Lesson 22: Optimization (Section 021 slides)

Section 4.5Optimization Problems

V63.0121.021, Calculus I

New York University

November 23, 2010

Announcements

I Turn in HW anytime between now and November 24, 2pmI No Thursday recitation this weekI Quiz 5 on §§4.1–4.4 next week in recitation

. . . . . .

Announcements

I Turn in HW anytimebetween now andNovember 24, 2pm

I No Thursday recitation thisweek

I Quiz 5 on §§4.1–4.4 nextweek in recitation

V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 2 / 31

. . . . . .

Objectives

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

I Solve optimizationproblems with calculus.

. . . . . .

Outline

Leading by Example

The Text in the Box

More Examples

. . . . . .

Leading by Example

Example

What is the rectangle of fixed perimeter with maximum area?

Solution

I Draw a rectangle.

. . . . . .

Leading by Example

Example

Solution

I Draw a rectangle.

. . . . . .

Leading by Example

Example

Solution

I Draw a rectangle.

. . . . . .

Leading by Example

Example

Solution

I Draw a rectangle.

. . . . . .

Solution Continued

I Let its length be ℓ and its width be w. The objective function isarea A = ℓw.

I This is a function of two variables, not one. But the perimeter isfixed.

I Since p = 2ℓ+ 2w, we have ℓ =p− 2w

A = ℓw =p− 2w

2· w =

12(p− 2w)(w) =

12pw− w2

I Now we have A as a function of w alone (p is constant).I The natural domain of this function is [0,p/2] (we want to make

sure A(w) ≥ 0).

. . . . . .

Solution Continued

A = ℓw =p− 2w

2· w =

12(p− 2w)(w) =

12pw− w2

sure A(w) ≥ 0).

. . . . . .

Solution Continued

A = ℓw =p− 2w

2· w =

12(p− 2w)(w) =

12pw− w2

sure A(w) ≥ 0).

. . . . . .

Solution Continued

A = ℓw =p− 2w

2· w =

12(p− 2w)(w) =

12pw− w2

sure A(w) ≥ 0).

. . . . . .

Solution Continued

A = ℓw =p− 2w

2· w =

12(p− 2w)(w) =

12pw− w2

I Now we have A as a function of w alone (p is constant).

I The natural domain of this function is [0,p/2] (we want to makesure A(w) ≥ 0).

. . . . . .

Solution Continued

A = ℓw =p− 2w

2· w =

12(p− 2w)(w) =

12pw− w2

sure A(w) ≥ 0).

. . . . . .

Solution Concluded

We use the Closed Interval Method for A(w) =12pw−w2 on [0,p/2].

I At the endpoints, A(0) = A(p/2) = 0.

I To find the critical points, we finddAdw

=12p− 2w.

I The critical points are when

0 =12p− 2w =⇒ w =

I Since this is the only critical point, it must be the maximum. In thiscase ℓ =

p4as well.

I We have a square! The maximal area is A(p/4) = p2/16.

. . . . . .

Solution Concluded

=12p− 2w.

0 =12p− 2w =⇒ w =

p4as well.

. . . . . .

Solution Concluded

=12p− 2w.

0 =12p− 2w =⇒ w =

p4as well.

. . . . . .

Solution Concluded

=12p− 2w.

0 =12p− 2w =⇒ w =

p4as well.

. . . . . .

Solution Concluded

=12p− 2w.

0 =12p− 2w =⇒ w =

p4as well.

. . . . . .

Outline

Leading by Example

The Text in the Box

More Examples

. . . . . .

Strategies for Problem Solving

1. Understand the problem2. Devise a plan3. Carry out the plan4. Review and extend

György Pólya(Hungarian, 1887–1985)

. . . . . .

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 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.

. . . . . .

The Text in the Box

2. Draw a diagram.

3. Introduce Notation.4. Express the “objective function” Q in terms of the other symbols5. If Q is a function of more than one “decision variable”, use the

. . . . . .

The Text in the Box

2. Draw a diagram.3. Introduce Notation.

4. Express the “objective function” Q in terms of the other symbols5. If Q is a function of more than one “decision variable”, use the

. . . . . .

The Text in the Box

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 thegiven information to eliminate all but one of them.

6. Find the absolute maximum (or minimum, depending on theproblem) of the function on its domain.

. . . . . .

The Text in the Box

given information to eliminate all but one of them.

6. Find the absolute maximum (or minimum, depending on theproblem) of the function on its domain.

. . . . . .

The Text in the Box

. . . . . .

Polya's Method in Kindergarten

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

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?

What numberdo I add to 3to make 10?

13iftill:ii ?11

ftI'•'«II

H 11M i l

U U U U> U U

. . . . . .

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 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

. . . . . .

Warning

Corollary

. . . . . .

Warning

Corollary

. . . . . .

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.

. . . . . .

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

Leading by Example

The Text in the Box

More Examples

. . . . . .

Another Example

Example (The Best Fencing Plan)

A rectangular plot of farmland will be bounded on one side by a riverand on the other three sides by a single-strand electric fence. With800m of wire at your disposal, what is the largest area you canenclose, and what are its dimensions?

I Known: amount of fence usedI Unknown: area enclosedI Objective: maximize areaI Constraint: fixed fence length

. . . . . .

Solution

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]

6. dQdw

= p− 4w, which is zero when w =p4.

Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

so the critical point is the absolute maximum.

. . . . . .

Another Example

. . . . . .

Another Example

I Known: amount of fence usedI Unknown: area enclosed

I Objective: maximize areaI Constraint: fixed fence length

. . . . . .

Another Example

. . . . . .

Solution

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

6. dQdw

Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

. . . . . .

Solution

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

6. dQdw

Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

. . . . . .

Diagram

A rectangular plot of farmland will be bounded on one side by a riverand 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?

. . . . . .

Solution

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

6. dQdw

Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

. . . . . .

Solution

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

6. dQdw

Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

. . . . . .

Diagram

. . . . . .

Diagram

. . . . . .

Solution

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

6. dQdw

Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

. . . . . .

Solution

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

6. dQdw

Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

. . . . . .

Solution

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

6. dQdw

Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

. . . . . .

Solution

Q(w) = (p− 2w)(w) = pw− 2w2

6. dQdw

Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

. . . . . .

Solution

Q(w) = (p− 2w)(w) = pw− 2w2

6. dQdw

= p− 4w, which is zero when w =p4. Q(0) = Q(p/2) = 0, but

)= p · p

4− 2 · p

8= 80,000m2

so the critical point is the absolute maximum.V63.0121.021, Calculus I (NYU) Section 4.5 Optimization Problems November 23, 2010 24 / 31

. . . . . .

Your turn

Example (The shortest fence)

A 216m2 rectangular pea patch is to be enclosed by a fence anddivided into two equal parts by another fence parallel to one of itssides. What dimensions for the outer rectangle will require the smallesttotal length of fence? How much fence will be needed?

SolutionLet the length and width of the pea patch be ℓ and w. The amount offence needed is f = 2ℓ+ 3w. Since ℓw = A, a constant, we have

f(w) = 2Aw

The domain is all positive numbers.

. . . . . .

Your turn

Example (The shortest fence)

A 216m2 rectangular pea patch is to be enclosed by a fence anddivided into two equal parts by another fence parallel to one of itssides. What dimensions for the outer rectangle will require the smallesttotal length of fence? How much fence will be needed?

SolutionLet the length and width of the pea patch be ℓ and w. The amount offence needed is f = 2ℓ+ 3w. Since ℓw = A, a constant, we have

f(w) = 2Aw

The domain is all positive numbers.

. . . . . .

Diagram

....ℓ

f = 2ℓ+ 3w A = ℓw ≡ 216

. . . . . .

Solution (Continued)

We need to find the minimum value of f(w) =2Aw

+ 3w on (0,∞).

I We havedfdw

= −2Aw2 + 3

which is zero when w =

√2A3.

I Since f′′(w) = 4Aw−3, which is positive for all positive w, thecritical point is a minimum, in fact the global minimum.

I So the area is minimized when w =

√2A3

= 12 and

ℓ =Aw

√3A2

= 18. The amount of fence needed is

(√2A3

)= 2 ·

√3A2

+ 3√

= 2√6A = 2

√6 · 216 = 72m

. . . . . .

+ 3w on (0,∞).I We have

= −2Aw2 + 3

√2A3.

√2A3

= 12 and

ℓ =Aw

√3A2

(√2A3

)= 2 ·

√3A2

+ 3√

= 2√6A = 2

√6 · 216 = 72m

. . . . . .

= −2Aw2 + 3

√2A3.

√2A3

= 12 and

ℓ =Aw

√3A2

(√2A3

)= 2 ·

√3A2

+ 3√

= 2√6A = 2

√6 · 216 = 72m

. . . . . .

= −2Aw2 + 3

√2A3.

√2A3

= 12 and

ℓ =Aw

√3A2

(√2A3

)= 2 ·

√3A2

+ 3√

= 2√6A = 2

√6 · 216 = 72m

. . . . . .

Try this one

Example

An advertisement consists of a rectangular printed region plus 1 inmargins on the sides and 1.5 in margins on the top and bottom. If thetotal area of the advertisement is to be 120 in2, what dimensions shouldthe advertisement be to maximize the area of the printed region?

AnswerThe optimal paper dimensions are 4

√5 in by 6

√5 in.

. . . . . .

Try this one

Example

An advertisement consists of a rectangular printed region plus 1 inmargins on the sides and 1.5 in margins on the top and bottom. If thetotal area of the advertisement is to be 120 in2, what dimensions shouldthe advertisement be to maximize the area of the printed region?

AnswerThe optimal paper dimensions are 4

√5 in by 6

√5 in.

. . . . . .

Solution

Let the dimensions of theprinted region be x and y, Pthe printed area, and A thepaper area. We wish tomaximize P = xy subject tothe constraint that

A = (x+ 2)(y+ 3) ≡ 120

Isolating y in A ≡ 120 gives

y =120x+ 2

− 3 which yields

P = x(

120x+ 2

− 3)

=120xx+ 2

The domain of P is (0,∞).

Lorem ipsum dolor sit amet,consectetur adipiscing elit. Namdapibus vehicula mollis. Proin nectristique mi. Pellentesque quisplacerat dolor. Praesent a nisl diam.Phasellus ut elit eu ligula accumsaneuismod. Nunc condimentumlacinia risus a sodales. Morbi nuncrisus, tincidunt in tristique sit amet,ultrices eu eros. Proin pellentesquealiquam nibh ut lobortis. Ut etsollicitudin ipsum. Proin gravidaligula eget odio molestie rhoncussed nec massa. In ante lorem,imperdiet eget tincidunt at, pharetrasit amet felis. Nunc nisi velit,tempus ac suscipit quis, blanditvitae mauris. Vestibulum ante ipsumprimis in faucibus orci luctus etultrices posuere cubilia Curae;.

1.5 cm.

1.5 cm

. . . . . .

Solution (Concluded)

I We want to find the absolute maximum value of P.dPdx

=(x+ 2)(120)− (120x)(1)

(x+ 2)2− 3 =

240− 3(x+ 2)2

(x+ 2)2I There is a single (positive) critical point when

(x+ 2)2 = 80 =⇒ x = 4√5− 2.

I The second derivative isd2Pdx2

=−480

(x+ 2)3, which is negative all

along the domain of P.I Hence the unique critical point x =

(4√5− 2

)cm is the absolute

maximum of P.I This means the paper width is 4

√5 cm, and the paper length is

1204√5= 6

√5 cm.

. . . . . .

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

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?

What numberdo I add to 3to make 10?

13iftill:ii ?11

ftI'•'«II

H 11M i l

U U U U> U U

Lesson 22: Optimization (Section 021 slides)

Technology

Transcript of Lesson 22: Optimization (Section 021 slides)

Lesson 24: Areas, Distances, the Integral (Section 021 slides)

Modeling and Optimization of Biorefineries - CEPACcepac.cheme.cmu.edu/pasi2008/slides/eden/library/slides/Eden_PASI... · Modeling and Optimization of Biorefineries ... • Application

Nonlinear Optimization for Optimal Control - Peoplepabbeel/cs287-fa12/slides/... · Nonlinear Optimization for Optimal Control Pieter Abbeel UC Berkeley EECS Many slides and figures

Sparse Optimization - Lecture 1: Review of Convex …wotaoyin/summer2013/slides/Lec01_Convex... · Lecture 1: Review of Convex Optimization ... A large number of online lecture slides,

Lesson 18: Maximum and Minimum Values (Section 021 slides)

Lesson 22: Optimization (Section 041 slides)

Optimization Slides

Lesson 26: The Fundamental Theorem of Calculus (Section 021 slides)

Lesson 22: Optimization II (Section 041 slides)

Lesson 13: Exponential and Logarithmic Functions (Section 021 slides)

5 seo-fundamentals-on page optimization (part 2)-slides

2008 - Multi-Objective Optimization Inspired by Nature (Slides)

4 seo-fundamentals-on page optimization (part 1)-slides

Lesson 19: The Mean Value Theorem (Section 021 slides)

SNP Optimization Training Slides

Lesson 22: Optimization Problems (slides)

3. Computational Complexity Combinatorial Optimization ...marco/DM63/Slides/dm63-lec1.pdf · COMBINATORIAL OPTIMIZATION PROBLEMS Lecture 1 Combinatorial Optimization Problems Marco

Solving composite optimization problems, with applications ...jduchi/projects/slides/DuchiRu17-pr.pdf · Composite optimization problems Methods for composite optimization ... Large

Introduction to Convex Optimization - Alex Smolaalex.smola.org/.../cmu2013-10-701/slides/2_Recitation_Optimization.pdfConvexity Unconstrained Convex Optimization Constrained Optimization

Lesson 20: Optimization (slides)