Page 92 - How To Solve Word Problems In Calculus

Step3
2
The given constraint is that the area must be 100 in .

xy = 100
100
y =
x

Now we can represent P as a function of x

100
P(x) = 2x + 2
x
200
= 2x +
x
= 2x + 200x −1

At this point the solution takes a different turn. Unlike Ex-
ample 2, we cannot allow x = 0. Not only would this cause
difficulty for P(x), but more fundamentally it is impossible to
have a rectangle with an area of 100 if one side has length 0.
On the other side of the spectrum, what is the largest
that x might be? A moment’s thought will convince you that
no matter how large you make x, you can always take y suffi-
ciently small so that xy = 100.
Since the domain of P(x) is the interval (0, ∞) the closed
interval method cannot be used here.

Step4
We find the critical value(s)

P (x) = 2 − 200x −2

200
0 = 2 −
x 2
200
= 2
x 2
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