Page 104 - Calculus Demystified

CHAPTER 3 Applications of the Derivative
Thus P is the quantity that we wish to maximize. Calculating the derivative, 91
we find that

P (x) = 60 − 2x.

Thus the only critical point for the problem is x = 30. Since P (x) ≡−2, we
find that x = 30 is a local maximum. Since the graph of P is a downward-
opening parabola, we can in fact be sure that x = 30 is a global maximum.
We conclude that the two numbers that add to 60 and maximize the product
are 30 and 30.

You Try It: A rectangular box is to be placed in the first quadrant {(x, y) : x ≥
0,y ≥ 0} in such a way that one side lies on the positive x-axis and one side lies
on the positive y-axis. The box is to lie below the line y =−2x + 5. Give the
dimensions of such a box having greatest possible area.

3.3 RelatedRates

If a tree is growing in a forest, then both its height and its radius will be increasing.
These two growths will depend in turn on (i) the amount of sunlight that hits the
tree, (ii) the amount of nutrients in the soil, (iii) the proximity of other trees. We
may wish to study the relationship among these various parameters. For example,
if we know that the amount of sunlight and nutrients are increasing at a certain rate
then we may wish to know how that affects the rate of change of the radius. This
consideration gives rise to related rates problems.
EXAMPLE 3.11
A toy balloon is in the shape of a sphere. It is being inflated at the rate
3
of 20 in. /min. At the moment that the sphere has volume 64 cubic inches,
what isthe rate of change of the radius?
SOLUTION
We know that volume and radius of a sphere are related by the formula
4π 3
V = r . (∗)
3
The free variable in this problem is time, so we differentiate equation (∗) with
respect to time t. It is important that we keep the chain rule in mind as we do
1
so. The result is
dV 4π 2 dr
= · 3r · . (∗∗)
dt 3 dt

1 The point is that we are not differentiating with respect to r.