Lesson 22: Optimization II (Section 021 slides)

Section 4.5
Optimization II
V63.0121.021, Calculus I
New York University
class supplement
Announcements
Quiz 5 on §§4.1–4.4 next week in recitation
Happy Thanksgiving!
. . . . . .

. . . . . .
Announcements
Quiz 5 on §§4.1–4.4 next
week in recitation
Happy Thanksgiving!
V63.0121.021, Calculus I (NYU) Section 4.5 Optimization II class supplement 2 / 25

. . . . . .
Objectives
Given a problem requiring
optimization, identify the
objective functions,
variables, and constraints.
Solve optimization
problems with calculus.

. . . . . .
Outline
Recall
More examples
Addition
Distance
Triangles
Economics
The Statue of Liberty

. . . . . .
Checklist for optimization problems
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.

. . . . . .
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.

. . . . . .
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.

. . . . . .
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).

. . . . . .
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.

. . . . . .
See Section 4.3
Let f, f′
, and f′′
(c) = 0.
If f′′
If f′′
Warning
If f′′
(c) = 0, the second derivative test is inconclusive (this does not
mean c is neither; we just don’t know yet).

. . . . . .
See Section 4.3
Let f, f′
, and f′′
(c) = 0.
If f′′
If f′′
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 f

. . . . . .
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

. . . . . .
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.

. . . . . .
Outline
Recall
More examples
Addition
Distance
Triangles
Economics

. . . . . .
Addition with a constraint
Example
Find two positive numbers x and y with xy = 16 and x + y as small as
possible.

. . . . . .
Example
possible.
Solution
Objective: minimize S = x + y subject to the constraint that
xy = 16

. . . . . .
Example
possible.
Solution
xy = 16
Eliminate y: y = 16/x so S = x + 16/x. The domain of
consideration is (0, ∞).

. . . . . .
Example
possible.
Solution
xy = 16
Find the critical points: S′
(x) = 1 − 16/x2
, which is 0 when x = 4.

. . . . . .
Example
possible.
Solution
xy = 16
(x) = 1 − 16/x2
Classify the critical points: S′′
(x) = 32/x3
, which is always
positive. So the graph is always concave up, 4 is a local min, and
therefore the global min.

. . . . . .
Example
possible.
Solution
xy = 16
(x) = 1 − 16/x2
Classify the critical points: S′′
(x) = 32/x3
, which is always
positive. So the graph is always concave up, 4 is a local min, and
therefore the global min.
So the numbers are x = y = 4, Smin = 8.

. . . . . .
Distance
Example
Find the point P on the parabola y = x2
closest to the point (3, 0).

. . . . . .
Distance
Example
Solution
.. x.
y
..
(x, x2
)
..
3

. . . . . .
Distance
Example
Solution
The distance between (x, x2
)
and (3, 0) is given by
f(x) =
√
(x − 3)2 + (x2 − 0)2
.. x.
y
..
(x, x2
)
..
3

. . . . . .
Distance
Example
Solution
)
f(x) =
√
(x − 3)2 + (x2 − 0)2
We may instead minimize the
square of f:
g(x) = f(x)2
= (x − 3)2
+ x4
.. x.
y
..
(x, x2
)
..
3

. . . . . .
Distance
Example
Solution
)
f(x) =
√
(x − 3)2 + (x2 − 0)2
We may instead minimize the
square of f:
g(x) = f(x)2
= (x − 3)2
+ x4
The domain is (−∞, ∞).
.. x.
y
..
(x, x2
)
..
3

. . . . . .
Distance problem
minimization step
We want to find the global minimum of g(x) = (x − 3)2
+ x4
.

. . . . . .
Distance problem
minimization step
+ x4
.
g′
(x) = 2(x − 3) + 4x3
= 4x3
+ 2x − 6 = 2(2x3
+ x − 3)

. . . . . .
Distance problem
minimization step
+ x4
.
g′
(x) = 2(x − 3) + 4x3
= 4x3
+ 2x − 6 = 2(2x3
+ x − 3)
If a polynomial has integer roots, they are factors of the constant
term (Euler)

. . . . . .
Distance problem
minimization step
+ x4
.
g′
(x) = 2(x − 3) + 4x3
= 4x3
+ 2x − 6 = 2(2x3
+ x − 3)
term (Euler)
1 is a root, so 2x3
+ x − 3 is divisible by x − 1:
f′
(x) = 2(2x3
+ x − 3) = 2(x − 1)(2x2
+ 2x + 3)
The quadratic has no real roots (the discriminant b2
− 4ac < 0)

. . . . . .
Distance problem
minimization step
+ x4
.
g′
(x) = 2(x − 3) + 4x3
= 4x3
+ 2x − 6 = 2(2x3
+ x − 3)
term (Euler)
1 is a root, so 2x3
f′
(x) = 2(2x3
+ x − 3) = 2(x − 1)(2x2
+ 2x + 3)
− 4ac < 0)
We see f′
(1) = 0, f′
(x) < 0 if x < 1, and f′
(x) > 0 if x > 1. So 1 is
the global minimum.

. . . . . .
Distance problem
minimization step
+ x4
.
g′
(x) = 2(x − 3) + 4x3
= 4x3
+ 2x − 6 = 2(2x3
+ x − 3)
term (Euler)
1 is a root, so 2x3
f′
(x) = 2(2x3
+ x − 3) = 2(x − 1)(2x2
+ 2x + 3)
− 4ac < 0)
We see f′
(1) = 0, f′
(x) < 0 if x < 1, and f′
(x) > 0 if x > 1. So 1 is
the global minimum.
The point on the parabola closest to (3, 0) is (1, 1). The minimum
distance is
√
5.

. . . . . .
Remark
We’ve used each of the methods (CIM, 1DT, 2DT) so far.
Notice how we argued that the critical points were absolute
extremes even though 1DT and 2DT only tell you relative/local
extremes.

. . . . . .
A problem with a triangle
..
Example
Find the rectangle of maximal area inscribed in a 3-4-5 right triangle with two
sides on legs of the triangle and one vertex on the hypotenuse.
Solution
..
3
.
4
.
5

. . . . . .
..
Example
Solution
Let the dimensions of the
rectangle be x and y.
..
3
.
4
.
5
.
y
.
x

. . . . . .
..
Example
Solution
Similar triangles give
y
3 − x
=
4
3
=⇒ 3y = 4(3 − x)
..
3
.
4
.
5
.
y
.
x

. . . . . .
..
Example
Solution
Similar triangles give
y
3 − x
=
4
3
=⇒ 3y = 4(3 − x)
So y = 4 −
4
3
x and
A(x) = x
(
4 −
4
3
x
)
= 4x−
4
3
x2 ..
3
.
4
.
5
.
y
.
x

. . . . . .
Triangle Problem
maximization step
We want to find the absolute maximum of A(x) = 4x −
4
3
x2
on the
interval [0, 3].

. . . . . .
Triangle Problem
maximization step
4
3
x2
on the
interval [0, 3].
A(0) = A(3) = 0

. . . . . .
Triangle Problem
maximization step
4
3
x2
on the
interval [0, 3].
A(0) = A(3) = 0
A′
(x) = 4 −
8
3
x, which is zero when x =
12
8
= 1.5.

. . . . . .
Triangle Problem
maximization step
4
3
x2
on the
interval [0, 3].
A(0) = A(3) = 0
A′
(x) = 4 −
8
3
12
8
= 1.5.
Since A(1.5) = 3, this is the absolute maximum.

. . . . . .
Triangle Problem
maximization step
4
3
x2
on the
interval [0, 3].
A(0) = A(3) = 0
A′
(x) = 4 −
8
3
12
8
= 1.5.
Since A(1.5) = 3, this is the absolute maximum.
So the dimensions of the rectangle of maximal area are 1.5 × 2.

. . . . . .
An Economics problem
Example
Let r be the monthly rent per unit in an apartment building with 100
units. A survey reveals that all units can be rented when r = 900 and
that one unit becomes vacant with each 10 increase in rent. Suppose
the average monthly maintenance costs per occupied unit is
$100/month. What rent should be charged to maximize profit?

. . . . . .
An Economics problem
Example
Let r be the monthly rent per unit in an apartment building with 100
units. A survey reveals that all units can be rented when r = 900 and
that one unit becomes vacant with each 10 increase in rent. Suppose
the average monthly maintenance costs per occupied unit is
$100/month. What rent should be charged to maximize profit?
Solution
Let n be the number of units rented, depending on price (the
demand function).
We have n(900) = 100 and
∆n
∆r
= −
1
10
. So
n − 100 = −
1
10
(r − 900) =⇒ n(r) = −
1
10
r + 190

. . . . . .
Economics Problem
Finishing the model and maximizing
The profit per unit rented is r − 100, so
P(r) = (r − 100)n(r) = (r − 100)
(
−
1
10
r + 190
)
= −
1
10
r2
+ 200r − 19000

. . . . . .
Economics Problem
P(r) = (r − 100)n(r) = (r − 100)
(
−
1
10
r + 190
)
= −
1
10
r2
+ 200r − 19000
We want to maximize P on the interval 900 ≤ r ≤ 1900. (Why?)

. . . . . .
Economics Problem
P(r) = (r − 100)n(r) = (r − 100)
(
−
1
10
r + 190
)
= −
1
10
r2
+ 200r − 19000
A(900) = $800 × 100 = $80, 000, A(1900) = 0

. . . . . .
Economics Problem
P(r) = (r − 100)n(r) = (r − 100)
(
−
1
10
r + 190
)
= −
1
10
r2
+ 200r − 19000
A(900) = $800 × 100 = $80, 000, A(1900) = 0
A′
(x) = −
1
5
r + 200, which is zero when r = 1000.

. . . . . .
Economics Problem
P(r) = (r − 100)n(r) = (r − 100)
(
−
1
10
r + 190
)
= −
1
10
r2
+ 200r − 19000
A(900) = $800 × 100 = $80, 000, A(1900) = 0
A′
(x) = −
1
5
r + 200, which is zero when r = 1000.
n(1000) = 90, so P(r) = $900 × 90 = $81, 000. This is the
maximum intake.

. . . . . .
Example
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?

. . . . . .
Seting up the model
Solution
The angle subtended by the
statue in the viewer’s eye can
be expressed as
θ = arctan
(
a + b
x
)
−arctan
(
b
x
)
.
a
b
θ
x
The domain of θ is all positive real numbers x.

. . . . . .
Finding the derivative
θ = arctan
(
a + b
x
)
− arctan
(
b
x
)
So
dθ
dx
=
1
1 +
(
a+b
x
)2
·
−(a + b)
x2
−
1
1 +
(
b
x
)2
·
−b
x2
=
b
x2 + b2
−
a + b
x2 + (a + b)2
=
[
x2 + (a + b)2
]
b − (a + b)
[
x2 + b2
]
(x2 + b2
)
[
x2 + (a + b)2
]

. . . . . .
Finding the critical points
dθ
dx
=
[
x2 + (a + b)2
]
b − (a + b)
[
x2 + b2
]
(x2 + b2
)
[
x2 + (a + b)2
]

. . . . . .
dθ
dx
=
[
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
)

. . . . . .
dθ
dx
=
[
x2 + (a + b)2
]
b − (a + b)
[
x2 + b2
]
(x2 + b2
)
[
x2 + (a + b)2
]
seek x such that
0 =
[
x2
+ (a + b)2
]
b − (a + b)
[
x2
+ b2
]
= a(ab + b2
− x2
)
The only positive solution is x =
√
b(a + b).

. . . . . .
dθ
dx
=
[
x2 + (a + b)2
]
b − (a + b)
[
x2 + b2
]
(x2 + b2
)
[
x2 + (a + b)2
]
seek x such that
0 =
[
x2
+ (a + b)2
]
b − (a + b)
[
x2
+ b2
]
= a(ab + b2
− x2
)
The only positive solution is 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, ∞).

. . . . . .
Final answer
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.

. . . . . .
Discussion
The length
√
b(a + b) is the geometric mean of the two distances
measured 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.

. . . . . .
Summary
Remember the checklist
Ask yourself: what is the
objective?
Remember your geometry:
similar triangles
right triangles
trigonometric functions
Name [_
Problem Solving Strategy
Draw a Picture
Kathy had a box of 8 crayons.
She gave some crayons away.
She has 5 left.
How many crayons did Kathy give away?
UNDERSTAND
•
What do you want to find out?
Draw a line under the question.
You can draw a picture
to solve the problem.
crayons
What number do I
add to 5 to get 8?
8 - = 5
5 + 3 = 8
CHECK
Does your answer make sense?
Explain.
Draw a picture to solve the problem.
Write how many were given away.
I. I had10 pencils.
I gave some away.
I have 3left. How many
pencils did I give away?
~7
What number
do I add to 3
to make 10?
13
i
ft
ill
:i
i ?
11
ft
I
'•'
«
I
I
ft A
H 11
M i l
U U U U> U U
I I

Lesson 22: Optimization II (Section 021 slides)

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