Page 124 - How To Solve Word Problems In Calculus

2
2
The combined area A = x + πr . Since the combined perimeter
of the two figures is equal to the length of the wire,
4x + 2πr = 100

2x + πr = 50
50 − 2x
r =
π
Substituting into the area equation,

2
50 − 2x
2
A(x) = x + π
π
1
2 2
= x + (50 − 2x)
π
The smallest allowable value of x is 0, in which case all the wire will
be used to form the circle. The largest value is x = 25, in which
case all the wire is used for the square. Hence 0 ≤ x ≤ 25.
We take the derivative and compute the critical value.

2

A (x) = 2x + (50 − 2x)(−2)
π
4
= 2x − (50 − 2x)
π
200 8
0 = 2x − + x
π π
0 = 2πx − 200 + 8x ← Multiply by π to eliminate fractions
200 = 2πx + 8x

100 = (π + 4)x
100
x =
π + 4

1
2 2
A(x) = x + (50 − 2x) . Checking the values at the endpoints
π
and critical value,
2500
x = 0 A(0) = ≈ 795.78 ← Absolute maximum
π
111