Page 154 - How To Solve Word Problems In Calculus

Solution

F

A θ
B
E 8
θ 4
H
D
C

G

Let θ = FAB as shown in the figure. By geometry, EDA = θ
as well. The distances |EA|= 4 sin θ, |AF|= 8 cos θ, |ED|=
4 cos θ, and |DG|=|FB|= 8 sin θ. The area of the circum-
scribed rectangle is the product (|ED|+|DG|)(|EA|+|AF|).
Hence

A(θ) = (4 cos θ + 8 sin θ)(4 sin θ + 8 cos θ)
2
2
= 32 cos θ + 32 sin θ + 80 sin θ cos θ
π
= 32 + 40 sin 2θ 0 ≤ θ ≤ sin θ + cos θ = 1
2
2
2
sin 2θ = 2 sin θ cos θ

A (θ) = 80 cos 2θ
0 = 80 cos 2θ
0 = cos 2θ
π
2θ =
2
π
θ =
4

Observe that A(θ) is a continuous function. Since A(0) =
A(π/2) = 32 and A(π/4) = 72, the maximum area of 72 in 2
occurs when θ = π/4.

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