Chapter 7 Optimization
One of the most important applications of the derivatives is in the optimization of a quantity. In realworld applications, a quantity will often need to be maximized or minimized. For example, a
manufacturing company may need to know what level of production will yield the maximum profit for
the company. A doctor may need to find out the minimum dose of a drug that will effect to her patient.
A farmer may be interested in finding the right dimension of his pasture in order to maximize the area or
to minimize the cost. Before we start to solve these real-world problems, we need to learn how to locate
the absolute (global) maximum and the absolute (global) minimum of a function.
.
Recall that a relative (local) maximum or a relative (local) minimum value of a function is a larger or
smaller value compared with nearby function values. However, an absolute maximum of a function is
the largest value in its domain. Similarly, an absolute minimum is the smallest value in its domain. A
precise definition of the absolute extrema (absolute maximum and absolute minimum) of a function
follows.
Table 7.1 Definition of Absolute Extrema
If f (c)  f ( x) for all x in the domain of f , then f (c) is called the absolute maximum value of f.
If f (c)  f ( x) for all x in the domain of f , then f (c) is called the absolute minimum value of f.
Absolute Extrema on a Closed Interval
The following theorem guarantees the existence of the extrema of a function that is continuous on a
closed interval [a, b].
Table 7.2 Extreme Value Theorem
If a function is continuous on a closed interval [a, b], then the function must have both an
absolute maximum and an absolute minimum on [a, b].

Figure 7.1 Absolute Extrema on a Closed Interval
●
Relative
maximum
●
Absolute
maximum
●
Relative
minimum
●
●
Absolute
minimum
a
b
From Figure 7.1, we see that if a function is continuous on a closed interval [a, b], then the extreme value
of f occurs either at a relative extreme or at one of the endpoints of the interval [a, b]. To find the
absolute extrema, or optimize the function, we take the following steps.
Table 7.3 Steps for Locating the Absolute Extrema of f on [a, b]
If f is continuous on [a, b], then to find the absolute extrema of f on [a, b],
1. Find all the critical values in (a, b).
2. Evaluate f(x) at the critical values and at the endpoints a and b.
3. The largest value found in step 2 is the absolute maximum of f on [a, b], and the smallest
value found in step 2 is the absolute minimum of f on [a, b].
Example 7.1 Find the absolute extrema of f ( x)  2x3  3x2  12x on each interval: [-4. 2] and [-1, 2].
Solution
Since f is a polynomial, f is continuous on each of the intervals. The Extreme Value
Theorem guarantees both maximum and minimum values on each of the intervals. Let us find them by
following the steps in Table 7.3.
Step 1. Find all critical values in (a, b).
f ( x)  2 x3  3x 2  12 x
f '( x)  6 x 2  6 x  12
 6( x 2  x  2)
 6( x  2)( x  1)
The first derivative is zero when x = -2 and x = 1. Hence, f(x) has two critical values: x = -2 and x = 1.
Step 2. Evaluate f(x) at the critical values and at the endpoints a and b.
Both critical values lie in the interval (-4. 2), so we need to compare four values: f (4) , f (2) , f (1) ,
and f (2) :
x
-4
-2
1
2
The left endpoint
The critical values
The right endpoint
f(x)
-32
20
-7
4
The absolute minimum of f
The absolute maximum of f
Step 3: In the above table, the largest number is 20 and the smallest number is -32. The absolute
maximum value of the function on [-4, 2] is f (2)  20 , and the absolute minimum value of the function
on [-4, 2] is f (4)  32 . Figure 7.2 confirms our answer.
 Figure 7.2 Absolute Extrema of f on [-4, 2]
(-2, 20)
(-4, -32)
Only one critical value, -2, lies in the interval [-1, 3], so we need to compare three values: f (1) , f (2) ,
and f (3) :
x
f(x)
The left endpoint
-1
The critical values
1
13
-7
The absolute minimum of f
The right endpoint
3
45
The absolute maximum of f
The absolute maximum is 45 and occurs at x = 3. The absolute minimum is -7 and occurs at x = 1. See
Figure 7.3

Figure 7.3 Absolute Extrema of f on [-1, 3]
(3, 45)
(1, -7)

Absolute Extrema on an Open Interval
The Extreme Value Theorem guarantees that a continuous function on the closed interval [a, b] has an
absolute maximum and an absolute minimum on [a, b]. However, what if the function is continuous on
an open interval (a, b) (excluding its endpoints), or on the entire real line which has no endpoints? Figure
7.4 demonstrates that a function that is continuous in an open interval may or may not have an absolute
extrema.
 Figure 7.4
○
b
a
○
No absolute extrema on (a, b).
Absolute maximum and absolute
minimum exist on (∞, ∞).
Absolute minimum exist, no
absolute maximum on (∞, ∞).
One of the best methods to find the absolute extremes of a continuous function on an open interval is to
find values of the function at all the critical values and sketch the graph (See Chapter 6—Curve
Sketching). However, if the function has only one critical value in the interval, the following Second
Derivative Test can quickly find the absolute extreme of the function.
Table 7.4 Second Derivative Test
When f is continuous on an interval I and c is the only critical value in I.
1. If f ''(c)  0 , then f(c) is an absolute
minimum of f on I.
(
2. If f ''(c)  0 , then f(c) is an absolute
maximum of f on I.
)
(
c
)
c
3. If f ''(c)  0 , then the test fails.
The Second Derivative Test may be used on any interval (closed, open, or half-closed) that contains only
one critical value.
Example 7.2 Find the absolute extreme on the interval (0, ∞) for f ( x)  2xe.2 x .
Solution To find the absolute extreme of a function, we first need to find the critical values.
f '( x)  2e.2 x  2 xe.2 x  .2 
 2e.2 x (1  .2 x)
Product rule and Chain rule
Factor out 2e.2 x
If f '( x)  0 , then 2e.2 x (1  .2x)  0 . Since e.2 x is always bigger than zero, we find that 1  .2 x  0 , or x
= 5. Thus, f(x) only has one critical value in (0, ∞). Also
f ''( x)  2e.2 x (.2)  2e.2 x (.2)  2 xe .2 x (.2) 2
 2e.2 x (.2)[1  1  x(.2)]
 .4e.2 x (2  .2 x)
f ''(5)  .4e.25 (2  .2  5)  .4e1
and e1  0 , we have f ''(5)  0 . Using the Second Derivative Test, the function has an absolute
maximum at x = 5, and the absolute maximum value is f (5)  3.679 . See the following figure.
 Figure 7.5 Graph of f ( x)  2xe.2 x
(5, 3.679)
f ( x)  2xe.2 x

Applications
We now turn our attention to illustrate how calculus can be used to solve real-world optimization
problems.
Example 7.3 Maximizing Area
Lisa has 50 yards of fencing with which she wants to fence in three sides of a rectangular vegetable
garden. The wall of the house will form the fourth side. Find the dimensions of the rectangle such that the
largest garden Lisa can fence.
Solution In order to get an idea of what is happening, Lisa might draw several rectangles that each
use 50 yards of fencing. Figure 7.7 (not to scale) shows three possible plans of her garden.
 Figure 7.6
14
18
30
16
16
10
10
House
House
Area  10  30  300 yards2
18
18
House
Area  16 18  288 yards2
Area  18 14  252 yards2
Unfortunately, there are infinitely many rectangles like those above. How can Lisa find which one has
the greatest area?
Figure 7.7 illustrates a general rectangle with x as the width and y as the length (in yards). The phrase
“largest garden” indicates that our object is to maximize the area, A, which is the product of width and
length. Hence, our objective equation is A  x  y . However, Lisa only has 50 yards of fencing, so we
have the constraint equation 2 x  y  50 .
 Figure 7.7
y
x
A
Hous
e
x
Therefore, our optimization problem can be formulated as:
A x y
50  2 x  y
Objective Equation
Constraint Equation
We cannot maximize A directly from the objective equation, because A is a function of two variables. But
from the constraint equation, we find y  50  2 x . By plugging y  50  2 x into the objective equation, A
is given as a function only of x:
A( x)  x  (50  2x)  50x  2x2 ,
0  x  25
Note that the domain of A(x) will be [0, 25] (otherwise A  0 ). Now, we can use the techniques we
learned from this chapter to maximize A(x).
A '( x)  50  4 x
A ''( x)  4  0
From A '( x)  0 , we find the critical value, x = 12.5, (only one) and that A ''( x)  0 for all x (A(x) is always
concave up). Therefore, A(x) will attain its absolute maximum at x = 12.5. Replacing x by 12.5 in
y  50  2 x , we find y = 25. Thus, if Lisa chooses the following dimensions:
x  12.5 yards
y  25 yards
the garden will attain the largest area: A  x  y  12.5  25  312.5 yards2 .
There are several alternative ways to determine the absolute extreme. In this problem, because A(x) is
continuous on the closed interval [0, 25], the maximum value of A exists and must occur either at a
critical value or at an endpoint of the interval by extreme value theorem. Since both A(0) and A(25) = 0,
A(12.5) = 312.5 must be the absolute maximum of A(x).
Also, since A '( x)  0 for all x < 12.5, means A(x) is increasing for all x < 12.5, and A '( x)  0 for all x >
12.5, means A(x) is decreasing for all x >12.5, A(12.5) must be the absolute maximum value of A. This is
called the first derivative test for absolute extrema.
From algebra, we know that A( x)  50 x  2 x2 is a quadratic function. See Figure 7.8.
 Figure 7.8 Graph of A( x)  50 x  2 x2
(12.5, 312.5)
For a quadratic function f ( x)  ax2  bx  c , f(x) has an absolute maximum value at x   b if a  0 . In
2a
this case, x   50  12.5 and A(12.5)  312.5 . This is the same answer we achieved using calculus.
2(2)

It is not possible to give an explicit procedure for solving optimization problems; However, Example 7.3
suggests a useful procedure.
Table 7.5 Suggestions for Solving Optimization Problems
Steps:
In Example 7.3:
1. Sketch a picture, if possible. Identify the
unknowns with a symbol.
Let x be the width and y be the length of the
garden.
2. Decide which quantity Q is to be maximized or
minimized.
The area of the garden, A, is to be
maximized.
3. Determine the objective equation that expresses
Q as a function of the variables involved in the
problem and the constraint equation(s) (if any)
that expresses limitations on related variables.
Objective equation: A  x  y
Constraint equation: 2 x  y  50
4. Use the constraint equation to simplify the
objective equation in such a way that Q becomes
a function of only one variable. Find the domain
of this function.
A is a function only for x:
A( x)  x  (50  2x)  50x  2x2 , 0  x  25
5. Find the absolute maximum (or minimum) of the
objective function by using the techniques we
have learned with derivatives.
6. Answer the question in the problem.
Example 7.4. Minimizing a Surface
See details from last page.
If Lisa chooses the dimensions:
x  12.5 yards and y  25 yards , the garden
will attain the largest area: A = 312.5 yards2.
A large soup can is to be designed so that the can will hold 50 cubic inches (about 28 ounces) of soup.
Find the dimensions that will minimize the cost of the metal to manufacture the can.
Solution Let us follow the suggested steps in Table 7.5.
Step 1. Sketch a picture and identify the unknowns.
 Figure 7.9
2π r
r
h
h
r
Side unrolled
Step 2. Let S be the total surface area of the cylinder (top, bottom, and side). In order to minimize the cost
of the metal, we need to minimize the total surface of the cylinder S.
Step 3. The objective equation is
S  top + bottom + side
  r 2   r 2  2 r  h .
 2 r 2  2 rh
The volume of the can must be 50 cubic inches, so the constraint equation is 50   r 2 h .
Therefore, our optimization problem can be formulated as:
S  2 r 2  2 rh
Objective Equation
50   r h
2
Constraint Equation
Step 4. To eliminate one variable from the objective equation, solve for h in terms of r using the
50
constraint equation. Since h  2 , we can substitute this into the expression for S to get
r
S (r )  2 r 2  2 r
50
 r2
 2 r 2 
100
r
r 0
Note that r cannot be zero, otherwise S(r) will be undefined. Moreover, in order to make a real can, r
must be positive.
Step 5. Find the minimized surface of the can using calculus.
To find the critical values, rewrite S(r) as S (r )  2 r 2  100 r -1 and differentiate it:
S '(r )  4 r 
100
r2
100
25
, so the only critical value is r  3
 2 . For any r > 0
2

r
200
25
S ''(r )  4  3  0 (S(r) is always concave up), S(r) has an absolute minimum at r  3
 2 . The

r
25
value of h corresponding to r  3
 2 is
Then note that S '(r )  0 when 4 r 

h
50
50
25
253
25
3


2

2

 2 3
 2r  4
2
2
2
3
2
3
3
25
25
25
 r  ( )
 ( )

 ( )
Step 6. Answer the question in the problem.
To minimize the cost of the can, the radius should be r  3
25

 2 inches, and the height should be twice

the radius: 4 inches.
 Example 7.5 Maximizing Revenue and Profit

 A company manufactures and sells x DVD players per week. Assume that the weekly cost and demand
equations are
C ( x)  20 x  10,000
p( x)  100  .01x
0  x  10,000
(a) Find the maximum revenue, the production level that will maximize the revenue, and the price that
the company should charge for each player to maximize the revenue.
(b) Find the maximum profit, the production level that will maximize the profit, and the price that the
company should charge for each player to maximize the profit.
Solution (a) The revenue received for selling x players at $p per player is R  x  p . Our object is to
maximize R while satisfying the price equation, p( x)  100  .01x . So the mathematical problem is
R  x p
p ( x)  100  .01x
Objective equation
Constraint equation
We should start with substituting p(x) into R,
R( x)  100x  .01x2
Now
0  x  10,000
R '( x)  100  .02 x
R ''( x)  .02  0
By setting R '( x)  0 , we have 100  .02 x  0 or x = 5000. Since x = 5000 is the only critical value for R(x)
and the second derivative is a negative constant, R(x) has an absolute maximum when x = 5000 and the
maximum revenue is
R(5000)  100  5000  .01 (5000)2  $250,000
To realize the maximum revenue, the price should be p(5000)  100  .01  5000  $50 .
(b) The profit received for selling x players is the difference between the revenue and the cost:
P( x)  R( x)  C ( x)
 (100 x  .01x 2 )  (20 x  10,000)
 80 x  .01x 2  10,000
Now we find
P '( x)  80  .02 x
P ''( x)  .02  0
By setting P( x)  0 , we have x = 4000. Since x = 4,000 is the only critical value and P ''( x)  0 for all x,
P(x) will be maximized when x = 4000.
By finding P(4000)  $150,000 and p(4000)  100  .01  4000  $60 , we know that if the company
produces 4,000 players and sold each one for $60 each week, the maximum weekly profit will be
$150,000.
From the results of (a) and (b), we notice that the maximum revenue and maximum profit don’t occur at
same production level. We also note that profit is maximized when
P '( x)  R '( x)  C '( x)  0
or R '( x)  C '( x)
Thus profit is maximized at a production level for which the marginal revenue equals marginal cost (the
slopes of the two curves are the same at this point). See the following figure.
 Figure 7.10
m = 20
C ( x)  20 x  10,000
Maximum
profit
Maximum
revenue
Example 7.6
R( x)  100x  .01x2
Maximizing Sustainable Harvest
A farmer estimates from past records that if 20 pecan trees are planted per acre, each tree on average will
yield 78 pounds of nuts per year. For each additional tree planted per acre (up to 10), the average yield
per tree drops 3 pounds due to overcrowding. How many trees should be planted to maximize the yield
per acre? What is the maximum yield?
Solution Let x be the number of pecan trees the farmer should plant per acre and y be the yield per
tree. Then, the yield per acre, denoted as Y, will be
Yield per acre = (Total number of tree per acre)  (Yield per tree)
Y  x y
Since each additional tree per acre causes the yield to drop 3 pounds per tree, the yield per tree, y, is a
linear function of the number of trees per acre. Using the point-slope formula of a line with the point (78,
20) and the slope -3, we have
y  78  3( x  20) or y  3x  138
Therefore, our optimization problem will be
Y  xy
y  3 x  138
Objective equation
Constraint equation
Plugging the constraint equation into the objective equation, the objective equation is reduced to a
function only of x:
Y ( x)  x  (3x  138)
 3x 2  138 x
0  x  30
Now we must find x such that Y(x) will be maximized. Since
Y '( x)  6 x  138
Y ''( x)  6  0 for all x
Y(x) has only one critical value when x = 23. The second derivative test implies that if 23 trees are planted
per acre, the yield will be maximized with a value of Y (23)  1,587 pounds per acre .

Inventory control is a very important question for many companies. For example, a furniture store expects
to sell 700 sofas next year. The manager faces the following problem: How often should she order sofas,
and how large should each order be to meet customer demand at all times and to minimize the cost of
delivery and storage at same time?
In such problems, we assume that the sofas sell steadily throughout the year and the new order arrives
when the stock is depleted. The manager could decide to order all 700 sofas at the beginning of the year.
This would certainly minimize the delivering cost but would result in very large storage cost. At the other
extreme, she could order 2 sofas (assume the store opens 350 days per year, then 700 / 350 = 2) daily.
This would minimize the storage cost but would result in very large delivery cost. Between these two
extremes, she could order these sofas at any times with different lot sizes. The inventory throughout the
year would look like the graph as follows. See Figure 7.11.
 Figure 7.11
Inventory
700
Inventory
Average
number in
storage 350
Average
number in
storage 87.5
350
175
12 months
One order of 700 sofas
3
6
9
12 months
Four orders of 175 sofas
Let’s determine the best lot size that minimizes the total inventory cost in Example 7.7.
Example 7.7
Minimizing Inventory Costs
A furniture store expects to sell 700 sofas next year. The cost for each delivery is $450 (a fixed cost per
delivery), and the storage cost for each sofa per year is around $30. How large should each order be and
how often should be ordered per year in order to minimize the inventory costs?
Solution Let x be the number of sofas in each order and y be the number of orders. Assuming that
each shipment arrives just as the previous shipment has been sold, then the average number of sofas in
store is x/2 over a year. See Figure 7.12.

Figure 7.12
Inventory
x
Average
number in
storage
x
2
1st order
arrives
2nd order
arrives
3rd order
arrives
Since it costs $30 to store a sofa, the total storage cost is 30 
x
, and each delivery costs $450, the total
2
delivery cost is 450y . Thus
the total Inventory Cost = delivery cost + storage cost
x
C = 450 y  30 
2
The store expects to sell 700 sofas in a year, so the constraint equation is xy  700 . Our mathematical
problem is
x
C = 450 y  30 
Objective equation
2
xy  700
Constraints equation
C is a function of two variables, but from the constraints equation, we find y 
700
. Substituting into the
x
objective equation gives
C ( x)  450 
700
 15 x  315000 x 1  15x
x
Now, C is a function only of x
C '( x)  315000 x 2  15  
Setting C '( x)  0 , we can solve it for x:
315000
 15
x2
0  x  700

315000
 15  0
x2
315000
 15
x2
x
315000
 144.9
15
- 144.9 is not a critical value, since 0  x  700 , leaving x = 144.9 as the only critical value of C(x). Next,
we find
C ''( x)  630000  0
The second derivative test implies that C(x) has an absolute maximum at the critical point x = 144.9 .
Since x represents the lot size, we round to the nearest integer 145. If there are 145 sofas per order, the
yearly total of 700 sofas would need 700 /145  4.83  5 orders per year. Therefore, to minimize
inventory costs, the store should place 5 orders throughout of the year with 145 sofas per order. In fact, if
the store place 5 orders a year with 140 sofas per order, then 140  5  700 . The difference is due to roundoff mistakes.
Sample Quiz
1. Find the absolute extrema of f ( x)  3x4  4x3  12 x2  10 on [-3, 2] if they exist.
2. Repeat the above problem for the interval [-1, 1].
3. Find the value(s) at which f ( x)  x2  2 attains absolute extreme on (-∞, ∞).
4. The quantity of a drug in the bloodstream t hours after injection is given, in mg, by
Q(t )  50(et  e3t )
When is the maximum quantity of the drug in the bloodstream?
5. A farmer wants to fence in a rectangular pasture adjacent to a river. The pasture must contain 180,000
square yards in order to provide enough grass for the herd. What dimensions should the farmer choose to
minimize the cost of fencing if no fencing is needed along the river?
6. At a price of $50 per person for a day trip, the company attracts 200 tourists. For every $1 decrease in
price, an additional 5 tourists are attracted. What price should the company charge to maximize the
revenue?
7. The total cost C(x) of producing x products is given by
C( x)  0.02x2  0.9x  5000
How many products should the company produce in order to minimize the average cost (Hint: the average
C ( x)
cost, C or AC, is defined as C 
).
x
8. A restaurant sells x hamburger per week. Assume that the weekly cost and demand equations are
C ( x)  1.1x  300
p( x)  5  .003x 0  x  1000
Find the number of hamburgers that the restaurant should sell and the price that the restaurant should
charge for each hamburger to maximize the profit.
9. A box with an open top is to be constructed from a square piece of cardboard, 5 ft wide, by cutting out
a square from each corner and folding up the sides. What should x be to maximize the volume?
x
x
5
5
10. A local garage estimates using the past records that 2000 tires will be steadily needed next year. The
delivery cost for each shipment of tires is $250 and the cost of storing each tire for a year is about $9.
Assuming that each shipment of tires is used up before the next shipment arrives, determine how many
orders the garage should place next year and how many tires should be in each shipment in order to
minimize the delivery and storage cost.
Answer to Sample Quiz
1. absolute maximum is f(2) = 42, absolute minimum is f(-2) = 22
2. absolute maximum is f(0) = 10, absolute minimum is f(-1) = 3
3. absolute minimum at x = 0, no absolute maximum
4. 0.55 hours. About half of an hour.
5. 600 yards × 300 yards
6. $45
7. 500
8. x = 650, price is $3.05
9. x = 5/6 ft
10. The garage should place 6 orders a year with 333 tires in each shipment.