15.2 contains the natural logarithms of the data:
Table 15.2
ln N and ln A
Island
ln A
ln N
Redonda
1.1
1.6
Saba
3.0
2.2
Montserrat
2.3
2.7
Puerto Rico
9.1
4.3
Jamaica
9.3
4.2
Hispaniola
11.2
4.8
Cuba
11.6
4.8
Using a least squares fit we find the line:
ln N = 1.20 + 0.32 ln A
This yields the power function:
N = e1.20 A0.32 = 3.32A0.32
Since 0.32 is pretty close to log 2 ≈ 0.301, the answer does agree with the biological rule.
30. Using Lagrange
multipliers,
let G = 2000 − 5x −10y = 0 be
the constraint.
2yx2 ~
xy 2 ~
yx2 ~
2xy 2 ~
i
+
2
+
j
=
1
+
i
+
2
+
j.
∇P = 1 +
2 · 108
2 · 108
108
108
∇G = −5~i − 10~j .
Now, ∇P = λ∇G, so
1+
Thus
xy 2
= −5λ
108
2+
Solving, we get 2y = x or x = 0 or y = 0.
and
2+
yx2
= −10λ.
108
yx2
2xy 2
=
2
+
.
108
108
y
200
Constraint
G=0
x
400
Figure 15.35
From G = 0 we have: when x = 0, y = 200, when y = 0, x = 400, and when x = 2y, x = 200, y = 100. So
(0,200), (400,0) and (200,100) are the critical points and they include the end points.
Substitute into P : P (0, 200) = 400, P (400, 0) = 400, P (200, 100) = 402 so the organization should buy 200 sacks of
rice and 100 sacks of beans.
31. The company wants to maximize f (x, y) = 500x0.6 y 0.3 given the constraint g(x, y) = 10x + 25y = 2000. Setting
grad f = λ grad g gives
500(0.6x−0.4 )y 0.3 = 10λ,
500x0.6 (0.3y −0.7 ) = 25λ.
From the first equation we have λ = 30y 0.3 /x0.4 , and from the second equation we have λ = 6x0.6 /y 0.7 . Setting these
equal gives
y = 0.2x.
1238
Substituting this into the constraint equation 10x + 25y = 2000 gives x = 133.33. Since y = 0.2x, the maximum value
occurs at x = 133.33 and y = 26.67.
(a) The company should purchase 133.33 units of chemical X and 26.67 units of chemical Y. With these purchases, the
company will be able to produce f (133.33, 26.67) = 25, 219 units of chemical Z.
(b) When x = 133.33 and y = 26.67, we see that λ = 11.348. If $1 is added to the budget, the company will be able to
produce about 11.348 additional units of chemical Z.
32. (a) Let c be the cost of producing the product. Then c = 10W + 20K = 3000. At optimum production,
∇q =
9
W
2
−1
4
K
1
4
Dividing yields K =
~i +
1
W,
6
3
W
2
3
4
K
−3
4
∇q = λ∇c.
~j , and ∇c = 10~i + 20~j . Equating we get
1
1
9
W−4 K 4
2
and
= λ10,
3
3
3
W 4 K− 4
2
= λ20.
so substituting into c gives
40
1
W =
W = 3000.
6
3
Thus W = 225 and K = 37.5. Substituting both answers to find λ gives
10W + 20
λ=
1
1
9
(225)− 4 (37.5) 4
2
10
3
4
= 0.2875.
1
We also find the optimum quantity produced, q = 6(225) (37.5) 4 = 862.57.
(b) At the optimum values found above, marginal productivity of labor is given by
1
1
∂q 9
= W − 4 K 4 = 2.875,
∂W (225,37.5)
2
(225,37.5)
and marginal productivity of capital is given by
3
3 3
∂q = 5.750.
= W 4 K − 4 ∂K (225,37.5)
2
(225,37.5)
The ratio of marginal productivity of labor to that of capital is
∂q
∂W
∂q
∂K
=
1
10
cost of a unit of L
=
=
.
2
20
cost of a unit of K
(c) When the budget is increased by one dollar, we substitute the relation K 1 = 16 W1 into 10W1 + 20K1 = 3001 which
gives 10W1 + 20( 61 W1 ) = 40
W1 = 3001. Solving yields W1 = 225.075 and K1 = 37.513, so q1 = 862.86 =
3
q + 0.29. Thus production has increased by 0.29 ≈ λ, the Lagrange multiplier.
33. (a) The problem is to maximize
V = 1000D 0.6 N 0.3
subject to the budget constraint in dollars
or (in thousand dollars)
40000D + 10000N ≤ 600000
40D + 10N ≤ 600
(b) Let B = 40D + 10N = 600 (thousand dollars) be the budget constraint. At the optimum
so
Thus
∇V = λ∇B,
∂V
∂B
=λ
= 40λ
∂D
∂D
∂V
∂B
=λ
= 10λ.
∂N
∂N
∂V /∂D
= 4.
∂V /∂N
Therefore, at the optimum point, the rate of increase in the number of visits with respect to an increase in the number
of doctors is four times the corresponding rate for nurses. This factor of four is the same as the ratio of the salaries.
1239
(c) Differentiating and setting ∇V = λ∇B yields
600D −0.4 N 0.3 = 40λ
300D 0.6 N −0.7 = 10λ
Thus, we get
600D −0.4 N 0.3
300D 0.6 N −0.7
=λ=
40
10
So
N = 2D.
To solve for D and N , substitute in the budget constraint:
600 − 40D − 10N = 0
600 − 40D − 10 · (2D) = 0
So D = 10 and N = 20.
600(10−0.4 )(200.3 )
≈ 14.67
40
Thus the clinic should hire 10 doctors and 20 nurses. With that staff, the clinic can provide
λ=
V = 1000(100.6 )(200.3 ) ≈ 9,779 visits per year.
(d) From part c), the Lagrange multiplier is λ = 14.67. At the optimum, the Lagrange multiplier tells us that about 14.67
extra visits can be generated through an increase of $1,000 in the budget. (If we had written out the constraint in
dollars instead of thousands of dollars, the Lagrange multiplier would tell us the number of extra visits per dollar.)
(e) The marginal cost, MC, is the cost of an additional visit. Thus, at the optimum point, we need the reciprocal of the
Lagrange multiplier:
1
1
MC = ≈
≈ 0.068 (thousand dollars),
λ
14.67
that is, at the optimum point, an extra visit costs the clinic 0.068 thousand dollars, or $68.
This production function exhibits declining returns to scale (e.g. doubling both inputs less than doubles output,
because the two exponents add up to less than one). This means that for large V , increasing V will require increasing
D and N by more than when V is small. Thus the cost of an additional visit is greater for large V than for small. In
other words, the marginal cost will rise with the number of visits.
34. (a) The solution to Problem 32 gives λ = 0.29. We recalculate λ with a budget of $4000.
The condition that grad q = λ grad(budget) in Problem 32 gives
9 −1/4 1/4
W
K
= λ(10)
2
and
3 3/4 −3/4
W K
= λ(20),
2
so K = 16 W . Substituting into the budget constraint after replacing the budget of $3000 by $4000 gives
40
1
W = 4000.
10W + 20( W ) =
6
3
Thus, W = 300 and K = 50 and q = 1150.098.
Multiplying the first equation by W and the second by K and adding gives
9
3
W ( W −1/4 K 1/4 ) + K( W 3/4 K −3/4 ) = W (10λ) + K(20λ).
2
2
So
9
3
+
W 3/4 K 1/4 = λ(10W + 20K)
2
2
6W 3/4 K 1/4 = λ(4000)
Thus,
λ=
Thus, the value of λ remains unchanged.
1150.098
6W 3/4 K 1/4
=
= 0.29
4000
4000
1240
(b) The solution to Problem 33 shows that λ = 14.67. We solve the problem again with a budget of $700,000.
The condition that grad V = λ grad B in Problem 33 gives
600D −0.4 N 0.3 = 40λ
300D 0.6 N −0.7 = 10λ
Thus, N = 2D. Substituting in the budget constraint after replacing the budget of 600 by 700 (the budget in measured
in thousands of dollars) gives
40D + 10(2D) = 700
so D = 11.667 and N = 23.337 and V = 11234.705. As in part a), we multiply the first equation by D and the
second by N and add:
D(600D −0.4 N 0.3 ) + N (300D 0.6 N −0.7 ) = D(40λ) + N (10λ),
so
(600 + 300)D 0.6 N 0.3 = λ(400 + 10N )
900D 0.6 N 0.3 = λ(700)
Since V = 1000D 0.6 N 0.3 = 11234.705, we have
λ=
900D 0.6 N 0.3
9 V
= (
) = 14.44.
700
7 1000
Thus, the value of λ has changed with the budget.
(c) We are interested in the marginal increase of production with budget (that is, the value of λ) and whether it is affected
by the budget.
Suppose $B is the budget. In part (a) we found
λ=
6W 3/4 K 1/4
B
and in part (b) we found
900D 0.6 N 0.3
.
B
In part (a), both W and K are proportional to B. Thus, W = c1 B and K = c2 B, so
λ=
6(c1 B)3/4 (c2 B)1/4
B
3/4 1/4
6c1 C2 B 3/4 B 1/4
=
B
3/4 1/4
= 6c1 c2 .
λ=
So we see λ is independent of B.
In part (b), both D and N are proportional to B, so D = c3 B and N = c4 B. Thus,
900(c3 B)0.6 (c4 B)0.3
B
900c30.6 C40.3 B 0.6 B 0.3
=
B
0.6 0.3 1
= 900c3 c4
.
B 0.1
λ=
So we see λ is not independent of B.
The crucial difference is that the exponents in Problem 32 add to 1, that is 3/4+1/4 = 1, whereas the exponents
in Problem 33 do not add to 1, since 0.6 + 0.3 = 0.9.
Thus, the condition that must be satisfied by the Cobb-Douglas production function
Q = cK a Lb
to ensure that the value of λ is not affected by production is that
a + b = 1.
This is called constant returns to scale.
1241
35. (a) Points A, B, C, D, E; that is, where a level curve of f and the constraint curve are parallel.
(b) Point F since the value of f is greatest at this point.
(c) Point D has the greatest f value of the points A, B, C, D, E.
g=c
R
A
E
F
13
D
12
B
11
10
C
9
36. We want to minimize the function h(x, y) subject to the constraint that
g(x, y) = x2 + y 2 = 1,0002 = 1,000,000.
Using the method of Lagrange multipliers, we obtain the following system of equations:
10x + 4y
= 2λx,
10,000
4x + 4y
= 2λy,
hy = −
10,000
x2 + y 2 = 1,000,000.
hx = −
Multiplying the first equation by y and the second by x we get
−y(10x + 4y)
−x(4x + 4y)
=
.
10,000
10,000
Hence:
2y 2 + 3xy − 2x2 = (2y − x)(y + 2x) = 0,
and so the climber either moves along the line x = 2y or y = −2x.
We must now choose one of these lines and the direction along that line which will lead to the point of minimum
height on the circle. To do this we find the points of intersection of these lines with the circle x 2 + y 2 = 1,000,000,
compute the corresponding heights, and then select the minimum point.
If x = 2y, the third equation gives
5y 2 = 1,0002 ,
√
so that y = ±1,000/ 5 ≈ ±447.21 and x = ±894.43. The corresponding height is h(±894.43, ±447.21) = 2400 m.
If y = −2x, we find that x = ±447.21 and y = ∓894.43. The corresponding height is h(±447.21, ∓894.43) =
2900 m. Therefore, she should travel along the line x = 2y, in either of the two possible directions.
37. The objective function is
f (x, y, z) =
and the constraint is
p
(x − a)2 + (y − b)2 + (z − c)2 ,
g(x, y, z) = Ax + By + Cz + D = 0.
1242
Partial derivatives of f and g are
fx =
fy =
fz =
1
2
· 2 · (x − a)
x−a
=
,
f (x, y, z)
f (x, y, z)
1
2
· 2 · (y − b)
y−b
=
,
f (x, y, z)
f (x, y, z)
1
2
· 2 · (z − c)
z−c
=
,
f (x, y, z)
f (x, y, z)
gx = A, gy = B, and gz = C.
Using Lagrange multipliers, we need to solve the equations
grad f = λ grad g
where grad f = fx~i + fy~j + fz ~k and grad g = gx~i + gy~j + gz ~k . This gives a system of equations:
x−a
= λA
f (x, y, z)
y−b
= λB
f (x, y, z)
z−c
= λC
f (x, y, z)
Ax + By + Cz + D = 0.
Now
x−a
A
=
y−b
B
=
z−c
C
= λf (x, y, z) gives
A
(y − b) + a,
B
C
z = (y − b) + c,
B
x=
Substitute into the constraint,
A
Hence
A
(y − b) + a + By + C
B
A2
C2
+B+
B
B
y=
y−b =
x−a =
=
z−c =
=
y=
C
(y − b) + c + D = 0,
B
A2
C2
b − Aa +
b − Cc − D.
B
B
(A2 + C 2 )b − B(Aa + Cc + D)
,
A2 + B 2 + C 2
−B(Aa + Bb + Cc + D)
A2 + B 2 + C 2
A
(y − b)
B
−A(Aa + Bb + Cc + D)
A2 + B 2 + C 2
C
(y − b)
B
−C(Aa + Bb + Cc + D)
A2 + B 2 + C 2
Thus the minimum f (x, y, z) is
f (x, y, z) =
=
p
(x − a)2 + (y − b)2 + (z − c)2
−A(Aa + Bb + Cc + D)
A2 + B 2 + C 2
2
+
−B(Aa + Bb + Cc + D)
A2 + B 2 + C 2
2
+
−C(Aa + Bb + Cc + D)
A2 + B 2 + C 2
|Aa + Bb + Cc + D|
√
.
A2 + B 2 + C 2
=
2
1243
1/2
The geometric meaning is finding the shortest distance from a point (a, b, c) to the plane Ax + By + Cz + D = 0.
38. We first express the revenue R in terms of the prices p1 and p2 :
R(p1 , p2 ) = p1 q1 + p2 q2
= p1 (517 − 3.5p1 + 0.8p2 ) + p2 (770 − 4.4p2 + 1.4p1 )
= 517p1 − 3.5p21 + 770p2 − 4.4p22 + 2.2p1 p2 .
At a local maximum we have grad R = 0, and so:
∂R
= 517 − 7p1 + 2.2p2 = 0,
∂p1
∂R
= 770 − 8.8p2 + 2.2p1 = 0.
∂p2
Solving these equations, we find that
p1 = 110 and
p2 = 115.
To see whether or not we have a found a local maximum, we compute the second-order partial derivatives:
∂2R
= −7,
∂p21
∂2R
= −8.8,
∂p22
∂2R
= 2.2.
∂p1 ∂p2
Therefore,
∂2R
∂2R ∂2R
−
= (−7)(−8.8) − (2.2)2 = 56.76,
2
2
∂p1 ∂p2
∂p1 ∂p2
and so we have found a local maximum point. The graph of P (p 1 , p2 ) has the shape of an upside down paraboloid. Since
P is quadratic in q1 and q2 , (110, 115) is a global maximum point.
D=
39. We want to minimize cost C = 100L + 200K subject to Q = 900L 1/2 K 2/3 = 36000. Using Lagrange multipliers, we
get
∇Q = 450L−1/2 K 2/3 ~i + 600L1/2 K −1/3 ~j .
∇C = 100~i + 200~j
∇C = λ∇Q gives
100 = λ450L−1/2 K 2/3
Since λ 6= 0 this gives
and
200 = λ600L1/2 K −1/3 .
450L−1/2 K 2/3 = 300L1/2 K −1/3 .
Solving, we get L = (3/2)K. Substituting into Q = 36,000 gives
900
h
Solving yields K = 40 ·
i
6/7
2 1/2
3
and L = 30 which gives C = $7, 000.
3
K
2
1/2
≈ 19.85, so L ≈
K 2/3 = 36,000.
3
(19.85)
2
= 29.78. We can thus calculate cost using K = 20
40. We wish to minimize the objective function
C(x, y, z) = 20x + 10y + 5z
subject to the budget constraint
Q(x, y, z) = 20x1/2 y 1/4 z 2/5 = 1, 200.
Therefore, we solve the equations grad C = λ grad Q and Q = 1, 200:
20 = 10λx−1/2 y 1/4 z 2/5
or λ = 2x1/2 y −1/4 z −2/5 ,
10 = 5λx1/2 y −3/4 z 2/5 ,
or λ = 2x−1/2 y 3/4 z −2/5 ,
5 = 8λx
1/2 1/4 −3/5
20x1/2 y 1/4 z 2/5 = 1, 200.
y
z
,
or λ = 0.625x−1/2 y −1/4 z 3/5 ,
1244
The first and second equations imply that
x = y,
while the second and third equations imply that
3.2y = z.
Substituting for x and z in the constraint equation gives
20y 1/2 y 1/4 (3.2y)2/5 = 1200
y ≈ 23.47,
and so
x ≈ 23.47
and z ≈ 75.1.
41. Cost of production, C, is given by
C = p1 W + p2 K = b. At the optimal point, ∇q = λ∇C.
Since ∇q = c(1 − a)W −a K a ~i + caW 1−a K a−1 ~j and ∇C = p1~i + p2~j , we get
c(1 − a)W −a K a = λp1 and caW 1−a K a−1 = λp2 .
∂q
Now, marginal productivity of labor is given by ∂W
= c(1 − a)W −a K a and marginal productivity of capital is given by
∂q
1−a a−1
= caW
K
, so their ratio is given by
∂K
∂q
∂W
∂q
∂K
=
c(1 − a)W −a K a
p1
λp1
=
=
caW 1−a K a−1
λp2
p2
which is the ratio of the cost of one unit of labor to the cost of one unit of capital.
42. (a) The objective function is the energy loss, i21 R1 + i22 R2 , and the constraint is i1 + i2 = I, where I is a constant. The
Lagrangian function is
L(i1 , i2 , λ) = i21 R1 + i22 R2 − λ(i1 + i2 − I).
We look for solutions to the system of equations we get from grad L = ~0 :
∂L
= 2i1 R1 − λ = 0
∂i1
∂L
= 2i2 R2 − λ = 0
∂i2
∂L
= −(i1 + i2 − I) = 0.
∂λ
Combining
∂L
∂L
∂L
−
= 2(i1 R1 − i2 R2 ) = 0 with
= 0 gives the two equation system
∂i1
∂i2
∂λ
i1 R 1 − i 2 R 2 = 0
i1 + i2 = I.
Substituting i2 = I − i1 into the first equation leads to
R2
I
R1 + R 2
R1
i2 =
I.
R1 + R 2
i1 =
(b) Ohm’s Law states that across a resistor
Voltage = Current · Resistance.
Since λ/2 = i1 · R1 = i2 · R2 , the Lagrange multiplier λ equals twice the voltage across the resistors.
1245
43. (a) We draw the level curves (parallel straight lines) of f (x, y) = ax + by + c. We can see that the level lines with the
maximum and minimum f -values which intersect with the disk are the level lines that are tangent to the boundary of
the disk. Therefore, the maximum and minimum occur at the boundary of the disk. See Figure 15.36.
f = max
f = min
M
f increases
Figure 15.36
f = max
f = max
f = min
f increases]
f increases
6
f = min
Figure 15.37
Figure 15.38
(b) Similar to part (a), we see the level lines with the largest and smallest f -values which intersect with the rectangle must
pass the corner of the rectangle. So the maximum and minimum occur at the corners of rectangle. See Figure 15.37.
When the level curves are parallel to a pair of the sides, then the points on the sides are all maximum or minimum, as
shown below in Figure 15.38.
(c) The graph of f is a plane. The part of the graph lying above a disk R is either a flat disk, in which case every point is
a maximum, or is a tilted ellipse, in which case you can see that the maximum will be on the edge. Similarly, the part
lying above a rectangle is either a rectangle or a tilted parallelogram, in which case the maximum will be at a corner.
44. The point P is the solution to the constraint optimization problem of maximizing the square of the distance function.
D = x2 + y 2 + x2
subject to the constraint
g(x, y, z) = f (x, y) − z = 0.
(We take the square of the distance between the point (x, y, z) and the origin, which is
Distance =
p
(x − 0)2 + (y − 0)2 + (z − 0)2 =
p
x2 + y 2 + z 2 ,
because it makes the calculations easier.) Therefore, at point P , we have ∇D = λ∇g, so ∇D is parallel to ∇g.
1246
Also
We know that ∇g is perpendicular to the surface g(x, y, z) = 0; that is, perpendicular to the surface z = f (x, y).
∇D = 2x~i + 2y~j + 2z~k .
At point P , whose position vector is p
~ = a~i + b~j + c~k , we have
∇D = 2(a~i + b~j + c~k ) = 2~
p.
Thus, p
~ is parallel to ∇D and therefore p
~ is also perpendicular to the surface.
45. You should try to anticipate your opponent’s choice. After you choose a value λ, your opponent will use calculus to
find the point (x, y) that maximizes the function f (x, y) = 10 − x2 − y 2 − 2x − λ(2x + 2y). At that point, we have
fx = −2x − 2 − 2λ = 0 and fy = −2y − 2λ = 0, so your opponent will choose x = −1 − λ and y = −λ. This gives a
value L(−1 − λ, −λ, λ) = 10 − (−1 − λ)2 − (−λ)2 − 2(−1 − λ) − λ(2(−1 − λ) + 2(−λ)) = 11 + 2λ + 2λ2 which you
want to make as small as possible. You should choose λ to minimize the function h(λ) = 11 + 2λ + 2λ 2 . You choose λ
so that h0 (λ) = 2 + 4λ = 0, or λ = −1/2. Your opponent then chooses (x, y) = (−1 − λ, −λ) = (−1/2, 1/2), giving
a final score of L(−1/2, 1/2, −1/2) = 10.5. No choice of λ that you can make can force the value of L below 10.5. But
your choice of λ = −1/2 makes it impossible for your opponent to force the value of L above 10.5.
46. The wetted perimeter of the trapezoid is given by the sum of the lengths of the three walls, so
p=w+
2d
sin θ
We want to minimize p subject to the constraint that the area is fixed at 50 m 2 . A trapezoid of height h and with parallel
sides of lengths b1 and b2 has
(b1 + b2 )
.
A = Area = h
2
In this case, d corresponds to h and b1 corresponds to w. The b2 term corresponds to the width of the exposed surface of
the canal. We find that b2 = w + (2d)/(tan θ). Substituting into our original equation for the area along with the fact that
the area is fixed at 50 m2 , we arrive at the formula:
Area =
d
2d
w+w+
2
tan θ
=d w+
d
tan θ
= 50
We now solve the constraint equation for one of the variables; we will choose w to give
w=
50
d
−
.
d
tan θ
Substituting into the expression for p gives
p=w+
2d
50
d
2d
=
−
+
.
sin θ
d
tan θ
sin θ
We now take partial derivatives:
∂p
1
50
2
=− 2 −
+
∂d
d
tan θ
sin θ
∂p
d
1
2d
· cos θ
=
·
−
∂θ
tan2 θ cos2 θ
sin2 θ
From ∂p/∂θ = 0, we get
1
2d
d · cos2 θ
·
· cos θ.
=
cos2 θ
sin2 θ
sin2 θ
Since sin θ =
6 0 and cos θ =
6 0, canceling gives
1 = 2 cos θ
so
cos θ =
1
.
2
π
, we get
2
Substituting into the equation ∂p/∂d = 0 and solving for d gives:
Since
0<θ<
θ=
2
−50
1
=0
−√ +√
d2
3
3/2
π
.
3
which leads to
d=
Then
r
1247
50
√ ≈ 5.37m.
3
d
50
5.37
50
−
≈
− √ ≈ 6.21 m.
d
tan θ
5.37
3
When θ = π/3, w ≈ 6.21 m and d ≈ 5.37 m, we have p ≈ 18.61 m.
Since there is only one critical point, and since p increases without limit as d or θ shrink to zero, the critical point
must give the global minimum for p.
w=
CAS Challenge Problems
47. (a) The partial derivatives of f are
√
√
a + x + −1 + a + x y
∂f
= √
2
√
√
∂x
2 a+x a+x+y 1+ a+x+y
√
1 − 2a − 2x + a + x − y
∂f
= √
2
√
∂y
2 a+x+y 1+ a+x+y
Solving ∂f /∂x = 0, ∂f /∂y = 0, we get x = 1/4 − a, y = 1. The discriminant at this point is D = −16/625.
Thus, by the second derivative test, the point is a saddle point.
(b) The y coordinate of the critical points stays the same and the x coordinate is a √
units to the left of its position when
x+y
√ by substituting x + a for
a = 0. The type is always a saddle point. This is because f is obtained from
1+y+ x
x, so that the graph is shifted a units in the negative x-direction but its shape remains the same.
48. (a) We have grad f = 2x~i + ~j and grad g = (2x + 2y)~i + (2x + 2y)~j . So the equations to be solved in the method
of Lagrange multipliers are
2x = λ(2x + 2y)
1 = λ(2x + 2y)
x2 + 2xy + y 2 − 9 = 0
Solving these with a CAS, we get two solutions:
x = 1/2, y = −7/2, λ = −1/6,
or x = 1/2, y = 5/2, λ = 1/6
Student A reasons that since f (1/2, −7/2) = −13/4 and f (1/2, 5/2) = 11/4 , the (global) maximum and
minimum values are 11/4 and 13/4, respectively. Student B graphs the constraint curve g = 0 and a contour diagram
of f . The constraint curve turns out to be two straight lines, since the constraint x 2 + 2xy + y 2 − 9 = 0, which can
be rewritten as (x + y)2 = 9, or x + y = ±3. The value of f goes to infinity on each of these straight lines. On the
line y = −x + 3, f (x, y) = x2 + y = x2 − x + 3, and on the line y = −x − 3, f (x, y) = x2 + y = x2 − x − 3.
Thus Student B is correct. The points Student A found are actually local maximum and local minimum values, not
global. Since the constraint is not bounded, there is no guarantee that there is a local maximum or minimum. See
Figure 15.39.

Solutions for Chapter 15 Review

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