Optimization I.
Dr. Julia Arnold
using Tan’s 5th edition Applied Calculus for the
managerial , life, and social sciences text
“Many real world applications call for finding the
absolute maximum value or the absolute minimum value
of a given function. For example, management is
interested in finding what level of production will yield
the maximum profit for a company; a farmer is
interested in finding the right amount of fertilizer to
maximize crop yield; a doctor is interested in finding
the maximum concentration of a drug in a patient’s
body and the time at which it occurs; and an engineer is
interested in finding the dimension of a container with
a specified shape and volume that can be constructed
at a minimum cost.”1
In the next two sections we will be learning how to find
the optimum shape and size of containers, or land, or
whatever the problem bestows on us to find.
1 From Optimization I, section 4.4, page 345 Tan’s 5th edition Applied Calculus for the managerial ,
life, and social sciences text
First we need to talk about Absolute Extrema
Definition: For a given function f(x), if f ( x)  f (c) for all
x in the domain of f, then f(c ) is called the absolute
maximum value of f.
If f ( x)  f (c) for all x in the domain of f, then f(c ) is
called the absolute minimum value of f.
y
4
3
2
1
-4
-3
-2
-1
1
2
3
4
x
5
-1
-2
-3
-4
Figure 1 shows an absolute
minimum for f(x) at the
origin.
All f(x)
values are
above (0,0)
in Figure 1.
y
4
3
2
1
-4
-3
-2
-1
1
2
3
4
x
5
-1
-2
-3
-4
Figure 2 shows an absolute
maximum for f(x) at (0,2).
All f(x)
values are
below (0,2)
in Figure 2.
Theorem :
If a function f is continuous on a closed interval [a,b], then f has
both an absolute maximum value and an absolute minimum value
on [a,b].
y
4
Continuous Function
3
Absolute Maximum
2
1
-4
-3
-2
[
-1
1
-1
Absolute Minimum
-2
-3
-4
]
2
3
4
x
5
Closed Interval
Theorem :
If a function f is continuous on a closed interval [a,b], then f has
both an absolute maximum value and an absolute minimum value
on [a,b].
This graph shows why the function has to be continuous.
Goes to positive
infinity, no absolute
maximum.
Function not continuous at x = 2
y
4
3
2
1
-4
-3
-2
[
-1
1
2
]
3
4
x
5
-1
-2
Closed Interval
-3
-4
Goes to negative infinity, no
absolute minimum.
Steps for finding the absolute extrema of a function f on a
Closed Interval [a,b].
1. Find the critical points of f that lie in (a,b) (open interval).
2. Compute the value of f at each critical point found in step 1
and also compute f(a) and f(b)
3.
The absolute maximum value and absolute minimum value of
f will correspond to the largest and smallest numbers,
respectively, found in step 2.
Example 1: Find the absolute extrema of the function f(x) = x2
defined on the closed interval [-1,2].
Step 1: Find the critical points.
f’(x) = 2x
0=2x
X = 0 is a critical point
Step 2: Calculate f(-1), f(0), and f(2) a=-1 and b=2, 0 is the
critical point.
f(-1)= 1
f(0) = 0
f(2) = 4
The point (0,0) is the lowest point on the interval, while (2,4) is
the highest point on the interval.
Thus (0,0) is the absolute minimum on [-1,2]
and (2,4) is the absolute maximum on [-1,2]
3
2
f
(
x
)

x
2
x
- 4x  4
Example 2: Find the absolute extrema for
on the interval [0,3]
Step 1:
Finding Critical Points
f ( x)  x 3 - 2 x 2 - 4 x  4
f ( x)  3 x 2 - 4 x - 4
0  (3 x  2)(x - 2)
3 x  2  0, x - 2  0
-2
x
,x  2
3
Step 2:
Substitution into f(x)
f(0),f(2),f(3)
Why not –2/3?
It’s not inside the closed interval.
f (0)  03 - 202 - 40  4  4
 
- 23  - 43  4  27 - 18 - 12  4  1
f (2)  23 - 2 2 2 - 42  4  8 - 8 - 8  4  -4
f (3)  33
2
(0,4) is the absolute maximum
(2,-4) is the absolute minimum
The Sonic Company’s total profit in dollars from
manufacturing and selling x units of their loudspeaker system
is given by
P( x)  -0.02x 2  300x - 200,000...(0  x  20,000)
How many units of the loudspeaker system must Sonic produce
to maximize its profits.
Solution:
P( x)  -0.02x 2  300x - 200,000...(0  x  20,000)
P( x)  -0.04x  300
0  -0.04x  300
0.04x  300
x  7500
Find f(0), f(7500), f(20,000)
 
P(0)  -0.02 0 2  3000 - 200,000  -200,000
P(7500)  -0.02(7500) 2  300(7500) - 200000 925,000
P(200000)  -0.02(20000) 2  300(20000) - 200000 -2,200,000
The largest profit occurs at the critical point x = 7500 and is
$925,000.
A Biology Problem:
When a person coughs, the trachea (windpipe) contracts,
allowing air to be expelled at a maximum velocity. It can be
shown that during a cough the velocity v of airflow is given by
the function v(r )  kr 2 R - r  where r is the trachea’s radius
(in centimeters), R is the trachea’s normal radius (in
centimeters), and k is a positive constant that depends on the
length of the trachea. Find the radius r for which the
velocity of airflow is greatest.
Since a trachea has a maximum size we can assume the closed
interval [0,R]
Solution:
v(r )  kr 2 R - r 
v(r )  kr 2 - 1  2krR - r 
Product Rule
v(r )  -kr 2  2krR - 2kr 2
Remove grouping symbols
0  2krR - 3kr 2
0  r 2kR - 3kr 
Combine like terms and set equal to 0
Factor out r.
r  0,2kR - 3kr  0
Set each factor equal to 0.
r  0,2kR  3kr
2kR
r
3k
2
r  0, R  r
3
r  0,
Solving for r.
2/3 R means that the radius is 2/3 the normal size of the trachea.
Find v(0), v(2R/3), v(R )
v(0)  k 0 R - 0  0
2
2 
2 
v R   k  R 
3 
3 
2
2  4kR 1R 4kR

 
 R - R 
3 
9
3
27

v ( R )  k R  R - R   0
2
minimum
2
3
maximum
minimum
The airflow is greatest when the cough contracts the trachea to a
radius of 2R/3 or when the contraction is approximately 1/3 of the
Radius.
Trachea at rest
Trachea during a cough.
2/3R
Maximizing Profits
The quantity demanded each month of the Walter
Serkin recording of Beethoven’s Moonlight Sonata,
manufactured by Phonola Record Industries, is related
to the price/compact disc. The equation
p  -0.00042 x  6 (0  x  12,000)
Where p denotes the unit price in dollars and x is the
number of discs demanded, relates the demand to the
price. The total monthly cost (in dollars) for pressing
and packaging x copies of this classical recording is
given by
C( x)  600  2x - 0.00002x2 (0  x  20,000)
To maximize its profits, how many copies should Phonola
produce each month?
Maximizing Profits Continued
The equation p is the relationship between the price we
are going to charge for the disc in terms of the quantity
of discs demanded.
So if 1000 discs are demanded, the price would be:
p  -0.00042 x  6 (0  x  12, 000)
p  -0.00042(1000)  6  $5.58 Per disc
To find the Revenue we multiply p by x. (price by number
of items to be sold)
R( x)  px  -0.00042 x 2  6 x (0  x  12, 000)
R( x)  px  -0.00042 x 2  6 x (0  x  12, 000)
C( x)  600  2x - 0.00002x2 (0  x  20,000)
Now that we have both Revenue and Cost we can find the Profit function:
P( x)  R( x) - C ( x)   -0.00042 x 2  6 x  -  600  2 x - 0.00002 x 2 
P( x)  -0.00042 x 2  6 x - 600 - 2 x  0.00002 x 2
P( x)  -0.0004 x 2  4 x - 600
Now that we have the profit function we can proceed to
find its maximum value which will lead us to the number of
discs we should make.
P( x)  -0.0004x2  4x - 600
To find the maximum profit, you find the first derivative, set it equal to 0
and solve for x.
P( x)  -0.0004 x 2  4 x - 600
P( x)  -0.0008 x  4
0  -0.0008 x  4
0.0008 x  4
x
4
 5000
.0008
X = 5000 discs made and sold will maximize the profits
If we look at the graph of the profit function we should see that x =
5000 is at the peak.
y
x
1
2
3
4
5




1
11
12
Self-Check
See if you can do the 3 problems below. If you need help click on the
appropriate link.
1. For the function f(x)  x - 2 x
a. Find the absolute extrema of f on [0,9]
b. Find the absolute extrema of f(x).
Hint
Answer
Completely Worked
2. Find the absolute extrema of f(x)  3x4  4x3  1 on [-2,1]
Hint
Answer
Completely Worked
3. The operating rate(expressed as a percentage) of factories, mines, and
utilities in a certain region of the country on the tth day of the year 2000 is given
by the function f(t)  80  1200t
(0  t  250)
t2  40,000
On which day of the first 250 days of 2000 was the manufacturing capacity
operating rate highest?
Hint
Answer
Completely Worked
End presentation
Hint:
Step 1 Take the first derivative of the function, then set it equal to 0
and solve.
Back to questions
Answers
1. a. (1,-1) absolute minimum and (9,3) absolute maximum
b. (1,-1) is the absolute minimum, no absolute maximum
Back to questions
Answers
2. (-1,0) absolute minimum and (-2,17) absolute maximum
Back to questions
Answers
3. The 200 th day
Back to questions
Complete Solution
1. a Step 1
f(x)  x - 2 x
f(x)  x - 2x
2
f(x)  1 - x
2
1
0 1x
Rewrite for the derivative
1
2
-1
2
1-
1
x
[0,9] and critical point x = 1
Use the power rule.
f(0)  0 - 2 0  0
f(1)  1 - 2 1  1 - 2  -1
Multiply by denominator
f(9)  9 - 2 9  9 - 6  3
(1,-1) Absolute minimum
(9,3) Absolute maximum
0  x -1
1 x
1x
Step 2
Square both sides.
y
4
3
1. B The domain over the
real numbers is x  0
(1,-1) is an absolute
minimum for the entire
function, but no max.
2
1
-4
-3
-2
-1
1
2
3
4
5




x
1
-1
-2
-3
-4
Back to questions
Complete Solution
2. Step 1
Step 2
f(x)  3x 4  4x3  1
f(x)  12x3  12x2
0  12x  12x
3
2
0  12x x  1
2
0  12x 2 , x  1  0
0  x , x  -1
Critical Values
Use the power rule.
[-2,1] and critical point x = -1,0
f(-2)  3- 2  4- 2  1  48 - 32  1  17
4
Factor GCF
Set equal to 0
3
f(-1)  3- 1  4- 1  1  3 - 4  1  0
4
3
f(0)  30   40   1  1
4
3
f(1)  31  41  1  3  4  1  8
(-1,0) Absolute minimum
(-2,17) Absolute maximum
4
3
Back to questions
Complete Solution
3. Step 1
f(x)  80 
f(x) 
t
2
1200t
t2  40000
Step 2

 400001200 - 1200t(2t)
t

2

 40000
2

Use the quotient
rule.
0  t2  400001200 - 1200t(2t) Only numerator
0  1200[t2  40000- 2t2 ]

0  1200 40000- t2
0  40000- t2

can be 0
Factor and divide by 1200
on both sides
t2  40000 Solve for t2
t  200
0,250] and critical point x = 200
12000 
 80
0  40000
1200200
f(200)  80 
 83
2002  40000
1200250
f(250)  80 
 82.92
2502  40000
f(0)  80 
2
(200,83) Absolute maximum
On the 200 th day of 2000 the
manufacturing capacity is highest.
Back to questions
If you have any questions or comments about
this presentation, please contact Dr. Julia Arnold
at jarnold@tcc.edu