Page 93 - How To Solve Word Problems In Calculus

2
2x = 200
The value x =−10 is disregarded
2
x = 100 since it falls outside the domain
of our function (0 < x < ∞).
x = 10

Step5
We apply the second derivative test to our critical value,
x = 10.

P (x) = 2 − 200x −2 (from step 4)
400

P (x) = 400x −3 =
x 3

P (10) > 0

We do not need to know the exact value of P (10). The fact

that it is positive tells us that we have a relative minimum at
x = 10. Since x = 10 is the only relative extremum, it must be
the absolute minimum. Since y = 100/x, y = 10 when x = 10.
The minimum perimeter = 2x + 2y = 40.

Note: Testing the critical number in a word problem (step 5)
may be considered to be an optional step. If you are
convinced that the problem has a solution and you find
only one critical point, then this point must be the so-
lution to the problem. However, intuition is sometimes
misleading, and, from a mathematical perspective, the
test is one way to justify the solution.

The next example illustrates the importance of checking
the endpoints of the interval.

EXAMPLE 4
A piece of wire 24 inches long is to be used to form a square
and/or a rectangle whose length is three times its width. De-
termine their maximum and minimum combined area.

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