Page 132 - How To Solve Word Problems In Calculus

The endpoints of the interval are r = 0 and r = 3. Since
V (0) = V (3) = 0, the maximum volume occurs when r = 2. The
10 20 10
corresponding value of h = 10 − r = 10 − = .
3 3 3
10 40π
2 2 3
The maximum volume is V = πr h = π · 2 · = in .
3 3
11.

x x

y
Since only three sides of the rectangle use fencing, P = 2x + y.
2
Since the enclosed area is to be 1800 ft ,
xy = 1800
1800
y =
x

Substituting into the equation for P, we obtain P as a function of x.
1800
P (x) = 2x + 0 < x < ∞
x
P (x) = 2x + 1800x −1 Clearly x cannot be negative, and x
cannot equal 0, since this would yield

P (x) = 2 − 1800x −2 a rectangle whose area is 0. Further-
more, x can be arbitrarily large, since
1800 it is always possible to find y suffi-
0 = 2 − ciently small to make xy = 1800.
x 2
1800
= 2
x 2
2
2x = 1800
2
x = 900
x = 30
1800 1800
The corresponding value of y = = = 60 and the
x 30
minimum amount of fencing needed is 2x + y = 120 feet.

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