124 D CHAPTER 16
I Let x be the width and y be the height of the page. Then 96 = xy, and (Fig. 16-10) the printed
area A = (x -§)(>- 1). Hence, 0 = xD xy + y, Dxy = -ylx. Now DXA = (x - |)D,y + y - 1 =
(x-l)(-y/x) + y-l=3y/2x-l.
Therefore,
D\A = l(xD f y - y)lx
2 = \(-2ylx
2 )
D A .y4=0, we obtain y = %x, 96 = x(lx), x
2 = 144, x=12, y = 8. Since the second derivative is negative, the unique critical number x - 12 yields an absolute maximum for A.
16.27
A paint manufacturer can produce anywhere from 10 to 30 cubic meters of paint per day. The profit for the day
(in hundreds of dollars) is given by P = (x - 15)
3
/1000 - 3(x - 15) /10 + 300, where x is the volume produced
and sold. What value of x maximizes the profit?
' D,P= T
2 = 100, x-15 = ±10, x = 25 or x = 5. Since
x — 5 is not within the permissible range, the only critical number is x = 25. Using the tabular method, we
find that the maximum profit is achieved when x = 10.
X
P
10
301.375
25
298
30
298.875
16.28
A printed page is to have a total area of 80 in
2 and margins of 1 inch at the top and on each side and of 1.5 inches at
the bottom. What should the dimensions of the page be so that the printed area will be a maximum?
I Let x be the width and y be the height of the page. Then 80 = xy, 0 = xD x y + y, D x y = —ylx. The
area of the printed page A = (x - 2)(y - 2.5), so D X A = (x -2) D,y + y -2.5 = (x -2)(-y/x) + y -
2.5=-y + 2y/x + y-2.5 = 2y/x-2.5. Also, D\A - 2(xD x y - y)/x
2 =2(-y - y)/x
2 = -4y/x
2 <0. Solving
D f A=0, we find y = 1.25*, 80=1.25*
2 , 64 = x
2 , x = 8, y = W. Since the second derivative is negative, this unique critical number yields an absolute maximum for A.
16.29
One side of an open field is bounded by a straight river. Determine how to put a fence around the other sides of a
rectangular plot in order to enclose as great an area as possible with 2000 feet of fence.
I Let x be the length of the side parallel to the river, and let y be the length of each of the other sides. Then
2y + A- = 2000. The area A = xy = X2000 - 2y) = 2000y - 2>>
2
, D y A = 2000 - 4y, and D \,A = -4. Solving D V A=0, we find the critical number y = 500. Since the second derivative is negative, this unique
critical number yields an absolute maximum, x = 2000 - 2(500) = 1000.
16.30
A box will be built with a square base and an open top. Material for the base costs $8 per square foot, while
material for the sides costs $2 per square foot. Find the dimensions of the box of maximum volume that can be
built for $2400.
I Let 5 be the side of the base and h be the height. Then V=s
2 h. We are told that 2400 = 8s
2 + 2(4fo),
so 300 = s
2 + hs, h = 3>00/s-s. Hence, V= s
2 (300/s - s) = 300s - s
3
. Then D S V= 300 -3s
2 , D S
2 V =
-6s. Solving D S V= 0, we find the critical number 5 = 10. Since the second derivative is negative, the
unique critical number yields an absolute maximum, h = ™ — 10 = 20.
16.31
Find the maximum area of any rectangle which may be inscribed in a circle of radius 1.
I Let the center of the circle be the origin. We may assume that the sides of the rectangle are parallel
to the coordinate axes. Let 2x be the length of the horizontal sides and 2y be the length of the vertical
sides. Then x
2 + v
2 = l, so 2x + 2yDv = 0, D v = -xly. The area A = (2x)(2v) = 4xv. So, DA =
Solving D f A=0, we find y — x, 2x
2 = l,
Since the second derivative is negative, the unique critical number vields an absolute
maximum. The maximum area is
16.32
A factory producing a certain type of electronic component has fixed costs of $250 per day and variable costs of
90*, where x is the number of components produced per day. The demand function for these components is
p(x) = 250 - x, and the feasible production levels satisfy 0 & x < 90. Find the level of production for
maximum profit.
Download
from Wow! eBook
I Let x be the width and y be the height of the page. Then 96 = xy, and (Fig. 16-10) the printed
area A = (x -§)(>- 1). Hence, 0 = xD xy + y, Dxy = -ylx. Now DXA = (x - |)D,y + y - 1 =
(x-l)(-y/x) + y-l=3y/2x-l.
Therefore,
D\A = l(xD f y - y)lx
2 = \(-2ylx
2 )
D A .y4=0, we obtain y = %x, 96 = x(lx), x
2 = 144, x=12, y = 8. Since the second derivative is negative, the unique critical number x - 12 yields an absolute maximum for A.
16.27
A paint manufacturer can produce anywhere from 10 to 30 cubic meters of paint per day. The profit for the day
(in hundreds of dollars) is given by P = (x - 15)
3
/1000 - 3(x - 15) /10 + 300, where x is the volume produced
and sold. What value of x maximizes the profit?
' D,P= T
2 = 100, x-15 = ±10, x = 25 or x = 5. Since
x — 5 is not within the permissible range, the only critical number is x = 25. Using the tabular method, we
find that the maximum profit is achieved when x = 10.
X
P
10
301.375
25
298
30
298.875
16.28
A printed page is to have a total area of 80 in
2 and margins of 1 inch at the top and on each side and of 1.5 inches at
the bottom. What should the dimensions of the page be so that the printed area will be a maximum?
I Let x be the width and y be the height of the page. Then 80 = xy, 0 = xD x y + y, D x y = —ylx. The
area of the printed page A = (x - 2)(y - 2.5), so D X A = (x -2) D,y + y -2.5 = (x -2)(-y/x) + y -
2.5=-y + 2y/x + y-2.5 = 2y/x-2.5. Also, D\A - 2(xD x y - y)/x
2 =2(-y - y)/x
2 = -4y/x
2 <0. Solving
D f A=0, we find y = 1.25*, 80=1.25*
2 , 64 = x
2 , x = 8, y = W. Since the second derivative is negative, this unique critical number yields an absolute maximum for A.
16.29
One side of an open field is bounded by a straight river. Determine how to put a fence around the other sides of a
rectangular plot in order to enclose as great an area as possible with 2000 feet of fence.
I Let x be the length of the side parallel to the river, and let y be the length of each of the other sides. Then
2y + A- = 2000. The area A = xy = X2000 - 2y) = 2000y - 2>>
2
, D y A = 2000 - 4y, and D \,A = -4. Solving D V A=0, we find the critical number y = 500. Since the second derivative is negative, this unique
critical number yields an absolute maximum, x = 2000 - 2(500) = 1000.
16.30
A box will be built with a square base and an open top. Material for the base costs $8 per square foot, while
material for the sides costs $2 per square foot. Find the dimensions of the box of maximum volume that can be
built for $2400.
I Let 5 be the side of the base and h be the height. Then V=s
2 h. We are told that 2400 = 8s
2 + 2(4fo),
so 300 = s
2 + hs, h = 3>00/s-s. Hence, V= s
2 (300/s - s) = 300s - s
3
. Then D S V= 300 -3s
2 , D S
2 V =
-6s. Solving D S V= 0, we find the critical number 5 = 10. Since the second derivative is negative, the
unique critical number yields an absolute maximum, h = ™ — 10 = 20.
16.31
Find the maximum area of any rectangle which may be inscribed in a circle of radius 1.
I Let the center of the circle be the origin. We may assume that the sides of the rectangle are parallel
to the coordinate axes. Let 2x be the length of the horizontal sides and 2y be the length of the vertical
sides. Then x
2 + v
2 = l, so 2x + 2yDv = 0, D v = -xly. The area A = (2x)(2v) = 4xv. So, DA =
Solving D f A=0, we find y — x, 2x
2 = l,
Since the second derivative is negative, the unique critical number vields an absolute
maximum. The maximum area is
16.32
A factory producing a certain type of electronic component has fixed costs of $250 per day and variable costs of
90*, where x is the number of components produced per day. The demand function for these components is
p(x) = 250 - x, and the feasible production levels satisfy 0 & x < 90. Find the level of production for
maximum profit.
Download
from Wow! eBook