1) The document is notes from a Calculus I class covering optimization problems. 2) It includes examples of maximizing the area of a rectangle with a fixed perimeter and maximizing the area that can be enclosed by a fence of fixed length. 3) The document reviews strategies for solving optimization problems, including identifying objectives and constraints, drawing diagrams, introducing variables, and using calculus techniques like finding critical points and applying the first and second derivative tests.

1. Section 4.4
Optimization Problems
V63.0121.021, Calculus I
New York University
November 23, 2010
Announcements
Turn in HW anytime between now and November 24, 2pm
No Thursday recitation this week
Quiz 4 on §§4.1–4.4 next week in recitation
Announcements
Turn in HW anytime
between now and November
24, 2pm
No Thursday recitation this
week
Quiz 4 on §§4.1–4.4 next
week in recitation
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 2 / 31
Objectives
Given a problem requiring
optimization, identify the
objective functions,
variables, and constraints.
Solve optimization problems
with calculus.
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 3 / 31
Notes
Notes
Notes
1
Section 4.4 : Optimization ProblemsV63.0121.021, Calculus I November 23, 2010

2. Outline
Leading by Example
The Text in the Box
More Examples
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 4 / 31
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solution
Draw a rectangle.
w
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 5 / 31
Solution Continued
Let its length be and its width be w. The objective function is area
A = w.
This is a function of two variables, not one. But the perimeter is fixed.
Since p = 2 + 2w, we have =
p − 2w
2
, so
A = w =
p − 2w
2
· w =
1
2
(p − 2w)(w) =
1
2
pw − w2
Now we have A as a function of w alone (p is constant).
The natural domain of this function is [0, p/2] (we want to make sure
A(w) ≥ 0).
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 6 / 31
Notes
Notes
Notes
2
Section 4.4 : Optimization ProblemsV63.0121.021, Calculus I November 23, 2010

3. Solution Concluded
We use the Closed Interval Method for A(w) =
1
2
pw − w2
on [0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
To find the critical points, we find
dA
dw
=
1
2
p − 2w.
The critical points are when
0 =
1
2
p − 2w =⇒ w =
p
4
Since this is the only critical point, it must be the maximum. In this
case =
p
4
as well.
We have a square! The maximal area is A(p/4) = p2
/16.
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 7 / 31
Outline
Leading by Example
The Text in the Box
More Examples
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 8 / 31
Strategies for Problem Solving
1. Understand the problem
2. Devise a plan
3. Carry out the plan
4. Review and extend
Gy¨orgy P´olya
(Hungarian, 1887–1985)
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 9 / 31
Notes
Notes
Notes
3
Section 4.4 : Optimization ProblemsV63.0121.021, Calculus I November 23, 2010

4. 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.
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 10 / 31
Recall: The Closed Interval Method
See Section 4.1
The Closed Interval Method
To find the extreme values of a function f on [a, b], we need to:
Evaluate f at the endpoints a and b
Evaluate f at the critical points x where either f (x) = 0 or f is not
differentiable at x.
The points with the largest function value are the global maximum
points
The points with the smallest/most negative function value are the
global minimum points.
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 12 / 31
Recall: The First Derivative Test
See Section 4.3
Theorem (The First Derivative Test)
Let f be continuous on (a, b) and c a critical point of f in (a, b).
If f changes from negative to positive at c, then c is a local
minimum.
If f changes from positive to negative at c, then c is a local
maximum.
If f does not change sign at c, then c is not a local extremum.
Corollary
If f < 0 for all x < c and f (x) > 0 for all x > c, then c is the global
minimum of f on (a, b).
If f < 0 for all x > c and f (x) > 0 for all x < c, then c is the global
maximum of f on (a, b).
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 13 / 31
Notes
Notes
Notes
4
Section 4.4 : Optimization ProblemsV63.0121.021, Calculus I November 23, 2010

5. Recall: The Second Derivative Test
See Section 4.3
Theorem (The Second Derivative Test)
Let f , f , and f be continuous on [a, b]. Let c be in (a, b) with
f (c) = 0.
If f (c) < 0, then f (c) is a local maximum.
If f (c) > 0, then f (c) is a local minimum.
Warning
If f (c) = 0, the second derivative test is inconclusive (this does not mean
c is neither; we just don’t know yet).
Corollary
If f (c) = 0 and f (x) > 0 for all x, then c is the global minimum of f
If f (c) = 0 and f (x) < 0 for all x, then c is the global maximum of
fV63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 14 / 31
Which to use when?
CIM 1DT 2DT
Pro – no need for
inequalities
– gets global extrema
automatically
– works on
non-closed,
non-bounded
intervals
– only one derivative
– works on
non-closed,
non-bounded
intervals
– no need for
inequalities
Con – only for closed
bounded intervals
– Uses inequalities
– More work at
boundary than CIM
– More derivatives
– less conclusive than
1DT
– more work at
boundary than CIM
Use CIM if it applies: the domain is a closed, bounded interval
If domain is not closed or not bounded, use 2DT if you like to take
derivatives, or 1DT if you like to compare signs.
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 15 / 31
Outline
Leading by Example
The Text in the Box
More Examples
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 16 / 31
Notes
Notes
Notes
5
Section 4.4 : Optimization ProblemsV63.0121.021, Calculus I November 23, 2010

6. Another Example
Example (The Best Fencing Plan)
A rectangular plot of farmland will be bounded on one side by a river and
on the other three sides by a single-strand electric fence. With 800m of
wire at your disposal, what is the largest area you can enclose, and what
are its dimensions?
Known: amount of fence used
Unknown: area enclosed
Objective: maximize area
Constraint: fixed fence length
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 17 / 31
Solution
1. Everybody understand?
2. Draw a diagram.
3. Length and width are and w. Length of wire used is p.
4. Q = area = w.
5. Since p = + 2w, we have = p − 2w and so
Q(w) = (p − 2w)(w) = pw − 2w2
The domain of Q is [0, p/2]
6.
dQ
dw
= p − 4w, which is zero when w =
p
4
. Q(0) = Q(p/2) = 0, but
Q
p
4
= p ·
p
4
− 2 ·
p2
16
=
p2
8
= 80, 000m2
so the critical point is the absolute maximum.
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 18 / 31
Diagram
A rectangular plot of farmland will be bounded on one side by a river and
on the other three sides by a single-strand electric fence. With 800 m of
wire at your disposal, what is the largest area you can enclose, and what
are its dimensions?
w
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 21 / 31
Notes
Notes
Notes
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Section 4.4 : Optimization ProblemsV63.0121.021, Calculus I November 23, 2010

7. 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
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 25 / 31
Diagram
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 26 / 31
Solution (Continued)
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 27 / 31
Notes
Notes
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Section 4.4 : Optimization ProblemsV63.0121.021, Calculus I November 23, 2010

8. Try this one
Example
An advertisement consists of a rectangular printed region plus 1 in margins
on the sides and 1.5 in margins on the top and bottom. If the total area of
the advertisement is to be 120 in2
, what dimensions should the
advertisement be to maximize the area of the printed region?
Answer
The optimal paper dimensions are 4
√
5 in by 6
√
5 in.
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 28 / 31
Solution
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 29 / 31
Solution (Concluded)
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 30 / 31
Notes
Notes
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Section 4.4 : Optimization ProblemsV63.0121.021, Calculus I November 23, 2010

9. Summary
Remember the checklist
Ask yourself: what is the
objective?
Remember your geometry:
similar triangles
right triangles
trigonometric functions
V63.0121.021, Calculus I (NYU) Section 4.4 Optimization Problems November 23, 2010 31 / 31
Notes
Notes
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Section 4.4 : Optimization ProblemsV63.0121.021, Calculus I November 23, 2010