What's the "best" design for a two-liter bottle? How do you get the best view of the Statue of Liberty?

1. Section 4.6
More Optimization Problems
Math 1a
Introduction to Calculus
March 21, 2008
Announcements
◮ Problem Sessions Sunday, Thursday, 7pm, SC 310 (not next
week)
◮ Office hours Tues, Weds, 2–4pm SC 323 (not next week)
. .
Image: Flickr user glassbeat
. . . . . .

2. Announcements
◮ Problem Sessions Sunday, Thursday, 7pm, SC 310 (not next
week)
◮ Office hours Tues, Weds, 2–4pm SC 323 (not next week)
. . . . . .

3. Outline
Modeling
The Text in the Box
More Examples
Shortest Fence
Norman Windows
Worksheet
Two-liter bottles
The Statue of Liberty
. . . . . .

4. The Modeling Process
. .
Real-World
.
. Mathematical
.
f
.ormulate
Problems Model
t
.est s
. olve
. .
Real-World
. p
. redict Mathematical
.
Predictions Conclusions
. . . . . .

5. Outline
Modeling
The Text in the Box
More Examples
Shortest Fence
Norman Windows
Worksheet
Two-liter bottles
The Statue of Liberty
. . . . . .

6. The Text in the Box
1. Understand the Problem. What is known? What is
unknown? What are the conditions?
2. Draw a diagram.
3. Introduce Notation.
4. Express the “objective function” Q in terms of the other
symbols
5. If Q is a function of more than one “decision variable”, use
the given information to eliminate all but one of them.
6. Find the absolute maximum (or minimum, depending on the
problem) of the function on its domain.
. . . . . .

7. Outline
Modeling
The Text in the Box
More Examples
Shortest Fence
Norman Windows
Worksheet
Two-liter bottles
The Statue of Liberty
. . . . . .

8. Your turn
Example (The shortest fence)
A 216m2 rectangular pea patch is to be enclosed by a fence and
divided into two equal parts by another fence parallel to one of
its sides. What dimensions for the outer rectangle will require the
smallest total length of fence? How much fence will be needed?
Solution
Let the length and width of the pea patch be ℓ and w. The
amount of fence needed is f = 2ℓ + 3w. Since ℓw = A, a
constant, we have
A
f(w) = 2 + 3w.
w
The domain is all positive numbers.
. . . . . .

9. . .
w
.
.
.
ℓ
f = 2ℓ + 3w A = ℓw ≡ 216
. . . . . .

10. Solution (Continued)
So
df 2A
=− 2 +3
dw w
√
2A
which is zero when w = .
3
Since f′′ (w) = 4Aw−3 , which is positive for all positive w, the
critical point is a minimum, in fact the global minimum.
√
2A
So the area is minimized when w = = 12 and
√ 3
A 3A
ℓ= = = 18. The amount of fence needed is
w 2
(√ ) √ √
2A 2A 2A √ √
f =2· +3 = 2 6A = 2 6 · 216 = 72m
3 2 3
. . . . . .

11. Example
A Norman window has the outline of a semicircle on top of a
rectangle. Suppose there is 8 + π feet of wood trim available.
Discuss why a window designer might want to maximize the area
of the window. Find the dimensions of the rectangle and
semicircle that will maximize the area of the window.
.
. . . . . .

12. Example
A Norman window has the outline of a semicircle on top of a
rectangle. Suppose there is 8 + π feet of wood trim available.
Discuss why a window designer might want to maximize the area
of the window. Find the dimensions of the rectangle and
semicircle that will maximize the area of the window.
.
Answer
The dimensions of the rectangular part are 2ft by 4ft.
. . . . . .

13. Solution
We have to maximize A = ℓw + (w/2)2 π subject to the constraint
that 2ℓ + w + π w = p. Solving for ℓ in terms of w gives
ℓ = 1 (p − w − π w)
2
So A = 1 w(p − w − π w) + 1 π w2 . Differentiating gives
2 4
πw 1 1
A′ (w) = + (−1 − π)w + (p − π w − w)
2 2 2
p
which is zero when w = . If p = 8 + 4π , w = 4. It follows
2+π
that ℓ = 2.
. . . . . .

14. Outline
Modeling
The Text in the Box
More Examples
Shortest Fence
Norman Windows
Worksheet
Two-liter bottles
The Statue of Liberty
. . . . . .

15. Worksheet
.
.
Image: Erick Cifuentes
. . . . . .

16. Two-liter bottles
A two-liter soda bottle is
roughly shaped like a
cylinder with a spherical
cap, and is made from a
plastic called polyethylene
terephthalate (PET). Its
volume is fixed at two liters.
What dimensions of the
bottle will minimize the cost
of production?
. . . . . .

17. Solution I
The volume of such a bottle is
V = π r 2 h + 2 π r3 ,
3
which is fixed. Thus
2
V − 3 π r3 V 2
h= 2
= 2 − 3 r.
πr πr
The objective function is the surface area (since the material is all
the same, cost is proportional to materials used)
( )
V − 2 π r2 2V 5 2
A = 2π rh + 3π r2 = 2π r 3
2
+ 3 π r2 = + πr .
πr r 3
. . . . . .

18. Solution II
The domain of this function is (0, ∞). To find the critical points
we need to find
dA 2V 10 −2V + 10 π r3
3
=− 2 + πr = .
dr r 3 r2
dA
The critical points are when = 0, or
dr
10 3
0 = −2V + πr
3
( )1 / 3
3V
=⇒ r = .
5π
. . . . . .

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

20. The Statue of Liberty
The Statue of Liberty stands on top of a pedestal which is on top
of on old fort. The top of the pedestal is 47 m above ground level.
The statue itself measures 46 m from the top of the pedestal to the
tip of the torch.
What distance should one stand away from the statue in order to
maximize the view of the statue? That is, what distance will
maximize the portion of the viewer’s vision taken up by the
statue?
. . . . . .

21. Model I
The angle subtended by the
statue in the viewer’s eye can a
be expressed as
( ) ( ) b
a+b b θ
θ = arctan −arctan .
x x
x
. . . . . .

22. Solution I
Maximizing θ with respect to x is a simple matter of
differentiation:
dθ 1 −(a + b) 1 −b
= ( )2 · − ( )2 · 2
dx a+b x2 b x
1+ x 1+ x
b a+b
= 2
−
x2 + b x2
+ (a + b)2
[ 2 ] [ ]
x + (a + b)2 b − (a + b) x2 + b2
=
(x2 + b2 ) [x2 + (a + b)2 ]
. . . . . .

23. Solution II
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 ),
or √
x= 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, ∞).
. . . . . .

24. Analysis and Discussion
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 of the island. Unfortunately, you’re not
allowed to walk on this part of the lawn.
√
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 two numbers is
always between them and greater than or equal to their average.
. . . . . .