15 optimization problems

8/12/2019 15 Optimization Problems

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Section 4.5Optimization Problems

V63.0121.002.2010Su, Calculus I

New York University

June 14, 2010

Announcements The midterm is graded! Quiz 4 Thursday on 4.14.4

Guest speaker on Thursday: Arjun Krishnan


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Announcements

The midterm is graded!

Quiz 4 Thursday on4.14.4

Guest speaker onThursday: Arjun Krishnan

V63.0121.002.2010Su, Calculus I (NYU) Section 4.5 Optimization Problems June 14, 2010 2 / 31


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Objectives

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

Solve optimizationproblems with calculus.



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Outline

Leading by Example

The Text in the Box

More Examples



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Leading by Example

ExampleWhat is the rectangle of fixed perimeter with maximum area?



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Leading by Example

ExampleWhat is the rectangle of fixed perimeter with maximum area?

Solution

Draw a rectangle.

w



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Solution Continued

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



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Solution Continued


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



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Solution Continued



Since p = 2 + 2w , we have = p 2w

2 ,



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Solution Continued




2 , so

A = w = p 2w

2 w = 12

( p 2w )(w ) = 12 pw w

2



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Solution Continued




2 , so

A = w = p 2w

2 w = 12

( p 2w )(w ) = 12 pw w

2

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 makesure A(w ) 0).



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Solution Concluded

We use the Closed Interval Method for A(w ) = 1

2 pw

w 2 on [0 , p/ 2].

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

To find the critical points, we find dAdw

= 12 p 2w .

The critical points are when

0 = 12 p 2w = w =

p4



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Solution Concluded

We use the Closed Interval Method for A(w ) = 1

2 pw

w 2 on [0 , p/ 2].

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

To find the critical points, we find dAdw

= 12 p 2w .

The critical points are when

0 = 12 p 2w = w =

p4

Since this is the only critical point, it must be the maximum. In thiscase = p4

as well.



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O li


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Outline

Leading by Example

The Text in the Box

More Examples


St t i f P bl S l i


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Strategies for Problem Solving

1. Understand the problem2. Devise a plan

3. Carry out the plan

4. Review and extend

Gyrgy Plya(Hungarian, 18871985)



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The Text in the Box


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The Text in the Box

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

2. Draw a diagram.


The Text in the Box


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The Text in the Box


2. Draw a diagram.

3. Introduce Notation.


The Text in the Box


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The Text in the Box


2. Draw a diagram.

3. Introduce Notation.4. Express the objective function Q in terms of the other symbols


The Text in the Box


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The Text in the Box


2. Draw a diagram.


5. If Q is a function of more than one decision variable, use thegiven information to eliminate all but one of them.


The Text in the Box


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The Text in the Box


2. Draw a diagram.


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.


Polya's Method in Kindergarten


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Polya s Method in Kindergarten

Name [_

Problem Solving StrategyDraw a Picture

Kathy had a box of 8 crayonsShe gave some crayons awayShe has 5 leftHow 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 I

add to 5 to get 8? - = 5

5 3 = 8

CHECKDoes your answer make sense?Explain

Draw a picture to solve the problemWrite how many were given aw ay

I I had 10 pencilsI gave some awayI have 3 left How manypencils did I give away?

What numberdo I ad d t o

to make 10?

3 ill i ?

ft A

i l

U U U U> U U


Recall: The Closed Interval Method


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Recall: The Closed Interval MethodSee 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 notdifferentiable at x .

The points with the largest function value are the global maximumpoints

The points with the smallest/most negative function value are theglobal minimum points.


Recall: The First Derivative Test


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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). 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.



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Recall: The Second Derivative Test


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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 notmean c is neither; we just dont know yet).


Recall: The Second Derivative Test


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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 notmean c is neither; we just dont 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?


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CIM 1DT 2DTPro no need for

inequalities gets globalextremaautomatically

works onnon-closed,non-boundedintervals only one derivative

works onnon-closed,non-boundedintervals no need for inequalities

Con only for closedbounded intervals

Uses inequalities More work atboundary than CIM

More derivatives less conclusivethan 1DT more work atboundary than CIM


Which to use when?


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CIM 1DT 2DTPro no need for

inequalities gets globalextremaautomatically

works onnon-closed,non-boundedintervals only one derivative

works onnon-closed,non-boundedintervals no need for inequalities

Con only for closedbounded intervals

Uses inequalities More work atboundary than CIM

More derivatives less conclusivethan 1DT more work atboundary 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 takederivatives, or 1DT if you like to compare signs.


Outline


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Leading by Example

The Text in the Box

More Examples


Another Example


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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. With800m of wire at your disposal, what is the largest area you can

enclose, and what are its dimensions?


Solution


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1. Everybody understand?


Another Example


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Another Example


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Known: amount of fence used

Unknown: area enclosed


Another Example


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Known: amount of fence used

Unknown: area enclosed Objective: maximize area

Constraint: fixed fence length


Solution


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1. Everybody understand?


Solution


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1. Everybody understand?2. Draw a diagram.


Diagram


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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?


Solution


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Solution


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3. Length and width are and w . Length of wire used is p.


Diagram


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Diagram


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w


Solution


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3. Length and width are and w . Length of wire used is p.



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Solution


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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 2w 2The domain of Q is [0 , p/ 2]


Solution


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6. dQdw

= p 4w , which is zero when w = p4

.


Solution


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6. dQdw

= p 4w , which is zero when w = p4

.



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Your turn


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Example (The shortest fence)

A 216m 2 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?


Your turn


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Example (The shortest fence)

A 216m 2 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?

Solution

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

f (w ) = 2 Aw + 3w .

The domain is all positive numbers.


Diagram


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w

f = 2 + 3w A = w 216


Solution (Continued)

W d fi d h i i l f f( ) 2 A

3 (0 )


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

+ 3w on (0 , ).



W d t fi d th i i l f f( ) 2 A

+ 3 (0 )


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+ 3w on (0 , ). We have

df dw

= 2 Aw 2 + 3

which is zero when w = 2 A3 .



W d t fi d th i i l f f( ) 2 A

+ 3 (0 )


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+ 3w on (0 , ). We have

df dw = 2 Aw 2 + 3

which is zero when w = 2 A3 . Since f

(w ) = 4 Aw 3

, which is positive for all positive w , thecritical point is a minimum, in fact the global minimum.



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Try this one


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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 in 2 , what dimensions shouldthe advertisement be to maximize the area of the printed region?


Try this one


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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 in 2 , what dimensions shouldthe advertisement be to maximize the area of the printed region?

Answer

The optimal paper dimensions are 4 5 in by 6 5in.


Solution

Let the dimensions of the


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Let the dimensions of theprinted region be x and y , P

the printed area, and A thepaper area. We wish tomaximize P = xy subject tothe constraint that

A = ( x + 2)(y + 3) 120Isolating y in A 120 givesy =

120 x + 2 3 which yields

P = x 120 x + 2 3 =

120 x x + 2 3 x

The domain of P is (0 , )

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primis in faucibus orci luctus etultrices posuere cubilia Curae;

1.5 cm

1.5 cm

1 c m

1 c m

x

y


Solution (Concluded)We want to find the absolute maximum value of P . Taking derivatives,


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dP

dx =

( x + 2)(120 ) (120 x )(1)( x + 2)2

3 = 240 3( x + 2)2

( x + 2)2

There is a single critical point when

( x + 2)2 = 80 = x = 4 5 2(the negative critical point doesnt count). The second derivative is

d 2 P dx 2

= 480( x + 2)3

which is negative all along the domain of P . Hence the unique criticalpoint x = 4 5 2 cm is the absolute maximum of P . This meansthe paper width is 4 5 cm, and the paper length is 120

4 5 = 6 5 cm .


Summary


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Remember the checklist Ask yourself: what is theobjective?

Remember your geometry:

similar trianglesright trianglestrigonometric functions

Name [_

Problem Solving StrategyDraw a Picture

Kathy had a box of 8 crayons

She gave some crayons awayShe has 5 leftHo w many crayons did Kathy give awa y?

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 I

add to 5 to get 8? - = 5

5 3 = 8

CHECK

Does your answer make sense?Explain

Draw a picture to solve the problemWrite how many were given away

I I had 10 pencilsI gave some awayI have 3 left How manypencils did I give away?

What number

do I ad d t oto make 10?

3

ill i ?

ft A

i l

U U U U> U U


15 optimization problems

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