Page 119 - How To Solve Word Problems In Calculus

x 2 y 2
17. A rectangle is to be inscribed in the ellipse + = 1.
200 50
Determine its maximum possible area.

Solutions to Supplementary Problems

1. The dimensions of the enclosure are labeled x and y as shown.

x x

y
Even though fencing is used on only three sides, the area is still the
product of the length and width of the enclosure.

A = xy

The constraint, however, involves only three sides of the rectangle.

2x + y = 500

y = 500 − 2x

Substituting this expression into the area equation, we obtain

A(x) = x(500 − 2x)

= 500x − 2x 2

Since x cannot be negative, x ≥ 0. Since y cannot be negative,
x ≤ 250. (Remember, y = 500 − 2x.) The domain of A(x)isthe
interval 0 ≤ x ≤ 250.
Next we determine the critical values of the function A(x).

A (x) = 500 − 4x
0 = 500 − 4x

4x = 500

x = 125 ft
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