Page 120 - How To Solve Word Problems In Calculus

The maximum area occurs either at the critical value or at an
endpoint of the interval. Since A(0) = 0 and A(250) = 0, the
maximum area occurs when x = 125. The corresponding value of y
is 500 − 2x = 500 − 250 = 250 ft. The maximum area
A max = xy = 125 × 250 = 31,250 ft 2

2. Let x and y represent the width and length, respectively, of the
enclosure. We will let the interior sections of fence run parallel to
the width. (The problem can also be solved by letting the interior
pieces run parallel to the length.)

y

x x x x

y
The area of the enclosure is still the product of its length and width.

A = xy

The 800 ft of fencing is to be divided into sixpieces, 4 of length x,
and 2 of length y.

4x + 2y = 800
2y = 800 − 4x
y = 400 − 2x

Replacing y by 400 − 2x in the area equation, A = xy, we obtain
the area function
A(x) = x(400 − 2x)
If x > 200, y < 0.
= 400x − 2x 2 0 ≤ x ≤ 200

The critical values of A(x) are determined by solving A (x) = 0.

A (x) = 400 − 4x

0 = 400 − 4x
4x = 400

x = 100 ft
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