Transcript Applied Max and Min Problems

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:
1) Problems that reduce to maximizing or minimizing
a continuous function over a finite closed interval.
2) 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 < x < 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
___+___|___-___
25
max
Example 1
•
•
•
•
___+___|___-___
25
max
We must now check the endpoints as well as the
critical numbers.
f(0) = 0
f(25) = 625
f(50) = 0
Example 1
•
•
•
•
___+___|___-___
25
max
We must now check the endpoints as well as the
critical numbers.
f(0) = 0
f(25) = 625 Maximum area of 625 ft2 @ x = 25
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 < x < 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 @ x = 10/3
• 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 x 2  25
• The distance from P to B is (8 – x)
• We are minimizing cost, so we need
a cost equation.


1
c  1 x  25  (8  x)
2
2
Example 3
• Now, we take the derivative to find the critical
numbers.
dc
x
1
1


c  1 x 2  25  (8  x)
2
dx
x  25 2
2


x
1

2
x  25 2
4 x  x  25
2
2
2 x  x 2  25
25
x 
3
2
5
x
3
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
5
 5 
• f    8.33 Min cost of 8.33 million @ x =
3
 3
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
V  r h
2
Example 4
• Again, we need to eliminate one of the variables. We
will do this with similar triangles.
10  h 10

r
6
5
h  10  r
3
V  r h
2
5 
5 3

2
V  r 10  r   10r  r
3 
3

2
Example 4
• Now, we take the derivative and find the critical
points.
5 3
V  10r  r
3
dV
 5r (4  r )
dr
2
dV
 20r  5r 2
dr
Example 4
• The domain of the radius is 0 < r < 6, so we check the
endpoints and the critical number.
dV
 5r (4  r )
dr
• f(0) = 0
• f(4) = 167.55 Max volume of 167.55 in3 @ x = 4
• 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
S  2r  2rh
2
1000  r 2 h
1000
h
r 2
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
S  2r  2rh
2
1000  r 2 h
1000
h
r 2
 1000 
S  2r  2r  2 
 r 
2
2000
S  2r 
r
2
Example 5
• Now we find the derivative and check the critical
number.
dS
2000
2000
2
 4r  2
S  2r 
dr
r
r
2000
r 
4
3
____-___|___+___
5.4
min
r  5.4cm
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: L  ( x  18) 2  ( y  0) 2
• Since y = x2, we can say that
L  ( x  18) 2  x 4
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: L  ( x  18) 2  ( y  0) 2
• Since y = x2, we can say that
L  ( x  18) 2  x 4
• We are told that the distance L and the square of the
distance L2 are minimized at the same value.
Example 6
• We will now take the derivative of L2 to find any
critical points
L  ( x  18) 2  x 4
dS
 4 x 3  2 x  36
dx
S  L  x  x  36 x  324
2
4
2
Example 6
• We will now take the derivative of L2 to find any
critical points
L  ( x  18) 2  x 4
S  L  x  x  36 x  324
2
4
2
dS
 4 x 3  2 x  36
dx
• 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
L  ( x  18) 2  x 4
S  L  x  x  36 x  324
2
4
2
dS
 4 x 3  2 x  36
dx
• This will factor to (x – 2)(2x2 + 4x + 9) [trust me!]
___-___|___+___ The closest point is (2, 4)
2
min
Homework
• Pages 318-319
• 1, 3, 5, 11, 17, 19, 21, 23