1. Applied Max and Min Problems
Objective: To use the methods of this chapter to solve applied optimization problems.

2. 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.

3. 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?

4. 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 < x < 50

5. 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 < x < 50

6. 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.

7. 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

8. 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

9. Example 1
We must now check the endpoints as well as the critical numbers.
f(0) = 0
f(25) = 625
f(50) = 0

10. Example 1
We must now check the endpoints as well as the critical numbers.
f(0) = 0
f(25) = 625 Maximum area of 625 x = 25 ft
f(50) = 0

11. 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 < x < 8

12. 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)

13. 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)

14. 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

15. 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) = Max volume of 726 x = 10/3 in
f(8) = 0

16. 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?

17. 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.

18. Example 3
Now, we take the derivative to find the critical numbers.

19. Example 3
Now, we take the derivative to find the critical numbers.

20. Example 3
We now need to check the endpoints and the critical number to find the min.
The domain is 0 < x < 8 so:
F(0) = 9
F(8) = 9.433
Min cost of 8.33 x = km

21. 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.

22. 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

23. Example 4
Again, we need to eliminate one of the variables. We will do this with similar triangles.

24. Example 4
Now, we take the derivative and find the critical points.

25. Example 4
The domain of the radius is 0 < r < 6, so we check the endpoints and the critical number.
f(0) = 0
f(4) = Max volume of x = 4 in
f(6) = 0

26. 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.

27. 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

28. 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

29. 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

30. Example 5
Let’s take a look at the graph of this equation.

31. Example 5
Now we find the derivative and check the critical number.

32. Example 5
Now we find the derivative and check the critical number.

33. Example 5 Now we find the derivative and check the critical number.
____-___|___+___
5.4
min

34. Example 6
Find a point on the curve y = x2 that is closest to the point (18, 0).

35. 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

36. 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?

37. Example 6
Let’s look at the two derivatives together.

38. Example 6
We will now take the derivative of L2 to find any critical points

39. 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

40. 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

41. Homework
Section 4.5
Pages
1, 3, 5, 11, 19, 21, 23