Page 117 - How To Solve Word Problems In Calculus

25
2
h =
3
5
h = √
3

50
The corresponding value of r is inches
3
√
2
(r = 25 − h ) and the maximum volume is

π 2 π 50 5
r h = √
3 3 3 3
250π
= √ in 3
9 3
Step5 (Optional)

π

V (h) = (−6h) =−2πh
3
5 10π

V √ =− √ < 0
3 3
By the second derivative test V has a relative maximum at
√
h = 5/ 3. Since this is the only relative extremum, it is the
location of the absolute maximum.

Supplementary Problems

1. An open field is bounded by a lake with a straight shoreline. A
rectangular enclosure is to be constructed using 500 ft of fencing
along three sides and the lake as a natural boundary on the fourth
side. What dimensions will maximize the enclosed area? What is
the maximum area?
2. Ryan has 800 ft of fencing. He wishes to form a rectangular
enclosure and then divide it into three sections by running two
lengths of fence parallel to one side. What should the dimensions of
the enclosure be in order to maximize the enclosed area?
3. 20 meters of fencing are to be laid out in the shape of a right
triangle. What should its dimensions be in order to maximize the
enclosed area?

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