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## Applied Max and Min Problems

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### Applied Max and Min Problems Objective: To use the methods of this chapter to solve applied optimization problems. – PowerPoint PPT presentation

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Title: Applied Max and Min Problems

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 lt x lt 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 lt x lt 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 ft2 _at_ 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 lt x lt 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) 726 Max volume of 726 in3 _at_ 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 lt x lt 8 so
• F(0) 9
• F(8) 9.433
• Min cost of 8.33
million _at_ 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.
• 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 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

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
• Lets 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
• Lets 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 283-284
• 1, 3, 5, 11, 19, 21, 23