Page 98 - How To Solve Word Problems In Calculus

Step2
The height of the rectangle is y and the width is 2x. The
rectangle’s area is A = 2xy.

Step3
Since the upper vertices of the rectangle lie on the para-
2
bola, their coordinates must satisfy the equation y = 12 − x .
The area of the rectangle then becomes

2
A(x) = 2x(12 − x )
√
= 24x − 2x 3 0 ≤ x ≤ 12

Step4
A (x) = 24 − 6x 2

0 = 24 − 6x 2
2
6x = 24
2
x = 4 x =−2 lies outside the
domain of the function.
x = 2

Step5
√
A(0) = 0, A(2) = 32, A( 12) = 0
2
If x = 2, y = 12 − x = 8. Since the width of the rectan-
gle is 2x, the dimensions of the rectangle are 4 × 8 and the
maximum area is 32.

EXAMPLE 7

A church window is in the shape of a rectangle surmounted
by a semicircle. If the perimeter of the window is 20 ft, what
is its maximum area?

85