# Optimization Problems using Derivatives Contributed by: In this section, we will learn:
How to solve problems involving maximization and minimization of factors.
1. 4
APPLICATIONS OF DIFFERENTIATION
2. APPLICATIONS OF DIFFERENTIATION
The methods we have learned in this
chapter for finding extreme values have
practical applications in many areas of life.
 A businessperson wants to minimize costs and
maximize profits.
 A traveler wants to minimize transportation time.
 Fermat’s Principle in optics states that light follows
the path that takes the least time.
3. APPLICATIONS OF DIFFERENTIATION
4.7
Optimization Problems
In this section, we will learn:
How to solve problems involving
maximization and minimization of factors.
4. OPTIMIZATION PROBLEMS
In this section (and the next), we solve
such problems as:
 Maximizing areas, volumes, and profits
 Minimizing distances, times, and costs
5. OPTIMIZATION PROBLEMS
In solving such practical problems, the
greatest challenge is often to convert the word
problem into a mathematical optimization
problem—by setting up the function that is
to be maximized or minimized.
6. OPTIMIZATION PROBLEMS
Let’s recall the problem-solving
principles discussed in Chapter 1 and
7. STEPS IN SOLVING OPTIMIZATION PROBLEMS
Thus, there are six steps involved in
solving optimization problems.
These are as follows.
8. 1. UNDERSTAND THE PROBLEM
Read the problem carefully until it is
clearly understood.
 What is the unknown?
 What are the given quantities?
 What are the given conditions?
9. 2. DRAW A DIAGRAM
In most problems, it is useful to draw
a diagram and identify the given and
required quantities on the diagram.
10. 3. INTRODUCE NOTATION
Assign a symbol to the quantity that
is to be maximized or minimized.
 Let’s call it Q for now.
11. 3. INTRODUCE NOTATION
Also, select symbols (a, b, c, . . . , x, y)
for other unknown quantities and label
the diagram with these symbols.
 It may help to use initials as suggestive symbols.
 Some examples are: A for area, h for height,
and t for time.
12. 4. EXPRESS Q IN TERMS OF THE VARIABLES
Express Q in terms of
some of the other symbols
from Step 3.
13. 5. EXPRESS Q IN TERMS OF ONE VARIABLE
If Q has been expressed as a function of
more than one variable in Step 4, use the
given information to find relationships—in the
form of equations—among these variables.
Then, use the equations to eliminate all but
one variable in the expression for Q.
14. 5. EXPRESS Q IN TERMS OF ONE VARIABLE
Thus, Q will be expressed as
a function of one variable x, say,
Q = f(x).
 Write the domain of this function.
15. 6. FIND THE ABSOLUTE MAX./MIN. VALUE OF f
Use the methods of Sections 4.1 and 4.3
to find the absolute maximum or minimum
value of f.
 In particular, if the domain of f is a closed interval,
then the Closed Interval Method in Section 4.1
can be used.
16. OPTIMIZATION PROBLEMS Example 1
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?
17. OPTIMIZATION PROBLEMS Example 1
In order to get a feeling for what
is happening in the problem, let’s
experiment with some special cases.
18. OPTIMIZATION PROBLEMS Example 1
Here are three
possible ways of
laying out the 2400 ft
of fencing.
19. OPTIMIZATION PROBLEMS Example 1
We see that when we try shallow, wide
fields or deep, narrow fields, we get
relatively small areas.
 It seems plausible that there is some intermediate
configuration that produces the largest area.
20. OPTIMIZATION PROBLEMS Example 1
This figure
the general case.
We wish to maximize the area A of
the rectangle.
 Let x and y be the depth and width of the rectangle
(in feet).
 Then, we express A in terms of x and y: A = xy
21. OPTIMIZATION PROBLEMS Example 1
We want to express A as a function of
just one variable.
 So, we eliminate y by expressing it in terms of x.
 To do this, we use the given information that
the total length of the fencing is 2400 ft.
 Thus, 2x + y = 2400
22. OPTIMIZATION PROBLEMS Example 1
From that equation, we have:
y = 2400 – 2x
This gives:
A = x(2400 – 2x) = 2400x - 2x2
 Note that x ≥ 0 and x ≤ 1200 (otherwise A < 0).
23. OPTIMIZATION PROBLEMS Example 1
So, the function that we wish to maximize
is: A(x) = 2400x – 2x2 0 ≤ x ≤ 1200
 The derivative is: A’(x) = 2400 – 4x
 So, to find the critical numbers, we solve: 2400 – 4x = 0
 This gives: x = 600
24. OPTIMIZATION PROBLEMS Example 1
The maximum value of A must
occur either at that critical number or
at an endpoint of the interval.
 A(0) = 0; A(600) = 720,000; and A(1200) = 0
 So, the Closed Interval Method gives the maximum
value as:
A(600) = 720,000
25. OPTIMIZATION PROBLEMS Example 1
Alternatively, we could have observed that
A’’(x) = –4 < 0 for all x
So, A is always concave downward
and the local maximum at x = 600 must be
an absolute maximum.
26. OPTIMIZATION PROBLEMS Example 1
Thus, the rectangular field should
 600 ft deep
 1200 ft wide
27. OPTIMIZATION PROBLEMS Example 2
A cylindrical can is to be made to
hold 1 L of oil.
 Find the dimensions that will minimize
the cost of the metal to manufacture the can.
28. OPTIMIZATION PROBLEMS Example 2
Draw the diagram as in
this figure, where
r is the radius and h the
height (both in
29. OPTIMIZATION PROBLEMS Example 2
To minimize the cost of
the metal, we minimize
the total surface area of
the cylinder (top, bottom,
30. OPTIMIZATION PROBLEMS Example 2
We see that the sides are made from
a rectangular sheet with dimensions
2πr and h.
31. OPTIMIZATION PROBLEMS Example 2
So, the surface
area is:
A = 2πr2 +
32. OPTIMIZATION PROBLEMS Example 2
To eliminate h, we use the fact that
the volume is given as 1 L, which we take
to be 1000 cm3.
 Thus, πr2h = 1000
 This gives h = 1000/(πr2)
33. OPTIMIZATION PROBLEMS Example 2
Substituting this in the expression for A gives:
2  1000  2 2000
A 2 r  2 r  2 
2 r 
 r  r
So, the function that we want to minimize is:
2 2000
A(r ) 2 r  r 0
r
34. OPTIMIZATION PROBLEMS Example 2
To find the critical numbers, we differentiate:
3
2000 4( r  500)
A '(r ) 4 r  2
 2
r r
 Then, A’(r) = 0 when πr3 = 500
 So, the only critical number is: r  3 500 / 
35. OPTIMIZATION PROBLEMS Example 2
As the domain of A is (0, ∞), we can’t use the
argument of Example 1 concerning endpoints.
 However, we can observe that A’(r) < 0 for r  3 500 / 
3
and A’(r) > 0 for r  500 / 
 So, A is decreasing for all r to the left of the critical
number and increasing for all r to the right.
 Thus, r  3 500 /  must give rise to an absolute
minimum.
36. OPTIMIZATION PROBLEMS Example 2
Alternatively, we could
argue that A(r) → ∞
as r → 0+ and A(r) → ∞
as r → ∞.
 So, there must be
a minimum value of A(r),
which must occur at
the critical number.
37. OPTIMIZATION PROBLEMS Example 2
The value of h corresponding to
3
r  500 /  is:
1000 1000 500
h 2  23
2 3 2r
r  (500 /  ) 
38. OPTIMIZATION PROBLEMS Example 2
Thus, to minimize the cost of
the can,
 The radius should be r  3 500 /  cm
 The height should be equal to twice the radius—
namely, the diameter
39. OPTIMIZATION PROBLEMS Note 1
The argument used in the example
to justify the absolute minimum is a variant
of the First Derivative Test—which applies
only to local maximum or minimum values.
 It is stated next for future reference.
40. FIRST DERIV. TEST FOR ABSOLUTE EXTREME VALUES
Suppose that c is a critical number of a
continuous function f defined on an interval.
a. If f’(x) > 0 for all x < c and f’(x) < 0 for all x > c,
then f(c) is the absolute maximum value of f.
b. If f’(x) < 0 for all x < c and if f’(x) > 0 for all x > c,
then f(c) is the absolute minimum value of f.
41. OPTIMIZATION PROBLEMS Note 2
An alternative method for solving
optimization problems is to use implicit
 Let’s look at the example again to illustrate
the method.
42. IMPLICIT DIFFERENTIATION Note 2
We work with the same equations
A = 2πr2 + 2πrh πr2h = 100
 However, instead of eliminating h,
we differentiate both equations implicitly
with respect to r :
A’ = 4πr + 2πh + 2πrh’ 2πrh + πr2h’ = 0
43. IMPLICIT DIFFERENTIATION Note 2
The minimum occurs at a critical
 So, we set A’ = 0, simplify, and arrive at
the equations
2r + h + rh’ = 0 2h + rh’ = 0
 Subtraction gives: 2r - h = 0 or h = 2r
44. OPTIMIZATION PROBLEMS Example 3
Find the point on the parabola
y2 = 2x
that is closest to the point (1, 4).
45. OPTIMIZATION PROBLEMS Example 3
The distance between
the point (1, 4) and 2 2
d  ( x  1)  ( y  4)
the point (x, y) is:
 However, if (x, y) lies on
the parabola, then x = ½ y2.
 So, the expression for d
becomes:
d  ( 12 y 2  1) 2  ( y  4) 2
46. OPTIMIZATION PROBLEMS Example 3
Alternatively, we could have
substituted y  2 x to get d in terms
of x alone.
47. OPTIMIZATION PROBLEMS Example 3
Instead of minimizing d, we minimize
its square:
2 2
d  f ( y )  y  1   y  4 
2 1
2
2
 You should convince yourself that the minimum of d
occurs at the same point as the minimum of d2.
 However, d2 is easier to work with.
48. OPTIMIZATION PROBLEMS Example 3
Differentiating, we obtain:
f '( y ) 2  y  1 y  2( y  4)  y  8
1
2
2 3
So, f’(y) = 0 when y = 2.
49. OPTIMIZATION PROBLEMS Example 3
Observe that f’(y) < 0 when y < 2 and f’(y) > 0
when y > 2.
So, by the First Derivative Test for Absolute
Extreme Values, the absolute minimum
occurs when y = 2.
 Alternatively, we could simply say that, due to
the geometric nature of the problem, it’s obvious that
there is a closest point but not a farthest point.
50. OPTIMIZATION PROBLEMS Example 3
The corresponding value
of x is:
x = ½ y2 = 2
Thus, the point on y2 = 2x
closest to (1, 4) is (2, 2).
51. OPTIMIZATION PROBLEMS Example 4
A man launches his boat
from point A on a bank of
a straight river, 3 km
and wants to reach point
(8 km downstream on
the opposite bank) as
quickly as possible.
52. OPTIMIZATION PROBLEMS Example 4
He could proceed in
of three ways:
 Row his boat directly across
the river to point C and then
run to B
 Row directly to B
 Row to some point D
between
C and B and then run to B
53. OPTIMIZATION PROBLEMS Example 4
If he can row 6 km/h and
run 8 km/h, where should
he land to reach B as
soon as possible?
 We assume that the speed
of
the water is negligible
compared with the speed at
which he rows.
54. OPTIMIZATION PROBLEMS Example 4
If we let x be the distance from C to D,
 The running distance is: |DB| = 8 – x
 The Pythagorean Theorem gives the rowing
distance as: |AD| = x 2  9
55. OPTIMIZATION PROBLEMS Example 4
distance
We use the equation time=
rate
 Then, the rowing time is: x 2  9 / 6
 The running time is: (8 – x)/8
 So, the total time T as a function of x is:
2
x 9 8 x
T ( x)  
6 8
56. OPTIMIZATION PROBLEMS Example 4
The domain of this function T is [0, 8].
 Notice that if x = 0, he rows to C, and if x = 8,
he rows directly to B.
 The derivative of T is: T '( x ) 
x 1

6 x2  9 8
57. OPTIMIZATION PROBLEMS Example 4
Thus, using the fact that x ≥ 0,
x 1
we have: T '( x) 0  
6 x 9 8
2
 4 x 3 x 2  9
2 2
 16 x 9( x  9)
2 9
 7 x 81  x 
7
 The only critical number is: 9 / 7
58. OPTIMIZATION PROBLEMS Example 4
To see whether the minimum occurs at
this critical number or at an endpoint of
the domain [0, 8], we evaluate T at all three
points: T (0) 1.5
 9  7
T  1  8 1.33
 7
73
T (8)  1.42
6
59. OPTIMIZATION PROBLEMS Example 4
Since the smallest of
these values of T 9/ 7
occurs when x = ,
the absolute minimum
value of T must occur
 The figure illustrates
this calculation by
showing the graph
of T.
60. OPTIMIZATION PROBLEMS Example 4
Thus, the man should
land 9/ 7
the boat at a point
(≈ 3.4 km) downstream
from his starting point.
61. OPTIMIZATION PROBLEMS Example 5
Find the area of the largest rectangle
that can be inscribed in a semicircle
62. OPTIMIZATION PROBLEMS E. g. 5—Solution 1
Let’s take the semicircle
to be the upper half of
the circle x2 + y2 = r2 with
center the origin.
 Then, the word
inscribed means
that the rectangle
has two vertices
on the semicircle
and two vertices
on the x-axis.
63. OPTIMIZATION PROBLEMS E. g. 5—Solution 1
Let (x, y) be the vertex
that lies in the first
 Then, the rectangle
has sides of
lengths 2x and y.
 So, its area is:
A = 2xy
64. OPTIMIZATION PROBLEMS E. g. 5—Solution 1
To eliminate y, we use the fact that (x, y)
lies on the circle x2 + y2 = r2.
2 2
 So, y r  x
2 2
 Thus, A 2 x r  x
65. OPTIMIZATION PROBLEMS E. g. 5—Solution 1
The domain of this function is 0 ≤ x ≤ r.
Its derivative is:
2 2 2
2 2 2x 2(r  2 x )
A ' 2 r  x  
2 2 2 2
r  x r  x
 This is 0 when 2x2 = r2, that is x = r / 2 ,
(since x ≥ 0).
66. OPTIMIZATION PROBLEMS E. g. 5—Solution 1
This value of x gives a maximum value of A,
since A(0) = 0 and A(r) = 0 .
Thus, the area of the largest inscribed
rectangle is: 2
 r  r 2 r 2
A  2 r  r
 2 2 2
67. OPTIMIZATION PROBLEMS Example 5
A simpler solution is possible
if we think of using an angle as
a variable.
68. OPTIMIZATION PROBLEMS E. g. 5—Solution 2
Let θ be the angle
shown here.
 Then, the area of the
rectangle is:
A(θ) = (2r cos θ)(r sin θ)
= r2(2 sin θ cos
θ)
= r2 sin 2θ
69. OPTIMIZATION PROBLEMS E. g. 5—Solution 2
We know that sin 2θ has a maximum
value of 1 and it occurs when 2θ = π/2.
 So, A(θ) has a maximum value of r2
and it occurs when θ = π/4.
70. OPTIMIZATION PROBLEMS E. g. 5—Solution 2
Notice that this trigonometric solution
doesn’t involve differentiation.
 In fact, we didn’t need to use calculus at all.
71. APPLICATIONS TO BUSINESS AND ECONOMICS
Let us now look at
and economics.
72. MARGINAL COST FUNCTION
In Section 3.7, we introduced the idea of
marginal cost.
 Recall that if C(x), the cost function, is the cost of
producing x units of a certain product, then the marginal
cost is the rate of change of C with respect to x.
 In other words, the marginal cost function is
the derivative, C’(x), of the cost function.
73. DEMAND FUNCTION
Now, let’s consider marketing.
 Let p(x) be the price per unit that the company
can charge if it sells x units.
 Then, p is called the demand function
(or price function), and we would expect it
to be a decreasing function of x.
74. REVENUE FUNCTION
If x units are sold and the price per unit
is p(x), then the total revenue is:
R(x) = xp(x)
 This is called the revenue function.
75. MARGINAL REVENUE FUNCTION
The derivative R’ of the revenue
function is called the marginal revenue
 It is the rate of change of revenue with respect
to the number of units sold.
76. MARGINAL PROFIT FUNCTION
If x units are sold, then the total profit
is P(x) = R(x) – C(x)
and is called the profit function.
The marginal profit function is P’,
the derivative of the profit function.
77. MINIMIZING COSTS AND MAXIMIZING REVENUES
In Exercises 53–58, you are asked to use
the marginal cost, revenue, and profit
functions to minimize costs and maximize
revenues and profits.
78. MAXIMIZING REVENUE Example 6
A store has been selling 200 DVD burners
a week at \$350 each. A market survey
indicates that, for each \$10 rebate offered to
buyers, the number of units sold will increase
by 20 a week.
 Find the demand function and the revenue function.
 How large a rebate should the store offer to maximize
its revenue?
79. DEMAND & REVENUE FUNCTIONS Example 6
If x is the number of DVD burners sold
per week, then the weekly increase in sales
is x – 200.
 For each increase of 20 units sold, the price
is decreased by \$10.
80. DEMAND FUNCTION Example 6
So, for each additional unit sold, the decrease
in price will be 1/20 x 10 and the demand
function is:
p(x) = 350 – (10/20)(x – 200)
= 450 – ½x
81. REVENUE FUNCTION Example 6
The revenue function is:
R(x) = xp(x)
= 450x – ½x2
82. MAXIMIZING REVENUE Example 6
Since R’(x) = 450 – x, we see that
R’(x) = 0 when x = 450.
 This value of x gives an absolute maximum
by the First Derivative Test (or simply by observing
that the graph of R is a parabola that opens
downward).
83. MAXIMIZING REVENUE Example 6
The corresponding price is:
p(450) = 450 – ½(450) = 225
The rebate is: 350 – 225 = 125
 Therefore, to maximize revenue, the store
should offer a rebate of \$125.