Page 125 - How To Solve Word Problems In Calculus

2 2
100 100 100 1 200
x = A = + 50 −
π + 4 π + 4 π + 4 π π + 4
2 2
100 1 50(π + 4) − 200
= +
π + 4 π π + 4
2 2
100 1 50π
= +
π + 4 π π + 4
10,000 1 2500π 2
= + ·
(π + 4) 2 π (π + 4) 2
10,000 + 2500π
=
(π + 4) 2
2500(4 + π)
=
(π + 4) 2
2500
= ≈ 350.06 ← Absolute maximum
π + 4
x = 25 A(25) = 625

Summary: (a) To maximize the combined area, use all the
wire to form the circle. (b) To minimize the area, cut the wire at a
400
distance from one end and form a square. (Remember, the
π + 4
perimeter of the square is 4x.) The remaining piece of wire is used
to form a circle.
5.

(x, y)

5 y

x x

We want to maximize the area of the rectangle. Since its
dimensions are 2x and y,
A = 2xy

The point (x, y) lies on the circle so its coordinates must satisfy the
equation of the circle. Solving for y,
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