Warm up 9/10/14 a) Find all relative extrema using the 2 nd derivative test: b) Find any points of inflection and discuss the concavity of the graph

3.7 Modeling and Optimization

Buffalo Bill’s Ranch, North Platte, NebraskaGreg Kelly, Hanford High School, Richland, WashingtonPhoto by Vickie Kelly, 1999

3.7 Optimization ProblemsOne of the most common applications of calculus involves the determination of minimum and maximum values. Consider how frequently you hear or read terms such as…

A Classic Problem

You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?

x x

40 2x

40 2A x x

240 2A x x

40 4A x

0 40 4x

4 40x

10x 40 2l x

w x 10 ftw

20 ftl

There must be a local maximum here, since the endpoints are minimums.

A Classic Problem

You have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?

x x

40 2x

40 2A x x

240 2A x x

40 4A x

0 40 4x

4 40x

10x

10 40 2 10A

10 20A

2200 ftA40 2l x

w x 10 ftw

20 ftl

To find the maximum (or minimum) value of a function:

1 Write it in terms of one variable.

2 Find the first derivative and set it equal to zero.

3 Check the end points if necessary.

Example 5: What dimensions for a one liter cylindrical can will use the least amount of material?

We can minimize the material by minimizing the area.

22 2A r rh area ofends

lateralarea

We need another equation that relates r and h:

2V r h

31 L 1000 cm21000 r h

2

1000h

r

22

10 02

02A r r

r

2 20002A r

r

2

20004A r

r

Example 5: What dimensions for a one liter cylindrical can will use the least amount of material?

22 2A r rh area ofends

lateralarea

2V r h

31 L 1000 cm21000 r h

2

1000h

r

22

10 02

02A r r

r

2 20002A r

r

2

20004A r

r

2

20000 4 r

r

2

20004 r

r

32000 4 r

3500r

3500

r

5.42 cmr

2

1000

5.42h

10.83 cmh

If the end points could be the maximum or minimum, you have to check those also.

Notes:

If the function that you want to optimize has more than one variable, use substitution to rewrite the function.

If you are not sure that the extreme you’ve found is a maximum or a minimum, you have to check by using either the 1st or 2nd derivative test.

You Try:

Find two positive numbers such that the sum of the first number squared and the second is 27 and the product is a maximum.

3.7 Optimization Problems

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

2V x h Primary Equa n tio2 4 108S x xh Secondary Equ n atio

(use the second

Express as a

ary equation t

function of

o write

one

i

variabl

n terms )

e

. of V x

V

2 4 108x xh 24 108xh x

2108

4

xh

x

22 108

4

xV x

x

3

274

xV x x x

x

h

Solution


Feasible Domain (real-world)?

must be positivex2 is less than 108bA x

0 108x ?dV

dx

2327

4

dV x

dx 0

2327

4

x

23 108x 2 36x

6x

(6) ?V 3108 in

If the domain was a closed interval,

we would also have to check for

extrema at its endpoints.

V

3

274

xV x x

Diagram

2

2

d y

dx 6x 3 6

9 02

Is this the maximum volume?

xx

h


2V x h2108 (6) h

3h

6" 6" 3"x x

6, 108, Solve for x V h

xx

h

Diagram is not drawn to scale


Which points on the graph of24y x

are closest to the point (0,2)?

2 2

The quantity to be minimized is distance:

0 2d x y

Work

2 2

2 1 2 1d x x y y


2 2( 2)d x y Primary

Eq n

uatio

24 Secondary Equ i n at oy x

2 224 2 ( )xd x 2 22 (2 )x x

2 2 4(4 4 )x x x

4 23 4x x

Use the secondary

equation to write

the primary equation

in terms of one variable.

Diagram


Find the minima of4 23 4d x x

because d is smallest when the radicand is smallest,

we simply need the minima of

4 2( ) 3 4f x x x 3'( ) 4 6f x x x

22 (2 3)x x 00 orx 22 3 0x

3

2x

Table


3 5 3 5Min. @ , & ,

2 2 2 2

Max. @ (0,4)

Closest points are 3 5 3 5, & ,

2 2 2 2

3'( ) 4 6f x x x

Double check with the 2nd Derivative Test:

2''( ) 12 6f x x

3'' 12 0

2f

3'' 12 0

2f

'' 0 6 0f


3 5- ,

2 2

3 5,

2 2

Note: 0 yields a relative maximum, there is no absolute maximum since the domain is the entire real line.

Homework:

Day 1: Pg. 223 5-21 odd & 29

Day 2 Problems

Two posts, one 12 feet high and the other 28 feet high, stand 30 feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. Where should the stake be placed to use the least wire?

The quantity to

be minimized is

length. From the

diagram you can

see that varies

between 0 and 30.

x

Domain?


Primary Equa n tioW y z 2 2 212y x 2 2 2(30 ) 28z x

2 2(30 28)xz 2900 60 784x x

2 60 1684x x 2 144y x

Secondary Equations!

Write y and z in terms of x.


Primary EquationW y z

in (0,30)x2 2144 60 1684W x x x 1 1

2 22 2( 144) ( 60 1684)x x x

1 12 22 2

2 60( )

( 144) ( 60 1684)2

x x

x x x

dw

dx

1 12 22 21 1

( 144) (2 ) ( 60 1684) (2 60)2 2x x x x x

Use the secondary equations to write the

primary equation in terms of .x


2 2

( 30)

144 60 1684

dw x x

dx x x x

0

dw

dx

2 2

( 30)0

144 60 1684

x x

x x x

2 2

30

144 60 1684

x x

x x x

2 260 168 (304 144)x xx x x

2 2 2 2( 60 1684) (30 ) ( 144)x x x x x

It could be the proportion

from &$##!

4 3 2 2 260 1684 (900 60 )( 144)x x x x x x


4 3 2 2 3 4 260 1684 900 60 129600 8640 144x x x x x x x x

2 2 21684 900 129600 8640 144x x x x 2 21684 1044 8640 129600x x x 2640 8640 129,600 0x x

2320(2 27 405) 0x x 320(2 45)( 9) 0x x 9, 22.5x

2 2 2 2( 60 1684) (30 ) ( 144)x x x x x

Obviously, 320 is a common factor

Factorable

Put in Quadratic Form! Combine Like Terms


2 2(9) 9 144 9 60(9) 1684w 50

The wire should be staked 9 feet from the 12

:

foot pole.

Conclusion


4 ' of wire is to be used to form a square and/or a circle.

How much of the wire should be used for the square and

how much should be used for the circle to enclose the

maximum total area?

2 2

The quantity to

be maximized is

area.

.A x r SolutionPrimary Equa n tio

4 ' of wire is to be used to form a square and/or a circle.

How much of the wire should be used for the square and

how much should be used for the circle to enclose the

maximum total area? Re member that an extreme value

can also occur at the endpoints of an interval.

2 2 Primary EquationA x r

4 4 2 Secondary Equationx r 4 4

2

xr

2 2x

2(1 )x

22 2(1 )x

A x

2

2 4(1 )xx

2 2(1 )4x x

2 21(1 2 )4x x x

2 21( 4 8 )4xx x

21(4 ) 8 4x x


Feasible Domain? in [0,1] x

The perimeter of the square could be as little as

zero or as much as 4.

(8 2 ) 80

dA

dx

x

(0) 1.273A 4

( ) 0.5604

A

(1) 1A The maximum area occurs

when 0. That is, x

Diagram

You could use all or

none of the wire for the

square.

21(4 ) 8 4x x

8 2 8 0

8 48 2 8

8 2 4

x x

x x

when all the wire is

used for the circle.

You Try:

A rancher has 200 feet of fencing with which to enclose two adjacent rectangular corrals. What dimensions should be used so that the enclosed area will be a maximum?


You must expect that real-life applications often

involve equations that are at least as complicated

as the primary equations seen in today's examples.

Remember, one of the main goals of this course is

to learn to use calculus to analyze equations that

initially seem formidable.

HW Day 2 MMM pgs 111-113

Documents

Warm up 9/10/14 a) Find all relative extrema using the 2 nd derivative test: b) Find any points of inflection and discuss the concavity of the graph