Transcript Section 3.7 – Optimization Problems

Section 3.7 – Optimization
Problems
Optimization Procedure
1. Draw a figure (if appropriate) and label all quantities
relevant to the problem.
2. Focus on the quantity to be optimized. Name it. Find a
formula for the quantity to be maximized or minimized.
3. Use conditions in the problem to eliminate variables in
order to express the quantity to be maximized or
minimized in terms of a single variable.
4. Find the practical domain for the variables in Step 3;
that is, the interval of possible values determined from
the physical restrictions in the problem.
5. If possible, use the methods of calculus to obtain the
required optimum value (typically a max/min on an
interval).
Example 1
River
x
Draw a Picture
x
y
What needs to be Optimized?
Area needs to be maximized:
A  x  2400  2x 
A  2400 x  2 x2
Find the Domain: 0  x  1200
x-intercepts
(no negative area)
A farmer has 2400 ft of fencing and wants to fence off a
rectangular field that borders a straight river (He needs
no fence along the river). What are the dimensions of
the field that has the largest area?
Use Calculus to Solve the Problem
A '  2400  4 x
0  2400  4x x  0 A  0
A  xy
Eliminate Variable(s) with other Conditions 4 x  2400
x  600 A  720000
x

600
P  2x  y
x  1200 A  0
2400  2x  y
y  2400  2  600
y  1200
y  2400  2 x
600ft x 1200 ft
Example 2
Find a positive number such that the sum of the number
and its reciprocal is as small as possible.
Draw a Picture
Find the Domain:
NOT APPLICABLE
Use Calculus to Solve the Problem
S '  1  x12
What needs to be Optimized?
The sum of a number and its
reciprocal needs to be
minimized:
S  x  1x
Eliminate Variable(s) with other Conditions
THE EQUATION ALREADY
HAS ONLY TWO
VARIABLES
It is a minimum!
x0
0  1  x12
x 0 S 
1
1
x 1 S  2
x2
x2  1 Since this is not a closed
interval , check to make
x  1, 1 sure the critical point is a
minimum.

1
x  0.5
1

x2
f '  0.5  3 f '  2  0.75
Example 3
V   45  2x  24  2x  x
V  4 x3  138x 2  1080 x
Find the Domain: 0  x  12
24 – 2x
Draw a Picture
45 – 2x
x
x
x
24
x
45
What needs to be Optimized?
Volume needs to be maximized:
V  lwh
x-intercepts
(no negative volume)
A carpenter wants to make an open-topped box out of a rectangular
sheet of tin 24 in. by 45 in. The carpenter plans to cut congruent
squares out of each corner of the sheet and then bend and solder
the edges of the sheet upward to firm the sides of the box. For what
dimensions does the box have the greatest possible volume?
Use Calculus to Solve the Problem
2
V
'

12
x
 276 x  1080
2
0  12 x  276 x  1080 x  0 V  0
0  x2  23x  90
x  5 V  2450
0   x 18 x  5
x  12 V  0
x  5,18
Eliminate Variable(s) with other Conditions
l  45  2  5  35 w  24  2  5  14
5 in x 14 in x 35 in
Example 4
Draw a Picture
r
h
What needs to be Optimized?
Volume needs to be maximized:


V r
V  150 r  2 r 3
Find the Domain: 0  r  5 3
2
150  2r 2
r
x-intercepts
(no negative volume)
We need to design a cylindrical can with radius r and height h. The top
and bottom must be made of copper, which will cost 2¢/in2. The
curved side is to be made of aluminum, which will cost 1 ¢/in2. We
seek the dimensions that will maximize the volume of the can. The
only constraint is that the total cost of the can is to be 300π cents.
Use Calculus to Solve the Problem
V '  1502  6 r 2
0  150  6 r r  0 V  0
V   r 2h
2
0

25

r
Eliminate Variable(s) with other Conditions
r  5 V  500
2
r  5, 5
C  2 r  2  2 rh 1
2
2
150

2
5


r  5 3V  0
300  4 r  2 rh
h

5
1502r 2
300 4 r 2
r = 5 in & h= 20 in
h  2 r  r
h  20
Example 5
A liquid antibiotic manufactured by a pharmaceutical firm is sold in bulk
at a price of $200 per unit. If the total production cost (in dollars) for x
units is:
C(x) = 500,000 + 80x + 0.003x2
And if the production capacity of the firm is at most 30,000 units in a
specified times, how many units of antibiotic must be manufactured and
sold in that time to maximize the profit?
Find the Domain: 0  x  30000
Use Calculus to Solve the Problem
Draw a Picture
NOT APPLICABLE
What needs to be Optimized?
The profit needs to be
maximized:
P '  x   200   80  0.006 x   120  0.006x
120  0.006 x  0
x  20000
P  x  R  x  C  x
P  x   200 x   50000  80 x  0.003 x 2 
Eliminate Variable(s)
The equation already has one
variable
x  0 P  500000
x  20000 P  700000
x  30000 P  400000
20,000 Units