Page 136 - How To Solve Word Problems In Calculus

We represent the dimensions of the page by x and y as shown. The
dimensions of the printed matter are then x − 3 and y − 2. (We
could let x and y be the dimensions of the printed area, but this
leads to a more difficult solution.)
The area to be maximized,

A = (x − 3)(y − 2)
= xy − 2x − 3y + 6

96
2
Since the area of the page is to be 96 in , xy = 96 and y = .
x
Substituting,

96
A(x) = 96 − 2x − 3 + 6
x
= 102 − 2x − 288x −1 0 < x < ∞

Next we find the critical number.
A (x) =−2 + 288x −2

288
0 =−2 +
x 2
288
2 =
x 2
2
2x = 288
2
x = 144
x = 12

96 96
The corresponding value of y is = = 8. The dimensions of
x 12
the page are 12 by 8 inches.
We can use the second derivative test to confirm a relative
maximum (optional).
A (x) =−2 + 288x −2

−576

A (x) =−576x −3 =
x 3

It is clear that A (12) < 0 so a relative maximum is confirmed. Since
x = 12 is the only relative extremum on the interval (0, ∞), it
yields the absolute maximum.
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