APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 123
16.22
16.23
I Let x be the length of the each side costing $1 per foot, and let y be the length of each side costing
$2 per foot. Then 400 = x.y. The cost C = 4y + 2x = 1600/x + 2x. Hence, D r C= -1600/*
2 + 2 and
D
2 C=1600/*
3 . Solving -1600/*
2 + 2 = 0, *
2 = 800, * = 20V2. Since the second derivative is positive,
this yields a relative minimum, which, by virtue of the uniqueness of the critical number, must be an absolute
minimum. Then the least cost is 1600/20V2 + 40V2 = 40V5 + 40V2 = 80V2, which is approximately $112.
A closed, right cylindrical container is to have a volume of 5000 in
3
. The material for the top and bottom of the
container will cost $2.50 per in
2
, while the material for the rest of the container will cost $4 per in
2
. How should
you choose the height h and the radius r in order to minimize the cost?
I 5000= 77T
2 /t. The lateral surface area is 2-rrrh. Hence, the cost C = 2(2.5Q)irr
2 + 4(2irrh) = 5irr
2 +
87r;7j = 5iTT
2 + 87rr(5000/?rr
2 ) = 57rr
2 +40,000/r. Hence, D r C= Wirr -40,000 /r
2
and D
2 C = WTT +
(80,000/r
3 ). Solving \Qirr- 40,000 /r
2 = 0, we find the unique critical number r = lO^hr. ' Since the
second derivative is positive, this yields a relative minimum, which, by virtue of the uniqueness of the critical
number, is an absolute minimum. The height h = 5000 lirr
2 = 25l3/2~TT.
The sum of the squares of two nonnegative numbers is to be 4.
their cubes is a maximum?
How should they be chosen so that the product of
I Let x and y be the numbers. Then x
2 + y
2 = 4, so Q The product of their cubes P = x
3 y
3 , so D X P = x\3y
2 D x y) + 3*
2
y
3 = x
3 [3y
2 (-x/y)] + 3x
2 y
3 = -3x*y +
3x
2 y
3 = 3*
2 y(—x
2 + y ). Hence, when D X P = Q, either jc = 0 or y = 0 (and x = 2), or y = x
(and then, by x
2 +y
2 = 4, x = V2). Thus, we have three critical numbers x = 0, x = 2, and x = V2.
Using the tabular method, with the endpoints 0 and 2, we find that the maximum value of P is achieved
when x = V2, y = V2.
16.24
16.25
Two nonnegative numbers are such that the first plus the square of the second is 10. Find the numbers if their
sum is as large as possible.
I Let x be the first and y the second number. Then x + y
2 = 10. Their sum S = x + y - 10 - y
2 + y.
Hence, D y S=—2y + l, D
2
y S = -2, so the critical number is y=\. Since the second derivative is
negative, this yields a relative maximum, which, by virtue of the uniqueness of the critical number, is an absolute
maximum. x = 10 — ( i )
2 = T •
Find two nonnegative numbers x and y whose sum is 300 and for which x
2 y is a maximum.
I x. + y = 300. The product P = x
2 y = *
2
(300 -x) = 300x
2 - x". D X P = 600* - 3x
2 , so the critical numbers are x = Q and A: = 200. Clearly, 0<^<300, so using the tabular method, we find that the absolute
maximum is attained when x = 200, y = 100.
16.26
A publisher decides to print the pages of a large book with | -inch margins on the top, bottom, and one side, and a
1-inch margin on the other side (to allow for the binding). The area of the entire page is to be 96 square inches.
Find the dimensions of the page that will maximize the printed area of the page.
Fig. 16-10