Solutions to Worksheet for Lesson 24 (Section 4.6)

Optimization II

Math 1a

November 21, 2007

1. A two-liter soda bottle is roughly shaped like a cylinder with a spherical cap, and is madefrom a plastic called polyethylene terephthalate (PET). Its volume is fixed at two liters. Whatdimensions of the bottle will minimize the cost of production?

Figure 1: A two-liter PET bottle

Solution. The volume of such a bottle is

V = πr2h+ 23πr

3,

which is fixed. Thus

h =V − 2

3πr3

πr2=

V

πr2− 2

3r.

Figure 2: Liberty Island including the Statue

The objective function is the surface area (since the material is all the same, cost is propor-tional to materials used)

A = 2πrh+ 3πr2 = 2πr(V − 2

3πr2

πr2

)+ 3πr2 =

2Vr

+53πr2.

The domain of this function is (0,∞). To find the critical points we need to find

dA

dr= −2V

r2+

103πr =

−2V + 103 πr

3

r2.

The critical points are whendA

dr= 0, or

0 = −2V +103πr3

=⇒ r =(

3V5π

)1/3

.

Substituting into our expression for h tells us (after a lot of algebra) that h = r. That’sdefinitely a much squatter bottle that we see in the stores. So it’s not material costs thatthey’re minimizing (unless our shape is too off).

2. The Statue of Liberty stands on top of a pedestal which is on top of on old fort. The top ofthe pedestal is 47 meters above ground level. The statue itself measures 46 meters from thetop of the pedestal to the tip of the torch. See Figure 2.

What distance should one stand away from the statue in order to maximize the view of thestatue? That is, what distance will maximize the portion of the viewer’s vision taken up bythe statue?

Solution. Setting up the Model

There is a little ambiguity in the wording, but the situation makes sense: If you stand at thebase of the statue and look straight up, you won’t see much. If you stand far away (say in

a

bθ

x

Figure 3: The statue after introducing variables

Battery Park at the lower tip of Manhattan Island), you also won’t see the statue very well.Somewhere in between the statue will appear as big as possible.

What do we mean by “appear” and “field of vision?” The top and bottom of the statue makean angle with the viewer’s eye at the vertex, and the measure of this angle determines how“big” it looks to the viewer. This is the angle we want to maximize, and it will depend onhow far we stand away from the statue. See Figure 3

We’ve labeled the important constants with letters rather than numbers: a for the height ofthe statue and b for the height of the pedestal. This tip makes algebra messy but arithmeticeasy—wait until the very end to substitute in the numerical values. Algebra is much easierto double-check than arithmetic, anyway. x is the distance the viewer stands from the baseof the statue.

The angle subtended by the statue in the viewer’s eye can be expressed as

θ = arctan(a+ b

x

)− arctan

(b

x

).

Solution

Maximizing θ with respect to x is a simple matter of differentiation:

dθ

dx=

1

1 +(

a+bx

)2 · −(a+ b)x2

− 1

1 +(

bx

)2 · −bx2

=b

x2 + b2− a+ b

x2 + (a+ b)2

=

[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

](x2 + b2) [x2 + (a+ b)2]

This derivative is zero if and only if the numerator is zero, so we seek x such that

0 =[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

]= a(ab+ b2 − x2),

orx =

√b(a+ b).

Using the first derivative test, we see that dθ/dx > 0 if 0 < x <√b(a+ b) and dθ/dx < 0 if

x >√b(a+ b). So this is definitely the absolute maximum on (0,∞).

If we substitute in the numerical dimensions given, we have

x =√

(46)(93) ≈ 66.1 meters

This distance would put you pretty close to the front of the old fort which lies at the base ofthe island. Unfortunately, you’re not allowed to walk on this part of the lawn.

Discussion

The length√b(a+ b) is the geometric mean of the two distances measure from the ground—

to the top of the pedestal (a) and the top of the statue (a+ b). The geometric mean is of twonumbers is always between them and greater than or equal to their average.