Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.

1. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
Notes
Sec on 4.5
Op miza on Problems
V63.0121.001: Calculus I
Professor Ma hew Leingang
New York University
April 18, 2011
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Notes
Announcements
Quiz 5 on Sec ons
4.1–4.4 April 28/29
Final Exam Thursday May
12, 2:00–3:50pm
I am teaching Calc II MW
2:00pm and Calc III TR
2:00pm both Fall ’11 and
Spring ’12
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Notes
Objectives
Given a problem
requiring op miza on,
iden fy the objec ve
func ons, variables, and
constraints.
Solve op miza on
problems with calculus.
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2. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
Notes
Leading by Example
Example
What is the rectangle of fixed perimeter with maximum area?
Solu on
Draw a rectangle.
w
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ℓ
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Notes
Solution
Solu on (Con nued)
Let its length be ℓ and its width be w. The objec ve func on is
area A = ℓw.
This is a func on of two variables, not one. But the perimeter is
fixed.
p − 2w
Since p = 2ℓ + 2w, we have ℓ = , so
2
p − 2w 1 1
A = ℓw = · w = (p − 2w)(w) = pw − w2
2 2 2
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Notes
Solution
Solu on (Con nued)
Now we have A as a func on of w alone (p is constant).
The natural domain of this func on is [0, p/2] (we want to
make sure A(w) ≥ 0).
1
We use the Closed Interval Method for A(w) = pw − w2 on
2
[0, p/2].
At the endpoints, A(0) = A(p/2) = 0.
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3. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
Notes
Solution
Solu on (Con nued)
dA 1
To find the cri cal points, we find = p − 2w.
dw 2
1 p
The cri cal points are when p − 2w − 0, or w = .
2 4
Since this is the only cri cal point, it must be the maximum. In
p
this case ℓ = as well.
4
We have a square! The maximal area is A(p/4) = p2 /16.
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Notes
Outline
The Text in the Box
More Examples
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Notes
Strategies for Problem Solving
1. Understand the problem
2. Devise a plan
3. Carry out the plan
4. Review and extend
György Pólya
(Hungarian, 1887–1985)
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4. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
Notes
The Text in the Box
1. Understand the Problem. What is known? What is unknown?
What are the condi ons?
2. Draw a diagram.
3. Introduce Nota on.
4. Express the “objec ve func on” Q in terms of the other
symbols
5. If Q is a func on of more than one “decision variable”, use the
given informa on to eliminate all but one of them.
6. Find the absolute maximum (or minimum, depending on the
problem) of the func on on its domain.
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The Closed Interval Method Notes
See Section 4.1
The Closed Interval Method
To find the extreme values of a func on f on [a, b], we need to:
Evaluate f at the endpoints a and b
Evaluate f at the cri cal points x where either f′ (x) = 0 or f is
not differen able at x.
The points with the largest func on value are the global
maximum points
The points with the smallest/most nega ve func on value are
the global minimum points.
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The First Derivative Test Notes
See Section 4.3
Theorem (The First Deriva ve Test)
Let f be con nuous on (a, b) and c a cri cal point of f in (a, b).
If f′ changes from nega ve to posi ve at c, then c is a local
minimum.
If f′ changes from posi ve to nega ve at c, then c is a local
maximum.
If f′ does not change sign at c, then c is not a local extremum.
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5. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
The First Derivative Test Notes
See Section 4.3
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).
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Recall: The Second Derivative Test Notes
See Section 4.3
Theorem (The Second Deriva ve Test)
Let f, f′ , and f′′ be con nuous 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 deriva ve test is inconclusive (this does not
mean c is neither; we just don’t know yet).
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Recall: The Second Derivative Test Notes
See Section 4.3
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 f
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6. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
Notes
Which to use when?
CIM 1DT 2DT
Pro
– no need for – works on – works on
inequali es non-closed, non-closed,
– gets global extrema non-bounded non-bounded
automa cally intervals intervals
– only one deriva ve – no need for
inequali es
Con
– only for closed – Uses inequali es – More deriva ves
bounded intervals – More work at – less conclusive than
boundary than CIM 1DT
– more work at
boundary than CIM
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Notes
Which to use when?
If domain is closed and bounded, use CIM.
If domain is not closed or not bounded, use 2DT if you like to
take deriva ves, or 1DT if you like to compare signs.
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Notes
Outline
The Text in the Box
More Examples
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7. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
Notes
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
Objec ve: maximize area
Constraint: fixed fence length
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Notes
Solution
Solu on
1. Everybody understand?
2. Draw a diagram.
3. Length and width are ℓ and w. Length of wire used is p.
4. Q = area = ℓw.
5. Since p = ℓ + 2w, we have ℓ = p − 2w and so
Q(w) = (p − 2w)(w) = pw − 2w2
The domain of Q is [0, p/2]
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Notes
Solution
Solu on
dQ p
6. = p − 4w, which is zero when w = .
dw 4
7. Q(0) = Q(p/2) = 0, but
(p) p p2 p2
Q =p· −2· = = 80, 000m2
4 4 16 8
so the cri cal point is the absolute maximum.
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8. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
Notes
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?
ℓ
w
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Notes
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?
Solu on
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Notes
Diagram
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9. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
Notes
Solution (Continued)
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Notes
Try this one
Example
An adver sement consists of a rectangular printed region plus 1 in
margins on the sides and 1.5 in margins on the top and bo om. If
the total area of the adver sement is to be 120 in2 , what
dimensions should the adver sement be to maximize the area of
the printed region?
Answer
√ √
The op mal paper dimensions are 4 5 in by 6 5 in.
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Notes
Solution
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10. . V63.0121.001: Calculus I
. Sec on 4.5: Op miza on Problems
. April 18, 2011
Notes
Solution (Concluded)
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Notes
Summary
Remember the checklist
Ask yourself: what is the
objec ve?
Remember your
geometry:
similar triangles
right triangles
trigonometric func ons
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Notes
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