APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 119
16.5
A printed page must contain 60 cm
2 of printed material. There are to be margins of 5 cm on either side and
margins of 3 cm on the top and bottom (Fig. 16-3). How long should the printed lines be in order to minimize the
amount of paper used?
Fig. 16-3
16.6
16.7
16.8
16.9
I Let x be the length of the line and let y be the height of the printed material. Then xy = 60. The amount
of paper A = (x + W)(y + 6) = (x + 10)(60/x + 6) = 6(10 + x + 100/x + 10) = 6(20 +x + 100/x). x can be
any positive number. Then D X A = 6(1 - 100/x
2 ) and D
2 /l = 1200/.X
3
. Solving 1 - 100/x
2 = 0, we see
that the only critical number is 10. Since the second derivative is positive, there is a relative minimum at
x = 10, and, since this is the only critical number, there is an absolute minimum at x = 10.
A farmer wishes to fence in a rectangular field of 10,000 ft
2 . The north-south fences will cost $1.50 per foot, while
the east-west fences will cost $6.00 per foot. Find the dimensions of the field that will minimize the cost.
I Let x be the east-west dimension, and let y be the north-south dimension. Then xy = 10,000. The cost
C = 6(2x) + 1.5(2v) = 12x + 3y = Ux + 3( 10,000Ix) = Ux + 30,000/x. x can be any positive number. D X C =
12 -30,000/x
2 . D
2 C = 60,000/x
3
. Solving 12 - 30,000/x
2 = 0, 2500 = x
2 , x = 50. Thus, 50 is the only
critical number. Since the second derivative is positive, there is a relative minimum at x = 50. Since this is the
only critical number, this is an absolute minimum. When x = 50, y = 200.
Find the dimensions of the closed cylindrical can that will have a capacity of k units of volume and will use the
minimum amount of material. Find the ratio of the height h to the radius r of the top and bottom.
I The volume k = irr
2 h. The amount of material M = 2irr
2 + 2irrh. (This is the area of the top and
bottom, plus the lateral area.) So M = 2irr
2 + 2irr(kiirr
2 ) = 2irr
2 + 2klr. Then D r M = 4trr - 2k/r
2 ,
D
2
r M = 477 + 4k/r*. Solving 47rr - 2Jt/r
2 = 0, we find that the only critical number is r = 3/kl2Tr. Since
the second derivative is positive, this yields a relative minimum, which, by the uniqueness of the critical number, is
an absolute minimum. Note that k = irr2h = Trr3(h/r) = ir(kl2it)(hlr). Hence, hlr = 2.
In Problem 16.7, find the ratio hlr that will minimize the amount of material used if the bottom and top of the can
have to be cut from square pieces of metal and the rest of these squares are wasted. Also find the resulting ratio
of height to radius.
I k=Trr
2 h. Now M = 8r
2 + 2-irrh = 8r
2 + 2irr(k/Trr
2 ) = 8r
2 + 2klr. D,M = 16r - 2k/r
2 . D
2 M = 16 +
4fc/r
3 . Solving for the critical number, r
3 = k/8, r = 3/~ki2. As before, this yields an absolute minimum.
Again, k = irr2h = Trr\h/r) = ir(k/8)(h/r). So, h/r = 8/ir.
A thin-walled cone-shaped cup (Fig. 16-4) is to hold 367T in
3 of water when full. What dimensions will minimize
the amount of material needed for the cup?
I Let r be the radius and h be the height. Then the volume 36TT = \irr
2 h. The lateral surface area
A = irrs,
where s is the slant height of the cone.
s
2 = r
2 + h
2
and
h = W8/r
2 .
Hence,
A =
Then,
16.5
A printed page must contain 60 cm
2 of printed material. There are to be margins of 5 cm on either side and
margins of 3 cm on the top and bottom (Fig. 16-3). How long should the printed lines be in order to minimize the
amount of paper used?
Fig. 16-3
16.6
16.7
16.8
16.9
I Let x be the length of the line and let y be the height of the printed material. Then xy = 60. The amount
of paper A = (x + W)(y + 6) = (x + 10)(60/x + 6) = 6(10 + x + 100/x + 10) = 6(20 +x + 100/x). x can be
any positive number. Then D X A = 6(1 - 100/x
2 ) and D
2 /l = 1200/.X
3
. Solving 1 - 100/x
2 = 0, we see
that the only critical number is 10. Since the second derivative is positive, there is a relative minimum at
x = 10, and, since this is the only critical number, there is an absolute minimum at x = 10.
A farmer wishes to fence in a rectangular field of 10,000 ft
2 . The north-south fences will cost $1.50 per foot, while
the east-west fences will cost $6.00 per foot. Find the dimensions of the field that will minimize the cost.
I Let x be the east-west dimension, and let y be the north-south dimension. Then xy = 10,000. The cost
C = 6(2x) + 1.5(2v) = 12x + 3y = Ux + 3( 10,000Ix) = Ux + 30,000/x. x can be any positive number. D X C =
12 -30,000/x
2 . D
2 C = 60,000/x
3
. Solving 12 - 30,000/x
2 = 0, 2500 = x
2 , x = 50. Thus, 50 is the only
critical number. Since the second derivative is positive, there is a relative minimum at x = 50. Since this is the
only critical number, this is an absolute minimum. When x = 50, y = 200.
Find the dimensions of the closed cylindrical can that will have a capacity of k units of volume and will use the
minimum amount of material. Find the ratio of the height h to the radius r of the top and bottom.
I The volume k = irr
2 h. The amount of material M = 2irr
2 + 2irrh. (This is the area of the top and
bottom, plus the lateral area.) So M = 2irr
2 + 2irr(kiirr
2 ) = 2irr
2 + 2klr. Then D r M = 4trr - 2k/r
2 ,
D
2
r M = 477 + 4k/r*. Solving 47rr - 2Jt/r
2 = 0, we find that the only critical number is r = 3/kl2Tr. Since
the second derivative is positive, this yields a relative minimum, which, by the uniqueness of the critical number, is
an absolute minimum. Note that k = irr2h = Trr3(h/r) = ir(kl2it)(hlr). Hence, hlr = 2.
In Problem 16.7, find the ratio hlr that will minimize the amount of material used if the bottom and top of the can
have to be cut from square pieces of metal and the rest of these squares are wasted. Also find the resulting ratio
of height to radius.
I k=Trr
2 h. Now M = 8r
2 + 2-irrh = 8r
2 + 2irr(k/Trr
2 ) = 8r
2 + 2klr. D,M = 16r - 2k/r
2 . D
2 M = 16 +
4fc/r
3 . Solving for the critical number, r
3 = k/8, r = 3/~ki2. As before, this yields an absolute minimum.
Again, k = irr2h = Trr\h/r) = ir(k/8)(h/r). So, h/r = 8/ir.
A thin-walled cone-shaped cup (Fig. 16-4) is to hold 367T in
3 of water when full. What dimensions will minimize
the amount of material needed for the cup?
I Let r be the radius and h be the height. Then the volume 36TT = \irr
2 h. The lateral surface area
A = irrs,
where s is the slant height of the cone.
s
2 = r
2 + h
2
and
h = W8/r
2 .
Hence,
A =
Then,