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Applied Max and Min Problems

- Objective To use the methods of this chapter to

solve applied optimization problems.

Classification

- The applied optimization problems that we will

consider in this section fall into the following

two categories - Problems that reduce to maximizing or minimizing

a continuous function over a finite closed

interval. - Problems that reduce to maximizing or minimizing

a continuous function over an infinite interval

or a finite interval that is not closed.

Example 1

- A garden is to be laid out in a rectangular area

and protected by a chicken wire fence. What is

the largest possible area of the garden if only

100 running feet of chicken wire is available?

Example 1

- A garden is to be laid out in a rectangular area

and protected by a chicken wire fence. What is

the largest possible area of the garden if only

100 running feet of chicken wire is available? - x length of the rectangle (ft)
- y width of the rectangle (ft)
- A area of the rectangle (ft2)
- The domain is 0 lt x lt 50

Example 1

- A garden is to be laid out in a rectangular area

and protected by a chicken wire fence. What is

the largest possible area of the garden if only

100 running feet of chicken wire is available? - x length of the rectangle (ft)
- y width of the rectangle (ft)
- A area of the rectangle (ft2)
- The domain is 0 lt x lt 50

Example 1

- x length of the rectangle (ft)
- y width of the rectangle (ft)
- A area of the rectangle (ft2) P perimeter of

fence - We are maximizing the area, so we need an

equation for the area in one variable. We will

use two separate equations and substitute to get

an area equation in one variable.

Example 1

- x length of the rectangle (ft)
- y width of the rectangle (ft)
- A area of the rectangle (ft2) P perimeter of

fence - We are maximizing the area, so we need an

equation for the area in one variable. We will

use two separate equations and substitute to get

an area equation in one variable. - A xy
- P 2x 2y 100 2x 2y y 50 - x

Example 1

- x length of the rectangle (ft)
- y width of the rectangle (ft)
- A area of the rectangle (ft2) P perimeter of

fence - A xy
- P 2x 2y 100 2x 2y y 50 x
- A x(50 x) 50x x2
- dA/dx 50 2x

Example 1

- We must now check the endpoints as well as the

critical numbers. - f(0) 0
- f(25) 625
- f(50) 0

Example 1

- We must now check the endpoints as well as the

critical numbers. - f(0) 0
- f(25) 625 Maximum area of 625 ft2 _at_ x 25 ft
- f(50) 0

Example 2

- An open box is to be made from a 16-inch by

30-inch piece of cardboard by cutting out squares

of equal size from the four corners and bending

up the sides. What size should the squares be to

obtain a box with the largest volume? - The largest the squares can be is 8 in (why?) so

the domain is 0 lt x lt 8

Example 2

- An open box is to be made from a 16-inch by

30-inch piece of cardboard by cutting out squares

of equal size from the four corners and bending

up the sides. What size should the squares be to

obtain a box with the largest volume? - We are maximizing the volume, so we need an

equation for volume. V x(30 2x)(16 2x)

Example 2

- We are maximizing the volume, so we need an

equation for volume. V x(30 2x)(16 2x) - V 480x 92x2 4x3
- dV/dx 480 184x 12x2
- 4(x 12)(3x 10)

Example 2

- We are maximizing the volume, so we need an

equation for volume. V x(30 2x)(16 2x) - V 480x 92x2 4x3
- dV/dx 480 184x 12x2
- 4(x 12)(3x 10)
- The only critical number in the domain is 10/3

so - f(0) 0
- f(10/3) 726
- f(8) 0

Example 2

- We are maximizing the volume, so we need an

equation for volume. V x(30 2x)(16 2x) - V 480x 92x2 4x3
- dV/dx 480 184x 12x2
- 4(x 12)(3x 10)
- The only critical number in the domain is 10/3

so - f(0) 0
- f(10/3) 726 Max volume of 726 in3 _at_ x

10/3 in - f(8) 0

Example 3

- An offshore oil well is located at a point W that

is 5km from the closest point A on a straight

shoreline. Oil is to be piped from W to a shore

point B that is 8km from A by piping it on a

straight line under water from W to some shore

point P between A B and then onto B via pipe

along the shoreline. If the cost of laying pipe

is 1,000,000/km under water and 500,000/km over

land, where should the point P be located to

minimize the cost of laying the pipe?

Example 3

- x the distance (in km) between A and P
- c cost (in millions) for the entire pipeline
- Using the Pythagorean Theorem, the distance from

W to P is - The distance from P to B is (8 x)
- We are minimizing cost, so we need
- a cost equation.

Example 3

- Now, we take the derivative to find the critical

numbers.

Example 3

- Now, we take the derivative to find the critical

numbers.

Example 3

- We now need to check the endpoints and the

critical number to find the min. - The domain is 0 lt x lt 8 so
- F(0) 9
- F(8) 9.433
- Min cost of 8.33

million _at_ x km

Example 4

- Find the radius and height of the right circular

cylinder of largest volume that can be inscribed

in a right circular cone with radius 6 in and

height 10 in.

Example 4

- Find the radius and height of the right circular

cylinder of largest volume that can be inscribed

in a right circular cone with radius 6 in and

height 10 in. - r radius of cylinder
- h height of cylinder
- V volume

Example 4

- Again, we need to eliminate one of the variables.

We will do this with similar triangles.

Example 4

- Now, we take the derivative and find the critical

points.

Example 4

- The domain of the radius is 0 lt r lt 6, so we

check the endpoints and the critical number. - f(0) 0
- f(4) 167.55 Max volume of 167.55 in3 _at_ x 4

in - f(6) 0

Example 5

- A closed cylindrical can is to hold 1000 cm3 of

liquid. How should we choose the height and

radius to minimize the amount of material needed

to manufacture the can? - There are no physical restraints on this problem,

so this is not a finite interval. We need to

confirm that our critical number is a minimum.

Example 5

- A closed cylindrical can is to hold 1000 cm3 of

liquid. How should we choose the height and

radius to minimize the amount of material needed

to manufacture the can? - h height (cm) of can
- r radius (cm) of can
- S surface area (cm2) of can

Example 5

- Again, we will find two equations and substitute

to get the surface area equation in one variable. - h height (cm) of can
- r radius (cm) of can
- S surface area (cm2) of can

Example 5

- Again, we will find two equations and substitute

to get the surface area equation in one variable. - h height (cm) of can
- r radius (cm) of can
- S surface area (cm2) of can

Example 5

- Lets take a look at the graph of this equation.

Example 5

- Now we find the derivative and check the critical

number.

Example 5

- Now we find the derivative and check the critical

number.

Example 5

- Now we find the derivative and check the critical

number. - ____-_________
- 5.4
- min

Example 6

- Find a point on the curve y x2 that is closest

to the point (18, 0).

Example 6

- Find a point on the curve y x2 that is closest

to the point (18, 0). - The distance L from (18, 0) and an arbitrary

point - (x, y) on the curve is
- Since y x2, we can say that

Example 6

- Find a point on the curve y x2 that is closest

to the point (18, 0). - The distance L from (18, 0) and an arbitrary

point - (x, y) on the curve is
- Since y x2, we can say that
- We are told that the distance L and the square of

the distance L2 are minimized at the same value.

Why?

Example 6

- Lets look at the two derivatives together.

Example 6

- We will now take the derivative of L2 to find any

critical points

Example 6

- We will now take the derivative of L2 to find any

critical points - This will factor to (x 2)(2x2 4x 9) trust

me! - ___-_________
- 2
- min

Example 6

- We will now take the derivative of L2 to find any

critical points - This will factor to (x 2)(2x2 4x 9) trust

me! - ___-_________ The closest point

is (2, 4) - 2
- min

Homework

- Section 4.5
- Pages 283-284
- 1, 3, 5, 11, 19, 21, 23