Lesson 22: Optimization II (Section 041 slides)

Section 4.5Optimization II

V63.0121.041, Calculus I

New York University

November 24, 2010

Announcements

I No recitation this weekI Quiz 5 on §§4.1–4.4 next week in recitationI Happy Thanksgiving!

. . . . . .

. . . . . .

Announcements

I No recitation this weekI Quiz 5 on §§4.1–4.4 next

week in recitationI Happy Thanksgiving!

V63.0121.041, Calculus I (NYU) Section 4.5 Optimization II November 24, 2010 2 / 25

. . . . . .

Objectives

I Given a problem requiringoptimization, identify theobjective functions,variables, and constraints.

I Solve optimizationproblems with calculus.


. . . . . .

Outline

Recall

More examplesAdditionDistanceTrianglesEconomicsThe Statue of Liberty


. . . . . .

Checklist for optimization problems

1. Understand the ProblemWhat is known? What is unknown?What are the conditions?

2. Draw a diagram3. Introduce Notation4. Express the “objective function” Q in terms of the other symbols5. 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 MethodSee Section 4.1

The Closed Interval MethodTo find the extreme values of a function f on [a,b], we need to:

I Evaluate f at the endpoints a and bI Evaluate f at the critical points x where either f′(x) = 0 or f is not

differentiable at x.I The points with the largest function value are the global maximum

pointsI The points with the smallest/most negative function value are the

global minimum points.


. . . . . .

Recall: The First Derivative TestSee 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).I If f′ changes from negative to positive at c, then c is a local

minimum.I If f′ changes from positive to negative at c, then c is a local

maximum.I If f′ does not change sign at c, then c is not a local extremum.

Corollary

I If f′ < 0 for all x < c and f′(x) > 0 for all x > c, then c is the globalminimum of f on (a,b).

I If f′ < 0 for all x > c and f′(x) > 0 for all x < c, then c is the globalmaximum of f on (a,b).


. . . . . .

Recall: The Second Derivative TestSee 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.I If f′′(c) < 0, then f(c) is a local maximum.I If f′′(c) > 0, then f(c) is a local minimum.

Warning

If f′′(c) = 0, the second derivative test is inconclusive (this does notmean c is neither; we just don’t know yet).

Corollary

I If f′(c) = 0 and f′′(x) > 0 for all x, then c is the global minimum of fI 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 2DTPro – no need for

inequalities– gets globalextremaautomatically

– works onnon-closed,non-boundedintervals– only one derivative

– works onnon-closed,non-boundedintervals– no need forinequalities

Con – only for closedbounded intervals

– Uses inequalities– More work atboundary than CIM

– More derivatives– less conclusivethan 1DT– more work atboundary than CIM

I Use CIM if it applies: the domain is a closed, bounded intervalI 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 examplesAdditionDistanceTrianglesEconomicsThe Statue of Liberty


. . . . . .

Addition with a constraint

Example

Find two positive numbers x and y with xy = 16 and x+ y as small aspossible.

Solution

I Objective: minimize S = x+ y subject to the constraint thatxy = 16

I Eliminate y: y = 16/x so S = x+ 16/x. The domain ofconsideration is (0,∞).

I Find the critical points: S′(x) = 1− 16/x2, which is 0 when x = 4.I Classify the critical points: S′′(x) = 32/x3, which is always

positive. So the graph is always concave up, 4 is a local min, andtherefore the global min.

I So the numbers are x = y = 4, Smin = 8.


. . . . . .


Example


Solution



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

I Classify the critical points: S′′(x) = 32/x3, which is alwayspositive. So the graph is always concave up, 4 is a local min, andtherefore the global min.



. . . . . .


Example


Solution



I Find the critical points: S′(x) = 1− 16/x2, which is 0 when x = 4.I Classify the critical points: S′′(x) = 32/x3, which is always

positive. So the graph is always concave up, 4 is a local min, andtherefore the global min.



. . . . . .

Distance

Example

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

Solution

The distance between (x, x2)and (3,0) is given by

f(x) =√

(x− 3)2 + (x2 − 0)2

We may instead minimize thesquare of f:

g(x) = f(x)2 = (x− 3)2 + x4

The domain is (−∞,∞).

.. x.

y

..

(x, x2)

..3


. . . . . .

Distance

Example


SolutionThe distance between (x, x2)and (3,0) is given by

f(x) =√

(x− 3)2 + (x2 − 0)2


g(x) = f(x)2 = (x− 3)2 + x4

The domain is (−∞,∞).

.. x.

y

..

(x, x2)

..3


. . . . . .

Distance

Example


SolutionThe distance between (x, x2)and (3,0) is given by

f(x) =√

(x− 3)2 + (x2 − 0)2


g(x) = f(x)2 = (x− 3)2 + x4

The domain is (−∞,∞). .. x.

y

..

(x, x2)

..3


. . . . . .

Distance problemminimization step

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

I g′(x) = 2(x− 3) + 4x3 = 4x3 + 2x− 6 = 2(2x3 + x− 3)I If a polynomial has integer roots, they are factors of the constant

term (Euler)I 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)I We see f′(1) = 0, f′(x) < 0 if x < 1, and f′(x) > 0 if x > 1. So 1 is

the global minimum.I The point on the parabola closest to (3,0) is (1,1). The minimum

distance is√5.


. . . . . .


We want to find the global minimum of g(x) = (x− 3)2 + x4.I g′(x) = 2(x− 3) + 4x3 = 4x3 + 2x− 6 = 2(2x3 + x− 3)

I If a polynomial has integer roots, they are factors of the constantterm (Euler)

I 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)



distance is√5.


. . . . . .


We want to find the global minimum of g(x) = (x− 3)2 + x4.I g′(x) = 2(x− 3) + 4x3 = 4x3 + 2x− 6 = 2(2x3 + x− 3)I If a polynomial has integer roots, they are factors of the constant

term (Euler)

I 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)



distance is√5.


. . . . . .




f′(x) = 2(2x3 + x− 3) = 2(x− 1)(2x2 + 2x+ 3)

The quadratic has no real roots (the discriminant b2 − 4ac < 0)

I We see f′(1) = 0, f′(x) < 0 if x < 1, and f′(x) > 0 if x > 1. So 1 isthe global minimum.

I The point on the parabola closest to (3,0) is (1,1). The minimumdistance is

√5.


. . . . . .




f′(x) = 2(2x3 + x− 3) = 2(x− 1)(2x2 + 2x+ 3)


the global minimum.

I The point on the parabola closest to (3,0) is (1,1). The minimumdistance is

√5.


. . . . . .




f′(x) = 2(2x3 + x− 3) = 2(x− 1)(2x2 + 2x+ 3)



distance is√5.


. . . . . .

Remark

I We’ve used each of the methods (CIM, 1DT, 2DT) so far.I Notice how we argued that the critical points were absolute

extremes even though 1DT and 2DT only tell you relative/localextremes.


. . . . . .

A problem with a triangle..Example

Find the rectangle of maximal area inscribed in a 3-4-5 right triangle with twosides on legs of the triangle and one vertex on the hypotenuse.

Solution

I Let the dimensions of therectangle be x and y.

I Similar triangles give

y3− x

=43

=⇒ 3y = 4(3− x)

I So y = 4− 43x and

A(x) = x(4− 4

3x)

= 4x− 43x2

..3

.

4

.

5


. . . . . .



Solution



y3− x

=43

=⇒ 3y = 4(3− x)


A(x) = x(4− 4

3x)

= 4x− 43x2

..3

.

4

.

5

.

y

.

x


. . . . . .



Solution



y3− x

=43

=⇒ 3y = 4(3− x)


A(x) = x(4− 4

3x)

= 4x− 43x2 ..

3.

4

.

5

.

y

.

x


. . . . . .

Triangle Problemmaximization step

We want to find the absolute maximum of A(x) = 4x− 43x2 on the

interval [0,3].

I A(0) = A(3) = 0

I A′(x) = 4− 83x, which is zero when x =

128

= 1.5.

I Since A(1.5) = 3, this is the absolute maximum.I So the dimensions of the rectangle of maximal area are 1.5× 2.


. . . . . .



interval [0,3].I A(0) = A(3) = 0


128

= 1.5.



. . . . . .



interval [0,3].I A(0) = A(3) = 0


128

= 1.5.

I Since A(1.5) = 3, this is the absolute maximum.

I So the dimensions of the rectangle of maximal area are 1.5× 2.


. . . . . .



interval [0,3].I A(0) = A(3) = 0


128

= 1.5.



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An Economics problem

Example

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

Solution

I Let n be the number of units rented, depending on price (thedemand function).

I We have n(900) = 100 and∆n∆r

= − 110

. So

n− 100 = − 110

(r− 900) =⇒ n(r) = − 110

r+ 190


. . . . . .

Economics ProblemFinishing the model and maximizing

I The profit per unit rented is r− 100, so

P(r) = (r− 100)n(r) = (r− 100)(− 110

r+ 190)

= − 110

r2 + 200r− 19000

I We want to maximize P on the interval 900 ≤ r ≤ 1900. (Why?)I A(900) = $800× 100 = $80,000, A(1900) = 0

I A′(x) = −15r+ 200, which is zero when r = 1000.

I n(1000) = 90, so P(r) = $900× 90 = $81,000. This is themaximum intake.


. . . . . .



P(r) = (r− 100)n(r) = (r− 100)(− 110

r+ 190)

= − 110

r2 + 200r− 19000

I We want to maximize P on the interval 900 ≤ r ≤ 1900. (Why?)

I A(900) = $800× 100 = $80,000, A(1900) = 0




. . . . . .



P(r) = (r− 100)n(r) = (r− 100)(− 110

r+ 190)

= − 110

r2 + 200r− 19000

I We want to maximize P on the interval 900 ≤ r ≤ 1900. (Why?)I A(900) = $800× 100 = $80,000, A(1900) = 0




. . . . . .

The Statue of Liberty

Example

The Statue of Liberty stands on top of a pedestal which is on top of onold fort. The top of the pedestal is 47m above ground level. The statueitself measures 46m from the top of the pedestal to the tip of the torch.

What distance should one stand away from the statue in order tomaximize the view of the statue? That is, what distance will maximizethe portion of the viewer’s vision taken up by the statue?


. . . . . .

The Statue of LibertySeting up the model

Solution

The angle subtended by thestatue in the viewer’s eye canbe expressed as

θ = arctan(a+ bx

)−arctan

(bx

).

a

bθ

xThe domain of θ is all positive real numbers x.


. . . . . .

The Statue of LibertyFinding the derivative

θ = arctan(a+ bx

)− arctan

(bx

)So

dθdx

=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

]


. . . . . .

The Statue of LibertyFinding the critical points

dθdx

=

[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

](x2 + b2)

[x2 + (a+ b)2

]

I This derivative is zero if and only if the numerator is zero, so weseek x such that

0 =[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

]= a(ab+ b2 − x2)

I The only positive solution is x =√

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


. . . . . .


dθdx

=

[x2 + (a+ b)2

]b− (a+ b)

[x2 + b2

](x2 + b2)

[x2 + (a+ b)2

]I 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)



0 < x <√




. . . . . .


dθdx

=

[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)


b(a+ b).

I Using the first derivative test, we see that dθ/dx > 0 if0 < x <

√b(a+ b) and dθ/dx < 0 if x >

√b(a+ b).

I So this is definitely the absolute maximum on (0,∞).


. . . . . .


dθdx

=

[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)



0 < x <√




. . . . . .

The Statue of LibertyFinal 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 fortwhich lies at the base of the island.

Unfortunately, you’re not allowed to walk on this part of the lawn.


. . . . . .

The Statue of LibertyDiscussion

I The length√

b(a+ b) is the geometric mean of the two distancesmeasured from the ground—to the top of the pedestal (a) and thetop of the statue (a+ b).

I The geometric mean is of two numbers is always between themand greater than or equal to their average.


. . . . . .

Summary

I Remember the checklistI Ask yourself: what is the

objective?I Remember your geometry:

I similar trianglesI right trianglesI trigonometric functions

Name [_

Problem Solving StrategyDraw 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 pictureto solve the problem.

crayons

What number do Iadd to 5 to get 8?

8 - = 55 + 3 = 8

CHECK

Does your answer make sense?Explain.

Draw a picture to solve the problem.Write how many were given away.I. I had 10 pencils.

I gave some away.I have 3 left. How manypencils did I give away?

~7

What numberdo I add to 3to make 10?

13iftill:ii ?11

ftI'•'«II

ft A

H 11M i l

U U U U> U U

I I


Lesson 22: Optimization II (Section 041 slides)

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