Page 100 - How To Solve Word Problems In Calculus

20 − 2r − πr 1
A(r) = 2r + πr 2
2 2
1
2 2 2
= 20r − 2r − πr + πr
2
1
2 2
= 20r − 2r − πr
2
Step4
1
2 2
A(r) = 20r − 2r − πr
2

A (r) = 20 − 4r − πr
= 20 − (4 + π)r

0 = 20 − (4 + π)r
(4 + π)r = 20

20
r =
4 + π

Step5 (optional)

A (r) =−4 − π. Since A (r) < 0 for all r, it is negative at
20
the critical value. Thus is a relative maximum. Since
4 + π
it is the only relative extremum, the absolute maximum area
occurs at this point.
The maximum area is

2 2
20 20 20 1 20
A = 20 − 2 − π
4 + π 4 + π 4 + π 2 4 + π
400 800 200π
= − −
4 + π (4 + π) 2 (4 + π) 2
400(4 + π) 800 200π
= − −
(4 + π) 2 (4 + π) 2 (4 + π) 2

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