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4-63 QC: OSO/OVY T1: OSO October 25, 2018 17:24 SECTION 4.7 • • Optimization 273
y given in Figure 4.87, the only zero appears to be slightly less than 1. Using x = 1 as
0
′
an initial guess in Newton’s method (applied to f (x) = 0) or using your calculator’s
300 solver, you should get the approximate root x ≈ 0.79728. We now compare
c
function values:
200 f(0) = 29, f(5) = 729 and f(x ) ≈ 24.6.
c
Thus, the minimum distance from (5, 11) to the parabola is approximately
100 √ 24.6 ≈ 4.96 and the closest point on the parabola is located at approximately
(0.79728, 8.364).
x Notice that in both Figures 4.83 and 4.85, the shortest path appears to be per-
1 2 3 4 5
pendicular to the tangent line to the curve at the point where the path intersects the
FIGURE 4.87 curve. We leave it as an exercise to prove that this is, in fact, true. This observation is
′
y = f (x) an important geometric principle that applies to many problems of this type.
REMARK 7.1

At this point you might be tempted to forgo the comparison of function values at
the endpoints and at the critical numbers. After all, in all of the examples we
have seen so far, the desired maximizer or minimizer (i.e., the point at which the
maximum or minimum occurred) was the only critical number in the interval
under consideration. You might just suspect that if there is only one critical
number, it will correspond to the maximizer or minimizer for which you are
searching. Unfortunately, this is not always the case. In 1945, two prominent
aeronautical engineers derived a function to model the range of an aircraft,
intending to maximize the range. They found a critical number of this function
(corresponding to distributing virtually all of the plane’s weight in the wings) and
reasoned that it gave the maximum range. The result was the famous “Flying
Wing” aircraft. Some years later, it was argued that the critical number they
found corresponded to a local minimum of the range function. In the engineers’
defense, they did not have easy, accurate computational power at their fingertips,
as we do today. Remarkably, this design strongly resembles the modern B-2
Stealth bomber. This story came out as controversy brewed over the production
of the B-2. The moral should be crystal clear: check the function values at the
critical numbers and at the endpoints. Do not simply assume (even by virtue of
having only one critical number) that a given critical number corresponds to the
extremum you are seeking.

Next, we consider an optimization problem that cannot be restricted to a closed
interval. We will use the fact that for a continuous function, a single local extremum
must be an absolute extremum. (Think about why this is true.)

EXAMPLE 7.5 Designing a Soda Can Using a Minimum
Amount of Material
r A soda can is to hold 12 fl oz. Find the dimensions that will minimize the amount
of material used in its construction, assuming that the thickness of the material
is uniform (i.e., the thickness of the aluminum is the same everywhere in
h the can).
Solution First, we draw and label a picture of a typical soda can. (See Figure 4.88.)
Copyright © McGraw-Hill Education FIGURE 4.88 r. Assuming uniform thickness of the aluminum, notice that we minimize the (7.2)
Here we are assuming that the can is a right circular cylinder of height h and radius
amount of material by minimizing the surface area of the can. We have

area = area of top + area of bottom + curved surface area
Soda can
2
= 2 r + 2 rh.

293