14.3 SOLUTIONS
217
22. To execute the gradient search we first evaluate the two first partial derivatives of C :
@C
@n
@C
@d
=
=
0:15 , 3
4d 1:8
n,2
,5:87
4d 0:8
4d 0:8
+ 1:44
+ 4:32
n,1
,11:688 45d
5
5
5
If we choose as our starting point for the gradient search a pipe diameter of 1 meter with three pumping
stations, after numerous iterations, we find the optimum values to be about 1:61 meter diameter pipe with six
pumping stations. This produces a minimum cost of 4.14 million dollars.
23. We have
fx = 2x(y + 1)3 = 0 only when x = 0 or y = ,1
fy = 3x2(y + 1)2 + 2y = 0 never when y = ,1 and only for y = 0 when x = 0
We conclude that fx = 0 and fy = 0 only when x = 0; y = 0, so f has only one critical point, namely (0; 0).
The second derivative test at (0; 0) gives
D = fxx fyy , (fxy )2 = 2(y + 1)3(6x2(y + 1) + 2) , (6x(y + 1)2)2
= 2(1)(2) , 0 > 0 when x = 0; y = 0
Since fxx > 0 at (0; 0), this means f has a local minimum at (0; 0).
[Alternatively, if we expand (y + 1)3 , then we can view f (x; y) as x2 + y2 + (terms of degree 3 or greater
in x and y), which means that f behaves likes x2 + y2 near (0; 0).]
Although (0; 0) is a local minimum, it cannot be a global minimum since for fixed x, say x = 1, the
function f (x; y) is a cubic polynomial in y and cubics take on arbitrarily large positive and negative values.
In the single-variable case, suppose a function f defined on the real line is differentiable and its derivative
is continuous. Then if f has only one critical point, say x = 0, then if that critical point is a local minimum,
it must also be a global minimum. This is because f 0 cannot change sign without f 0 = 0 so we must have
f 0 < 0 for x < 0 and f 0 > 0 for x > 0. Thus f is decreasing for all x < 0 and increasing for all x > 0,
which makes x = 0 the global minimum for f .
Solutions for Section 14.3
1. Our objective function is f (x; y) = x + y and our equation of constraint is g(x; y) = x2 + y2 = 1. To
optimize f (x; y) with Lagrange multipliers, we solve rf (x; y) = rg(x; y) subject to g(x; y) = 1. The
gradients of f and g are
rf (x; y) = ~i + ~j ;
rg(x; y) = 2x~i + 2y~j :
So the equation rf
=
rg becomes
~i + ~j
Solving for gives
=
(2x~i + 2y~j )
= 21x = 21y ;
218
CHAPTER FOURTEEN /SOLUTIONS
which tells us that x
for y
=
y. Going back to our equation of constraint, we use the substitution x = y to solve
g(y; y) = y2 + y2 = 1
2y 2 = 1
y2 = 12
Since x = y, our critical points are (
p 2 p2
maximum value is f ( 2 ;
2. Our objective function is
Their gradients are
p
2
2 ) =
(
) =
f x; y
p2 p2
2
;
)
2
r
y=
and (,
p2
2
;,
p
2
:
2
=
).
Evaluating f at these points we find that the
p2
2
1
2
p
p
p
and the minimum value is f (, 22 ; , 22 ) = , 2.
3x , 2y and our equation of constraint is g(x; y)
=
x2 + 2y2
=
44.
rf (x; y) = 3~i , 2~j ;
rg(x; y) = 2x~i + 4y~j :
So the equation rf = rg becomes 3~i , 2~j = (2x~i + 4y~j ). Solving for gives us
= 3 = ,2 ;
which we can use to find x in terms of y:
2x
4y
,2
3
=
2x
4y
,4x = 12y
x = ,3y:
Using this relation in our equation of constraint, we can solve for y:
3.
x2 + 2y2 = 44
2
2
(,3y ) + 2y = 44
9y2 + 2y2 = 44
11y2 = 44
y2 = 4
y = 2:
Thus, the critical points are (6; 2) and (6; ,2). Evaluating f at these points, we find that the maximum is
f (6; ,2) = 18 + 4 = 22 and the minimum value is f (,6; 2) = ,18 , 4 = ,22.
Our objective function is f (x; y) = x2 + y and our equation of constraint is g(x; y) = x2 , y2 = 1. Their
gradients are
Thus rf
=
rg gives
rf (x; y) = 2x~i + ~j ;
rg(x; y) = 2x~i , 2y~j :
2 x = 2 x
1 = 2 y
14.3 SOLUTIONS
But x cannot be zero, since the constraint equation, ,y2
equation rf = rg becomes
=
219
1, would then have no real solution for y. So the
= 22xx = ,12y
1
,2y
,2 y = 1
y = , 12 :
1=
Substituting this into our equation of constraint we find
g(x; , 12 ) = x2 ,
1 2
,2
=
1
x2 = 54
p
x=
p
p
5
:
2
p
; , 21 ) and (, 25 ; , 12 ). Evaluating f at these points we find f ( 25 ; , 12 ) =
p5 1
f (, 2 ; , 2 ) = 54 , 12 = 34 . This is the minimum value for f (x; y) constrained to g(x; y) = 1. To see that f
has no maximum on g(x; y) = 1, note that as x and y increase, f increases and that on the part of the graph
of g(x; y) = 1 in quadrant I x ! 1 and y ! 1.
So the critical points are
( 25
Figure 14.13: Graph of
x2 y 2 = 1
,
4. Our objective function is
gradients are
f (x; y)
=
xy and our equation of constraint is g(x; y) = 4x2 + y2
rf (x; y) = y~i + x~j ;
rg(x; y) = 8x~i + 2y~j :
So the equation rf
Multiplying, we get
=
rg becomes y~i + x~j = (8x~i + 2y~j ). This gives
8x = y and 2y = x:
8x2 = 2y2 :
=
8. Their
220
CHAPTER FOURTEEN /SOLUTIONS
If = 0, then x = y
=
0, which doesn’t satisfy the constraint equation. So 6= 0 and we get
2y 2
y
y
2
8x2
2
= 4x
= 2x:
=
To find x, we substitute for y in our equation of constraint.
4x2 + y2
4x2 + 4x2
5.
=
8
=8
=1
x2
x = 1
So our critical points are (1; 2), (1; ,2), (,1; 2) and (,1; ,2). Evaluating f (x; y) at these points we have
f (1; 2) = f (,1; ,2) = 2
f (1; ,2) = f (1; ,2) = ,2:
So the maximum value of f on g(x; y) = 8 is 2, and the minimum value is ,2.
The objective function is f (x; y) = x2 + y2 and the equation of constraint is g(x; y) = x4 + y4 = 2. Their
gradients are
rf (x; y) = 2x~i + 2y~j ;
rg(x; y) = 4x3~i + 4y3~j :
rg becomes 2x~i + 2y~j
(4x3~i + 4y3~j ). This tells us that
2x = 4x3 ;
2y = 4y3 :
Now if x = 0, the first equation is true for any value of . In particular, we can choose 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 and find
= 42xx3 = 42yx3
So the equation rf
=
=
1
2x2
=
1
2y 2
y2 = x2
y = x:
Going back to our equation of constraint, we find
p
g (0 ; y ) = 0 4 + y 4 = 2 ) y = 2
p
g(x; 0) = x4 + 04 = 2 ) x = 2
g(x; x) = x4 + (x)4 = 2 ) x = 1:
p
p
Thus, the critical points are (0; 2), ( 2; 0), (1; 1) and (,1; 1). Evaluating f at these points we find
f (1; 1) = f (1; ,1) = f (,1; 1) = f (,1; ,1) = 2;
p
p
p
p
p
f (0; 2) = f (0; , 2) = f ( 2; 0) = f (, 2; 0) = 2:
p
So the minimum value of f (x; y) on g(x; y) = 2 is 2 and the maximum value is 2.
4
4
4
4
4
4
4
4
14.3 SOLUTIONS
6. The objective function is f (x; y)
The gradients of f and g are
=
221
x2 , xy + y2 and the equation of constraint is g(x; y) = x2 , y2 = 1.
rf (x; y) = (2x , y)~i + (,x + 2y)~j ;
rg(x; y) = 2x~i , 2y~j :
Therefore the equation rf (x; y) = rg(x; y) gives
2x , y = 2x
,x + 2y = ,2y
x2 , y2 = 1.
Let us suppose that = 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 6= 0. Now let’s multiply the first two equations
,2y(2x , y) = 2x(,x + 2y):
As 6= 0, we can cancel it in the equation above and after doing the algebra we get
p
p
x2 , 4xy + y2 = 0
(2 + 3)y or x = (2 , 3)y .
which gives x = p
If x = (2 + 3)y, the third equation gives
(2 +
so y
p
3)2 y2 , y2
=
1
0:278 and
p x 1:038. These give the solutions (1:038; 0:278), (,1:038; ,0:278).
If x = (2 , 3)y, from the third equation we get
But (2 ,
(2
p
,
p
3)2 y2 , y2
=
1:
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 x ! 1 and y
! 1 on the constraint, rewriting f as
2
f (x; y) = x , y2 + 34 y2
shows that f has no maximum on the constraint. The minimum value of f is 0:866.
7. The objective function is f (x; y; z ) = x + 3y + 5z and the equation of constraint is g(x; y; z ) = x2 + y2 + z 2
1. Their gradients are
rf (x; y; z ) = ~i + 3~j + 5~k ;
rg(x; y; z ) = 2x~i + 2y~j + 2z~k :
So the equation rf
=
rg becomes ~i
+
3y~j
+
5~k
=
(2x~i + 2y~j
= 21x = 23y = 25z :
+2
z~k ). Solving for we find
=
222
CHAPTER FOURTEEN /SOLUTIONS
Which provides us with the equations
2y = 6x
10x = 2z:
Solving the first equation for y gives us y = 3x. Solving the second equation for
Substituting these into the equation of constraint, we can find x:
x2 + (3x)2 + (5x)2 = 1
x2 + 9x2 + 25x2 = 1
35x2 = 1
1
x2 = 35
x=
Since y
=
3x and z
=
r
1
35
=
z
gives us
z
=
5x.
p
35
:
35
p p
p
5x, the critical points are at ( 3535 ; 3 3535 ; 735 ). Evaluating f at these two points, we
p
p35 p35
p
p
p 35
p 35 p
35
35
35
find the maximum is f ( 35 ; 3 35 ; 7 ) = 35 35 = 35, and the minimum value is f (, 35 ; ,3 35 ; , 7 ) =
p
, 35.
8. Our objective function is f (x; y; z ) = 2x + y + 4z and our equation of constraint is g(x; y; z ) = x2 + y + z 2 =
16. Their gradients are
rf (x; y; z ) = 2~i + 1~j + 4~k ;
rg(x; y; z ) = 2x~i + 1~j + 2z~k :
So the equation rf
=
~k
rg becomes 2~i + 1~j
+4
=
(2x~i + 1~j
+
2z~k ). Solving for we find
= 22x = 11 = 24z
Which tells us that x = 1 and z
= x1 = 1 = 2z :
=
2. Going back to our equation of constraint, we can solve for y.
g(1; y; 2) = 16
1 + y + 22 = 16
y = 11:
So our one critical point is at (1; 11; 2). The value of f at this point is f (1; 11; 2) = 2 + 11 + 8 = 21. This is
the maximum value of f (x; y; z ) on g(x; y; z ) = 16. To see this, look at a graph of x2 + y + z 2 = 16. This
shape is called a paraboloid. The important thing to note is that as you move farther back on the y-axis, the
graph opens up more and more; specifically, x ! ,1; y ! ,1; z ! ,1 on the surface. Now f (x; y; z )
decreases for decreasing x, y and z , so there is no minimum value for f (x; y; z ) on g(x; y; z ) = 16.
Our objective function is f (x; y; z ) = x2 , y2 , 2z and our equation of constraint is g(x; y; z ) = x2 + y2 ,
z = 0. To optimize f (x; y; z ) with Lagrange multipliers, we solve rf (x; y; z ) = rg(x; y; z ) subject to
g(x; y; z ) = 0. The gradients of f and g are
rf (x; y; z ) = 2x~i , 2y~j , 2~k ;
rg(x; y; z ) = 2x~i + 2y~j , ~k :
2
9.
14.3 SOLUTIONS
223
We get
2x = 2x
,2y = 2y
10.
, 2 = ,
x2 + y 2 = z .
The third equation gives = 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.
a2 + 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.
Our objective function is f (x; y; z ) = x2 , 2y + 2z 2 and our equation of constraint is g(x; y; z ) = x2 + y2 +
z 2 , 1 = 0. To optimize f (x; y; z ) with Lagrange multipliers, we solve rf (x; y; z ) = rg(x; y; z ) subject
to g(x; y; z ) = 0. The gradients of f and g are
rf (x; y; z ) = 2x~i , 2~j + 4z~k ;
rg(x; y) = 2x~i + 2y~j + 2z~k :
We get,
x = x
,1 = y
2z = z
2
2
x + y + z 2 = 1.
From the first equation we get x = 0 or = 1.
If x = 0 we have
,1 = y
2z = z
y 2 + z 2 = 1:
From the second equation z = 0 or = 2. So if z = 0, we have y = 1 and we get the solutions
p
(0; 1; 0),(0; ,1; 0). If z 6= 0 then = 2 and y = , 12 . So z 2 = 34 which gives the solutions (0; , 12 ; 23 ),
p
(0; , 21 ; , 23 ).
If x 6= 0, then = 1, so y = ,1, which implies, from the equation x2 + y2 + z 2 = 1, that x = 0, which
contradicts the assumption.
p
Therefore, evaluating f at these points we get f (0; 1; 0) = ,2, f (0; ,1; 0) = 2 and f (0; , 12 ; 23 ) =
p
f (0; , 12 ; , 23 ) = 4. So the maximum value of f is 4 and the minimum is ,2.
11. Our objective function is f (x; y; z ) = x + y + z and our equations of constraint are g(x; y; z ) = x2 + y2 +
z 2 , 1 = 0 and h(x; y; z ) = x , y , 1 = 0. To optimize f (x; y; z ) with Lagrange multipliers, we solve
rf (x; y; z ) = 1 rg(x; y; z ) + 2 rh(x; y; z ) subject to g(x; y; z ) = 0 and h(x; y; z ) = 0. The gradients of
f , g and h are
rf (x; y; z ) = ~i + ~j + ~k ;
rg(x; y; z ) = 2x~i + 2y~j + 2z~k
rh(x; y; z ) = ~i , ~j
224
CHAPTER FOURTEEN /SOLUTIONS
Therefore we get
1 = 21 x + 2
1 = 21 y , 2
x
2
1 = 21 z
2
+y +z =1
x,y = 1
2
Adding the first two equations we get
21 (x + y)
=
2:
From this and the third equation we get
41 z
As 1
=
21 (x + y):
6= 0, we can cancel it and get x + y = 2z . From this and x , y = 1 we get
x = 2z 2+ 1 ,
y = 2z , 1 .
2
Substituting into the equation x2 + y2 + z 2
and ( 12 , p16 ; , 21 , p16 ; , p16 ).
Evaluating f at these points gives
=
1 gives z 2
1
1
= 6 . So we get the solutions ( 2 +
p16 ; , 12 + p16 ; p16 )
f ( 12 + p16 ; , 12 + p16 ; p16 ) = p36
f ( 12 , p16 ; , 12 , p16 ; , p16 ) = , p36 :
Therefore the maximum value of f is
p36 and the minimum is , p36 .
12. The region x2 + y2 4 is the shaded disk of radius 2 centered at the origin (including the circle x2 + y2
shown in Figure 14.14.
4
=
4)
y
16
8
1
,4
4
4
x
,4
Figure 14.14
We will first find the relative maxima and minima in the interior of the disk. So we need to find the
extrema of
f (x; y) = x2 + 2y2 in the region x2 + y2 < 4:
14.3 SOLUTIONS
225
For this we compute the critical points:
fx = 2x = 0
fy = 4 y = 0
So the critical point is (0; 0). As fxx (0; 0) = 2, fyy (0; 0) = 4 and fxy (0; 0) = 0 we have
D = fxx (0; 0) fyy (0; 0) , (fxy (0; 0))2 = 8 > 0
and
fxx (0; 0) = 2 > 0:
Therefore (0; 0) is a minimum point and f (0; 0) = 0.
Now let’s find the relative extrema of f on the boundary of the disk, hence this time we have to solve a
constraint problem. We want the extrema of f (x; y) = x2 + 2y2 subject to g(x; y) = x2 + y2 , 4 = 0. We
use Lagrange multipliers:
grad f = grad g and x2 + y2 = 4;
which give
2x = 2x
x
4y
2
+
y
2
2y
= 4.
=
From the first equation we have x = 0 or = 1. If x = 0, from the last equation y2 = 4 and therefore
(0; 2) and (0; ,2) are solutions.
If x 6= 0 then = 1 and from the second equation y = 0. Substituting this into the third equation we get
x2 = 4 so (2; 0) and (,2; 0) are the other two solutions.
Therefore, as f (0; 2) = f (0; ,2) = 8 and f (2; 0) = f (,2; 0) = 4, (0; 2) and (0; ,2) are global
maxima and (0; 0) is the global minimum on the whole region x2 = y2 4. The maximum value of f is 8
and the minimum value of f is 0.
13. The domain x2 + 2y2
in Figure 14.15.
1 is the shaded interior of the ellipse x2 + 2y2 = 1 including the boundary, shown
,0:5
,0:3
, p12
,0:1
y
:
01
,1
:
05
03
:
1
x
, p12
Figure 14.15
First we want to find the relative maxima and minima of f in the interior of the ellipse. So we need to
find the extrema of
f (x; y) = xy; in the region x2 + 2y2 < 1:
For this we compute the critical points:
fx = y = 0
and
fy = x = 0:
226
CHAPTER FOURTEEN /SOLUTIONS
So there is one critical point, (0; 0). As fxx (0; 0) = 0, fyy (0; 0) = 0 and fxy (0; 0) = 1 we have
D = fxx (0; 0) fyy (0; 0) , (fxy (0; 0))2 = ,1 < 0
so (0; 0) is a saddle and f doesn’t have relative extrema in the interior of the ellipse.
Now let’s find the relative extrema of f on the boundary, hence this time we’ll have a constraint problem.
We want the extrema of f (x; y) = xy subject to g(x; y) = x2 + 2y2 , 1 = 0. We use Lagrange multipliers:
grad f
=
grad g
and
x2 + 2y2 = 1
which give
y = 2x
x = 4y
x2 + 2 y 2 = 1
From the first two equations we get
xy = 82 xy:
So x = 0 or y = 0 or 82 = 1.
p
p
p
If x = 0, from the last equation 2y2 = 1 so y = 22 and we get the solutions (0; 22 ) and (0; , 22 ).
If y = 0, from the last equation we get x2 = 1 and so the solutions are (1; 0) and (,1; 0).
1
1
If x 6= 0 and y 6= 0 then 82 = 1, hence = 2p
. For = 2p
2
2
x=
and plugging into the third equation gives 4y2
1
For = , 2p
we get
2
=
p
2y
1 so we get the solutions (
p2
2
p
x = , 2y
p
and plugging into the third equation gives 4y2 = 1, and the solutions ( 22 ; , 12 ) and (,
p
p
p
p
have the solutions: (1; 0), (,1; 0), ( 22 ; 12 ), (, 22 ; , 21 ), ( 22 ; , 12 ), (, 22 ; 12 ).
Evaluating f at these points gives:
p
p
f (0; 22 ) = f (0; , 22 ) = f (1; 0) = f (,1; 0) = 0
p
p
p
2 1
2 1
f ( 2 ; 2 ) = f (, 2 ; , 2 ) = 42
p
p
p
2 1
2 1
f ( 2 ; , 2 ) = f (, 2 ; 2 ) = , 42 .
Hence the maximum value of f is
p2
4
p
; 12 ) and (, 22 ; , 12 ).
and the minimum value of f is ,
p2
4
.
p2
2
; 12 ). So finally we
14.3 SOLUTIONS
14. The region x2 y is the shaded region in Figure 14.16 which includes the parabola y
=
227
x2 .
y
,70
,50,30
70
30
,10
50
10
x
Figure 14.16
We first want to find the relative maxima and minima of f in the interior of our region. So we need to
find the extrema of
f (x; y) = x2 , y2 ;
in the region
x2 > y:
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 relative 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 , y2 subject to g(x; y) = x2 , y = 0. We use Lagrange multipliers:
grad f
=
grad g
and
x2 = y:
This gives
2x = 2x
2y = x2 = y:
From the first equation we get x = 0 or = 1.
If x = 0, from the third equation we get y = 0, so one solution is (0; 0). If x 6= 0, then = 1 and from
the second equation we get y = 12 . This gives x2 = 12 so the solutions ( p12 ; 12 ) and (, p12 ; 12 ).
So f (0; 0) = 0 and f ( p12 ; 12 ) = f (, p12 ; 12 ) = 14 . From Figure 14.16 showing the level curves of f and
the region x2 y, we see that (0; 0) is a local minimum of f on x2 = y, but not a global minimum and that
( p1 ; 12 ) and (, p1 ; 12 ) are global maxima of f on x2 = y but not global maxima of f on the whole region
2
2
x2 y.
So there are no global extrema of f in the region x2 y.
228
CHAPTER FOURTEEN /SOLUTIONS
15. The region x2 + y2 2 is the shaded disk of radius
as shown in Figure 14.17.
p
2 centered at the origin (including the circle x2 + y2
=
2)
p y
2
5
3
,
p x
1
p
,1
2
2
,3
,5 ,p2
Figure 14.17
We first find the relative maxima and minima of f in the interior of our disk. So we need to find the
extrema of
f (x; y) = x + 3y; in the region x2 + y2 < 2:
As
fx = 1
fy = 3
f doesn’t have critical points. Now let’s find the relative extrema of f on the boundary of the disk. We want
to find the extrema of f (x; y) = x + 3y subject to the constraint g(x; y) = x2 + y2 , 2 = 0. We use Lagrange
multipliers
grad f = grad g and x2 + y2 = 2;
which give
1 = 2x
3 = 2y
2
x + y2 = 2.
As cannot be zero, we solve for x and y in the first two equations and get
into the third equation gives
82 = 10
so = p5
2
x = 21 and y = 23 . Plugging
and we get the solutions ( p15 ; p35 ) and (, p15 ; , p35 ). Evaluating f at these points gives
f ( p1 ; p3
)=
p
2 5 and
5
5
p
1
3
f (, p ; , p ) = ,2 5
5
5
Therefore ( p15 ; p35 ) is a global maximum of
region x2 + y2 2.
f
and
, p15 ; , p35 ) is a global minimum of f on the whole
(
14.3 SOLUTIONS
16. The region x + y
229
1 is the shaded half plane (including the line x + y = 1) shown in Figure 14.18.
y
3
2
1
,3
,5
,1
3
5
1
1
2
3
x
Figure 14.18
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
grad f = grad g and x + y = 1;
which give
3x2 = 1=
x + y = 1.
From the first two equations we get 3x2 = 1, so the solutions are
(
p1 ; 1 , p1
3
3
)
and
, p1 ; 1 + p1 ):
(
3
3
Evaluating f at these points we get
f ( p1 ; 1 , p1
)=
1,
p2
3
3
3 3
1
1
f (, p ; 1 + p ) = 1 + p2 .
3
3
3 3
So ( p13 ; 1 , p13 ) is a local minimum and (, p13 ; 1 + p13 ) 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) = x3 + 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 14.18.
230
CHAPTER FOURTEEN /SOLUTIONS
17. The region x2 + y2 1 is the shaded disk of radius 1 centered at the origin (including the circle x2 + y2
shown in Figure 14.19.
Let’s first compute the critical points of f in the interior of the disk. We have
=
1)
fx = 3x2 = 0
fy = ,2y = 0,
whose solution is x = y
fxy (0; 0) = 0,
=
0. So the only one critical point is (0; 0). As fxx (0; 0) = 0, fyy (0; 0) = ,2 and
D = fxx (0; 0) fyy (0; 0) , (fxy (0; 0))2 = 0
which doesn’t tell us anything about the nature of the critical point (0; 0).
But, if we choose x,y very small in absolute value and such that x3 > y2 , then f (x; y) > 0. If we choose
x,y very small in absolute value and such that x3 < y2 , then f (x; y) < 0. As f (0; 0) = 0, we conclude that
(0; 0) is a saddle point.
We can get the same conclusion looking at the level curves of f around (0; 0), as shown in Figure 14.20.
y
y
1
1
1
,0 :7
,0:3
,0:1
x
0
,1
: :
03 07
1
x
,1
Figure 14.20: Level curves of f
Figure 14.19
So, f doesn’t have extrema in the interior of the disk.
Now, let’s find the relative extrema of f on the circle x2 + y2 = 1. So we want the extrema of
f (x; y) = x3 , y2 subject to the constraint g(x; y) = x2 + y2 , 1 = 0. Using Lagrange multipliers we get
grad f
=
grad g
and
x2 + y2 = 1;
which gives
3x2
,2 y
x2 + y 2
2x
= 2y
= 1.
=
From the second equation y = 0 or = ,1.
If y = 0, from the third equation we get x2 = 1, which gives the solutions (1; 0), (,1; 0).
If y 6= 0 then = ,1 and from the first equation we get 3x2 = ,2x, hence x = 0 or x = , 23 . If x = 0,
from the third equation we get y2 = 1, so the solutions (0; 1),(0; ,1). If x = , 23 , from the third equation we
p
get y2 = 59 , so the solutions (, 23 ; 35 ), (, 23 ; ,
Evaluating f at these points we get
p5
3 ).
f (1; 0) = 1; f (,1; 0) = f (0; 1) = f (0; ,1) = ,1
14.3 SOLUTIONS
p!
and
f
, 23 ; , 35
=
p!
, 23 ; 35
f
=
, 23
:
27
Therefore the maximum value of f is 1 and the minimum value is ,1.
18. We wish to minimize the objective function
C (x; y; z ) = 20x + 10y + 5z
subject to the budget constraint
Q(x; y; z ) = 20x1=2y1=4 z 2=5 = 1; 200:
Therefore, we solve the equations grad C = grad Q and Q = 1; 200:
20 = 10x,1=2y1=4 z 2=5 or = 2x1=2y,1=4 z ,2=5 ;
10 = 5x1=2y,3=4 z 2=5 ; or = 2x,1=2 y3=4 z ,2=5 ;
5 = 8x1=2y1=4 z ,3=5 ; or = 0:625x,1=2y,1=4 z 3=5 ;
20x1=2 y1=4 z 2=5 = 1; 200:
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
20y1=2 y1=4 (3:2y)2=5
=
1200
y 23:47;
and so
x 23:47
and
z 75:1:
19. We want to minimize
C = f (q1 ; q2) = 2q12 + q1q2 + q22 + 500
subject to the constraint q1 + q2 = 200 or g(q1 ; q2 ) = q1 + q2 , 200 = 0.
Since rf = (4q1 + q2 )~i + (2q2 + q1 )~j and rg = ~i + ~j , rf = rg gives
4q1 + q2 = 2q2 + q1 = :
Solving we get
So
We want
Therefore
4q1 + q2 = 2q2 + q1 :
3q1
q:
= 2
q1 + q2 = 200
q1 + 3q1 = 4q1 = 200:
q1 = 50 units, q2 = 150 units:
231
232
CHAPTER FOURTEEN /SOLUTIONS
20. Constraint is G = P1 x + P2 y , K
Since rQ = rG, we have
=
0.
caxa,1yb = P1 and cbxayb,1 = P2:
caxa,1yb = P1 , or simplifying, ay = P1 . Hence, y = bP1 x.
Dividing the two equations yields
cbxayb,1 P2
aP2
a +bxb P2
bP
1
Substitute into the constraint to obtain P1 x + P2
aP x = P1 a x = K , giving
2
x = (a +aKb)P
y = (a +bKb)P :
and
1
2
We now check that this is indeed the maximization point. Since x; y
0, possible maximization points
K K
aK ; bK ). Since Q = 0 for the first two points and Q is positive for
are (0; ), ( ; 0), and (
P2 P1
(a + b)P1 (a + b)P2
aK ; bK ) gives the maximal value.
the last point, it follows that (
(a + b)P1 (a + b)P2
21.
(a)
The curves are shown in Figure 14.21.
s (income)
1500
III
II
I
1000
(50
500
; 500)
s = 1000 , 10l
20
40
60
80
100
l
(leisure hours)
Figure 14.21
(b) The income equals $10/hour times the number of hours of work:
s = 10(100 , l) = 1000 , 10l:
(c) The graph of this constraint is the straight line in Figure 14.21.
(d) For any given salary, curve III allows for the most leisure time, curve I the least. Similarly, for any
amount of leisure time, any curve III also has the greatest salary, and curve I the least. Thus, any point
on curve III is preferable to any point on curve II, which is preferable to any point on curve I. We prefer
to be on the outermost curve that our constraint allows. We want to choose the point on s = 1000 , 10l
which is on the most preferable curve. Since all the curves are concave up, this occurs at the point where
s = 1000 , 10l is tangent to curve II. So we choose l = 50, s = 500, and work 50 hours a week.
14.3 SOLUTIONS
22.
233
(a) The gradient vectors, rf , point inward around a local maximum. See the two points marked A in
Figure 14.22.
(b) Some of the gradient vectors around a saddle are pointing inward toward the point; some are pointing
outward away from the point. See the point marked B in Figure 14.22.
(c) The critical points on g = 1 are at points where rf is perpendicular to the curve g = 1. There are four
of them, all marked with a dot in Figure 14.22. Imagine the level surfaces of f sketched in everywhere
perpendicular to rf ; the maximum value of f is at the point marked C in Figure 14.22
(d) Again imagine level curves of f . The minimum value of f is at the point marked D.
C
B
A
A
D
Figure 14.22
(e) At C , the maximum on g = 1, the vector rg points outwards (because it points towards g = 2), while
rf points inwards. The Lagrange multiplier, , is defined so that rf = rg, so must be negative.
23.
r
6
-
h
?
Figure 14.23
Let V be the volume and S be the surface area of the container. Then
V
=
r2 h
and
S = 2rh + 2r2
where h is the height and r is the radius as shown in Figure 14.23. We have V
Since
and
=
100 cm3 as our constraint.
rS = (2h + 4r)~i + 2r~j = ((2h + 4r)~i + 2r~j )
rV = 2rh~i + r2~j = (2rh~i + r2~j );
234
CHAPTER FOURTEEN /SOLUTIONS
at the optimum
rS = rV; we have
~
((2h + 4r)i + 2r~j ) = (2rh~i + r2~j );
that is
2h + 4r = 2rh and 2r = r2 ;
We assume
obtain:
hence
= 2r :
r 6= 0 or else we have a very awkward cylinder. Then, plug = 2=r into the first equation to
2h + 4r
=
2h + 4r
=
2
2
4h
h = 2r:
r rh
Finally, solve for r and h using the constraint:
V
Solving for h, we obtain h = 2r
24.
(a)
(b)
r
=
3
r2 h = 100
r2 (2r) = 100
r3 = 50
=
r=
r
3
:
50
50
.
We want to minimize C subject to g = x + y = 39. Solving rC = rg gives
10x + 2y = 2x + 6y = so y = 2x. Solving with x + y = 39 gives x = 13; y = 26; = 182. Therefore C = 4349.
Since = 182, increasing production by 1 will cause costs to increase by approximately $182 (bekr C k = rate of change of C with g). Similarly, decreasing production by 1 will save
cause =
kr gk
2
approximately $182.
25. We want to minimize the function h(x; y) subject to the constraint that
g(x; y) = x2 + y2 = 1;0002 = 1;000;000:
Using the method of Lagrange multipliers, we obtain the following system of equations:
4y
hx = , 1010x;+
= 2x;
000
x + 4y = 2y;
hy = , 410
;000
2
x + y2 = 1;000;000:
Multiplying the first equation by y and the second by x we get
,y(10x + 4y) = ,x(4x + 4y) :
10;000
10;000
14.3 SOLUTIONS
Hence:
2y2 + 3xy , 2x2
235
y , x)(y + 2x) = 0;
and so the climber either moves along the line x = 2y or y = ,2x.
= (2
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
x2 + y2 = 1;000;000, compute the corresponding heights, and then select the minimum point.
If x = 2y, the third equation gives
5y2 = 1;0002;
p
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.
26.
(a) Let c be the cost of producing the product. Then c = 10W
9
3
+
20K
=
3000: At optimum production,
rq = rc:
rq = 2 W , K ~i + 2 W K , ~j , and rc = 10~i + 20~j : Equating we get
1
4
1
4
3
4
9
2
Dividing yields K
= 16
3
4
W, K
1
4
1
4
=
225 and K
10;
3
2
and
W K,
3
4
3
4
=
20:
W , so substituting into c gives
10W
Thus W
=
=
+ 20
1 6
W
40
W
3
=
3000:
=
37:5. Substituting both answers to find gives
=
, 14 (37:5) 41
9
(225)
2
=
10
0:2875:
We also find the optimum quantity produced, q = 6(225) 4 (37:5) 4 = 862:57:
(b) At the optimum values found above, marginal productivity of labor is given by
3
@q 9 ,
@W (225;37:5) = 2 W K
1
4
1
4
@q 3
,
@K (225;37:5) = 2 W K
3
4
225;37:5
(
and marginal productivity of capital is given by
3
4
1
225;37:5
(
=
2:875;
=
5:750:
)
)
The ratio of marginal productivity of labor to that of capital is
@q
@W
@q
@K
=
1
2
=
10
20
=
cost of a unit of L
:
cost of a unit of K
(c) When the budget is increased by one dollar, we substitute the relation K1 = 16 W1 into 10W1 + 20K1 =
3001 which gives 10W1 + 20( 16 W1 ) = 40
3 W1 = 3001: Solving yields W1 = 225:075 and K1 = 37:513;
so q1 = 862:86 = q + 0:29. Thus production has increased by 0:29 ; the Lagrange Multiplier.
236
27.
CHAPTER FOURTEEN /SOLUTIONS
(a)
The problem is to maximize
V
=
1000D0:6N 0:3
subject to the budget constraint in dollars
40000D + 10000N
or (in thousand dollars)
(b) Let B
=
40D + 10N
=
40D + 10N
600000
600
600 (thousand dollars) be the budget constraint. At the optimum
rV = rB;
so
Thus
(c)
@V
@D
@V
@N
@B
@D
@B
=
@N
@V
@D = 4:
@V
=
=
40
=
10:
@N
Therefore, at the optimum point, the rate of increase in the number of visits to 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.
Differentiating and setting rV = rB yields
600D,0:4 N 0:3 = 40
300D0:6 N ,0:7 = 10
Thus, we get
600D,0:4 N 0:3
40
So
=
0:6 ,0:7
= 300D 10N
N = 2D:
To solve for D and N , substitute in the budget constraint:
600 , 40D , 10N
So D
=
10 and N
=
0
600 , 40D , 10 (2D) = 0
=
20.
,0:4)(200:3)
14:67
40
Thus the clinic should hire 10 doctors and 20 nurses. With that staff, the clinic can provide
= 600(10
V
=
1000(100:6)(200:3) 9;779 visits per year.
(d) From part c), the Lagrange multiplier is = 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.)
14.3 SOLUTIONS
237
(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:
MC =
1
14:67 0:068 (thousand dollars)
1
i.e. 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.
28. 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:
Partial derivatives of f and g are
1
2 (x , a)
fx = 2 f (x; y; z ) = f (xx;,y;az ) ;
1
fy = 2 f2(x; (y;y ,z )b) = f (yx;,y;bz ) ;
1
2 (z , c)
fz = 2 f (x; y; z ) = f (zx;,y;cz ) ;
gx = A; gy = B;
and gz
=
C:
Using Lagrange multipliers, we need to solve the equations
grad f
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 = A
f (x; y; z )
y,b
f (x; y; z ) = B
z , c = C
f (x; y; z )
Ax + By + Cz + D = 0:
a
Now x,
A
=
y,b
B
=
z,c
C
=
=
f (x; y; z ) gives
A (y , b) + a;
x= B
z = BC (y , b) + c;
238
CHAPTER FOURTEEN /SOLUTIONS
Substitute into the constraint,
A
C
A B
(y , b) + a + By + C
(y , b) + c + D = 0;
B
A2
C 2 y = A2 b , Aa + C 2 b , Cc , D:
+B +
B
B
B
B
Hence
2
2
+ Cc + D )
y = (A + C A)b2 ,+ BB(2Aa
;
+ C2
+ Bb + Cc + D )
y , b = ,B (Aa
A2 + B 2 + C 2
A (y , b)
x,a= B
,A(Aa + Bb + Cc + D)
=
A2 + B 2 + C 2
C (y , b)
z,c= B
,C (Aa + Bb + Cc + D)
=
A2 + B 2 + C 2
Thus the minimum f (x; y; z ) is
p
(x , a)2 + (y , b)2 + (z , c)2
,A(Aa + Bb + Cc + D) 2 + ,B(Aa + Bb + Cc + D) 2
=
A2 + B 2 + C 2
A2 + B 2 + C 2
,C (Aa + Bb + Cc + D) 2 1=2
f (x; y; z ) =
+
A2 + B 2 + C 2
jAap+ Bb + Cc + Dj :
=
A2 + B 2 + C 2
The geometric meaning is finding the shortest distance from a point (a; b; c) to the plane Ax + By + Cz + D
0.
=
14.3 SOLUTIONS
29.
239
(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 14.24.
f = max
f = min
M
f increases
Figure 14.24
f = max
f = max
f = min
f increases]
f increases
6
Figure 14.25
f = min
Figure 14.26
(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 14.25. 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 14.26.
(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.
30.
(a) The solution to Problem 26 gives = 0:29. We recalculate with a budget of $4000.
The condition that grad q = grad(budget) in Problem 26 gives
so K
=
9 ,1=4 1=4
3 3=4 ,3=4
W
K
= (10) and
W K = (20);
2
2
1
W . Substituting into the budget constraint after replacing the budget of $3000 by $4000 gives
6
10W
Thus, W
=
300 and K
1
40
W
)=
W
6
3
= 50 and q = 1150:098.
+ 20(
=
4000:
240
CHAPTER FOURTEEN /SOLUTIONS
Multiplying the first equation by W and the second by K and adding gives
W ( 92 W ,1=4K 1=4) + K ( 32 W 3=4K ,3=4) = W (10) + K (20):
So
9
2
+
3
2
W 3=4K 1=4 = (10W + 20K )
6W 3=4K 1=4
Thus,
=
(4000)
3=4 1=4
= 6W4000K
=
1150:098
4000
=
0:29
Thus, the value of remains unchanged.
(b) The solution to Problem 27 shows that = 14:67. We solve the problem again with a budget of
$700;000.
The condition that grad V = grad B in Problem 27 gives
600D,0:4 N 0:3 = 40
300D0:6 N ,0:7 = 10
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,0:4N 0:3) + N (300D0:6N ,0:7) = D(40) + N (10);
so
(600 +
Since V
=
300)D0:6 N 0:3 = (400 + 10N )
900D0:6 N 0:3 = (700)
1000D0:6 N 0:3 = 11234:705, we have
0:6 0:3
V ) = 14:44:
= 900D700N = 97 ( 1000
(c)
Thus, the value of has changed with the budget.
We are interested in the marginal increase of production with budget (that is, the value of ) and whether
it is affected by the budget.
Suppose $B is the budget. In part (a) we found
3=4 1=4
= 6W BK
and in part (b) we found
:
:
= 900DB N :
06
03
SOLUTIONS TO REVIEW PROBLEMS FOR CHAPTER FOURTEEN
In part (a), both W and K are proportional to B . Thus, W
241
c B and K = c2 B , so
= 1
3 =4
1 =4
= 6(c1 B ) B(c2B )
3 =4 1 =4
6c1 C2 B 3=4 B 1=4
=
B
3 =4 1 =4
= 6c1 c2 :
So we see is independent of B .
In part (b), both D and N are proportional to B , so D
:
c B and N = c4 B . Thus,
= 3
:
= 900(c3B )B (c4 B )
900c03:6 C40:3B 0:6 B 0:3
=
B
0:6 0:3 1
= 900c3 c4
B 0:1 :
06
03
So we see is not independent of B .
The crucial difference is that the exponents in Problem 26 add to 1, that is 3=4 + 1=4 = 1, whereas
the exponents in Problem 27 do not add to 1, since 0:6 + 0:3 = 0:9.
Thus, the condition that must be satisfied by the Cobb-Donglas production function
Q = cK a Lb
to ensure that the value of is not affected by production is that
a + b = 1:
This is called constant returns to scale.
Solutions for Chapter 14 Review
1. The partial derivatives are
fx = cos x + cos (x + y):
fy = cos y + cos (x + y):
Setting fx
=
0 and fy
=
0 gives
For 0 < x < and 0 < y
cos x = cos y
< , cos x = cos y only if x = y. Then, setting fx = 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 = =3 or x = . For the given domain 0 < x < , 0 < y < , we
only consider the solution when x = =3 then y = x = =3. Therefore, the critical point is ( 3 ; 3 ).