Page 140 - How To Solve Word Problems In Calculus

Since A (5) > 0 (exact calculation is unnecessary), r = 5 yields a

relative minimum. Since it is the only relative extremum on (0, ∞),
it gives the absolute minimum as well.
17.

P (x, y)
x x
y

y

Let (x, y) represent the point P in the first quadrant where the
rectangle meets the ellipse. Then the dimensions of the rectangle
are 2x by 2y so its area A = 4xy. Since P lies on the ellipse, its
x 2 y 2
coordinates must satisfy the equation of the ellipse, + = 1.
200 50
We solve for y:

y 2 x 2
= 1 −
50 200
x 2
2
y = 50 −
4
200 − x 2
=
4
√
200 − x 2
y =
2

The area may be expressed as a function of x by substitution into
A = 4xy:
√
200 − x 2
A(x) = 4x
2
√ √
= 2x 200 − x 2 0 ≤ x ≤ 200

Rather than deal with this function directly, it is more convenient to
consider its square. We can call it F (x). The value of x that
maximizes F (x) will maximize A(x).
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