Lecture Notes 4—CHAPTER 9
Optimization
Any economic activity is undertaken because of the benefit gained from that action. But, at the same time, the
economic activity requires using up resources. That is, the activity entails costs. Net benefit is the difference between
total benefit and total cost of that activity.
When we talk of an economic activity, this activity is one among many different choices or alternatives before the
entity undertaking such activities. Thus, choices are made according to the criterion: which alternative provides the
“maximum net benefit”. Of course, at any given moment, the economic entity is undertaking or planning many
different activities. Thus, to simplify the analysis of choice, we start by looking at the process one decision at a time.
The firm is an economic entity. The principal objective of the firm is to maximize profit by producing the “right”
quantity of output. The right quantity is the optimum quantity, the output level which maximizes profit, with profit
defined as the difference between total revenue earned from producing and selling a given quantity minus the cost of
producing that quantity. The term optimum in this situation implies maximum. Optimum may also imply something
that is being minimized. For example the optimum combination of inputs, say labor and capital, to produce a given
level of output implies the least (minimum) cost combination of inputs.
1. Optimizing an Objective Function Using Derivatives
In the discussion of optimization, the objective function is that in which the dependent variable is the object of
optimization and the independent variable(s) constitute the choice variables. Let’s consider the firm’s profit as our
objective function. To simplify the discussion we will use a numeric example. Profit is defined as the difference
between total revenue and total cost of a given level of output. In functional form, we write the profit function as:
𝜋(𝑄) = 𝑅(𝑄) − 𝐶(𝑄)
Let the firm’s total cost function be,
𝐶 = 𝑄3 − 18𝑄2 + 141𝑄 + 100
Assume the firm is an imperfectly competitive firm (or a monopoly) facing a downward sloping demand curve
represented by,
𝑃 = 150 – 5𝑄
The revenue function is then obtained by multiplying both sides of the demand function by Q
𝑅 = 𝑃𝑄
𝑅 = 150𝑄– 5𝑄2
The profit function is then, 𝜋 = 𝑅 − 𝐶
π = −𝑄3 + 13𝑄2 + 9𝑄 − 100
The graph of this function is shown in Figure 1. Note that profit is maximized at 𝑄 = 6. At the point (6, 260) on the
curve the slope of the line tangent to the curve is zero. We know that the derivative of a function evaluated at a given
point on the curve is the slope of the line tangent to the curve at that point.
Thus, to find the output level at which the profit is maximized, take the first derivative of the profit function with
respect to 𝑄 and set it equal to zero.
𝑑π
= −3𝑄2 + 26𝑄 + 9 = 0
𝑑𝑄
1
3𝑄2 − 26𝑄 − 9 = 0
Using the quadratic formula, we have,
𝑄=
26 + √262 − 4(3)(−9)
=9
2(3)
π = −93 + 13(92 ) + 9(9) − 100
Figure 1
The Profit Function
π = –Q3 + 13Q2 + 9Q – 100
$
9, 305
0
1
2
3
4
5
6
7
8
9 10 11 12 13 14
Q
1.1. Alternative Way of Finding the Optimum Output Using Derivate Functions
Figure 2-a shows the cost and revenue curves drawn in the same graph. Computing the total revenue and total cost
for 𝑄 = 9, we have:
𝑅 = 150(9) − 5(92 ) = 945
𝐶 = 93 − 18(92 ) + 14(9) + 100 = 640
𝜋 = 945 − 640 = 305
2
Figure 2-a
Total Revenue and Total Cost
$
C
R
945
640
Q
9
Figure 2-b
Marginal Revenue and Marginal Cost
$
MC
60
MR
9
Q
Note that in Figure 2-a the lines tangent to 𝑅 and 𝐶 are parallel at 𝑄 = 9. This implies that the slope of the two
functions are equal at the optimum output level, that is, the derivatives of the two functions evaluated at 𝑄 = 9 are
equal.
Thus, to find the optimum output find the derivatives of the cost and revenue functions and then find the 𝑄 at which
the derivatives are equal. The derivatives of the cost and revenue functions are, respectively, the marginal cost (𝑀𝐶)
function and the marginal revenue (𝑀𝑅) function.
𝑀𝐶 =
𝑑𝐶
= 3𝑄2 − 36𝑄 + 141
𝑑𝑄
𝑀𝑅 =
𝑑𝑅
= 150 − 10𝑄
𝑑𝑄
3
𝑀𝐶 = 𝑀𝑅
3Q2 − 36Q + 141 = 150 – 10Q
3Q2 − 26Q – 9 = 0
Q=9
Figure 2-b shows the graphs of 𝑀𝐶 and 𝑀𝑅. They intersect at 𝑄 = 9.
Example
Graph the firm’s average variable cost and show the output level at which 𝐴𝑉𝐶 is minimized.
𝑇𝑉𝐶 = 𝑄3 − 18𝑄2 + 141𝑄
𝐴𝑉𝐶 =
𝑇𝑉𝐶
= 𝑄2 − 18𝑄 + 141
𝑄
Figure 3 shows the graph of 𝐴𝑉𝐶.
Figure 3
𝐴𝑉𝐶 = 𝑄2 − 18𝑄 + 141
$
AVC
60
9
Q
Note, again, that at 𝑄 = 9, the slope of the tangent is zero, implying that the derivative of the average cost function at
𝑄 = 9 is zero. Here the derivative is zero where the function is at its minimum.
𝑑𝐴𝑉𝐶
= 2𝑄 − 18 = 0
𝑑𝑄
𝑄=9
2. The First Derivative Test of Relative Extremum
Note the word “extremum”. Extremum means either a maximum or a minimum. In either case, to determine the
extremum of a function 𝑦 = 𝑓(𝑥) we use the first derivative of the function and find the 𝑥 value at which the first
derivative is equal to zero.
Let’s first graph the following function and its first derivative (Figure 4),
𝑦 = 𝑓(𝑥) = 𝑥 3 − 18𝑥 2 + 90𝑥 + 100
4
𝑦 ′ = 𝑓 ′ (𝑥) = 3𝑥 2 − 36𝑥 + 90
To find the 𝑥 value at which the function reaches the relative maximum at point 𝐴 and the relative minimum at point
𝐵 (Figure 4-a), set the first derivative equal to zero.
3𝑥 2 − 36𝑥 + 90 = 0
𝑥1 , 𝑥2 =
36 ± √362 − 4(3)(90)
= 3.55, 8.45
6
Note that here 𝑥 has two non-negative values. The question is, without drawing the graph, how can we tell which 𝑥
value leads to a relative maximum and which to a relative minimum of the primitive function 𝑦?
First, let’s use the graph of the derivative function (Figure 4-b) to find the mathematical clue to distinguish between
the relative maximum and the relative minimum of 𝑦 = 𝑓(𝑥). Note that in Figure 4-b the graph of the derivative
function crosses the x-axis at 𝑥 = 3.55 (corresponding to the relative maximum point 𝐴 on the primitive function)
from above, and at 𝑥 = 8.45 (corresponding to the relative minimum point 𝐵 on the primitive function) it crosses the
x-axis from below.
The mathematical clue is then, the value of the primitive function at a given 𝑥 value is: 1) a relative maximum, if the
derivative function 𝑦′ changes its sign from positive to negative from the immediate left of the point 𝑥 to its immediate
right; 2) a relative minimum, if the derivative function 𝑦′ changes its sign from negative to positive from the immediate
left of the point 𝑥 to its immediate right.
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Figure 4-a
𝑦 = 𝑥 3 − 18𝑥 2 + 90𝑥 + 100
y
A
B
3.55
8.45
x
Figure 4-b
𝑦 ′ = 3𝑥 2 − 36𝑥 + 90
y'
3.55
8.45
x
At 𝑥 = 3 (immediate left of 𝑥 = 3.55), 𝑦′ = 9, and at 𝑥 = 4 (immediate right of 𝑥 = 3.55), 𝑦′ = −6. Thus, at 𝑥 = 3.55,
the value of 𝑦 is a relative maximum. Similarly, At 𝑥 = 8 (immediate left of 𝑥 = 8.45), 𝑦′ = −6, and at 𝑥 = 9
(immediate right of 𝑥 = 8.455), 𝑦′ = 9. Thus, at 𝑥 = 8.45, the value of 𝑦 is a relative minimum.
3. The Derivative Being Equal to Zero is a Necessary, But Not Sufficient, Condition
for a Relative Extremum
In some cases the derivative of a function may equal to zero at a given 𝑥 value, but that does not necessarily imply that
the primitive function itself is at a relative extremum. Consider, for example, the function 𝑦 = 𝑥³ − 18𝑥 2 + 108𝑥 and
its derivative 𝑦 ′ = 3𝑥² − 36𝑥 + 108. The graph of the primitive function is shown in Figure 5-a and that of the
derivative in 5-b. Note that at 𝑥 = 6 the derivative is equal to zero. However the y value at 𝑥 = 6 (point 𝐴) is neither a
maximum nor a minimum.
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The sufficient condition for the extremum on the primitive function is that the sign of the derivative at the x value
change from the immediate left to the immediate right of the 𝑥 value. Here this condition is not satisfied, for in either
side of 𝑥 = 6, 𝑦′ > 0.
Any point such as 𝐴 on the primitive function at which the derivative function reaches an extremum is called an
inflection point. In economic theory the inflection point of an objective function conveys an important economic
“message”.
Figure 5-a
𝑦 = 𝑥³ − 18𝑥 2 + 108𝑥
y
A
6
x
Figure 5-b
𝑦 ′ = 3𝑥² − 36𝑥 + 108
y'
6
x
4. The Inflection Point on the Short-Run Production Function
In the short-run, holding one input fixed, output is a function of the variable input. Denoting output as 𝑄 and the
variable input as 𝐿 (for labor), the short-run production function is generally denoted as 𝑄 = 𝑓(𝐿). The principal
feature of the short-run production function is that it reflects the law of diminishing returns. The point of diminishing
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returns in the production process is shown as the inflection point in the production function. Consider the following
short-run production function, followed by its derivative function. The graphs are shown in Figure 6.
𝑄 = −𝐿3 + 15𝐿2
𝑄′ =
𝑑𝑄
= −3𝐿2 + 30𝐿
𝑑𝐿
In Figure 6-a point 𝐷 represents the inflection point of the production function. Up to that point the production
function is rising at an increasing rate. This means that the marginal product of each additional unit of 𝐿 is increasing.
This is also shown as the rising marginal product in Figure 6-b.
At the inflection point 𝐷, where 𝐿 = 5, the marginal product reaches the maximum. After point 𝐷 the production
function rises at a decreasing rate, also indicated by the decreasing marginal product in the lower diagram. The
decreasing marginal product reflects the well-known law of diminishing returns. Thus, the inflection point 𝐷 in Figure
6-a, which corresponds to the peak of the marginal product curve in Figure 6-b, is the point of diminishing returns.
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Figure 6-a
Production Function: 𝑄 = −𝐿3 + 15𝐿2
Q
D
5
L
Figure 6-b
Marginal Product Function: 𝑄′ = −3𝐿2 + 30𝐿
Q
5
L
5. The Inflection Point on the Short-Run Cost Function
Consider the short-run (total) cost function and its derivative (marginal cost) function below:
𝐶 = 𝑄3 − 18𝑄2 + 141𝑄 + 100
𝐶′ =
𝑑𝐶
= 3𝑄2 − 36𝑄 + 141
𝑑𝑄
In Figure 7-a the cost function rises at a decreasing rate until point 𝐹. Beyond output 𝑄 = 6 costs begin to rise at an
increasing rate. When costs rise at a decreasing rate, marginal cost decreases, and when costs begin to rise at an
increasing rate (after the inflection point 𝐹) 𝑀𝐶 begins to rise. Thus, the inflection point on the 𝐶 curve corresponds
to the minimum point 𝑀 on the 𝑀𝐶 curve.
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Figure 7-a
Total Cost Function: 𝐶 = 𝑓(𝑄) = 𝑄3 − 18𝑄2 + 141𝑄 + 100
$
C
F
6
Q
Figure 7-b
Marginal Cost: 𝑀𝐶 = 𝑓′(𝑄) = 3𝑄2 − 36𝑄 + 141
$
MC
M
6
Q
6. Derivative of a Derivative—Second Derivative
Let us go back to the cost function above. For ease of presentation, let’s use y in place of C for the dependent variable,
and 𝑥 in place of 𝑄 for the independent variable.
𝑦 = 𝑓(𝑥) = 𝑥 3 − 18𝑥 2 + 141𝑥 + 100
The (first) derivative of this function is
𝑦 ′ = 𝑓 ′ (𝑥) =
𝑑𝑦
= 3𝑥 2 − 36𝑥 + 141
𝑑𝑥
Now take the derivative of the derivative function. The second derivative is
𝑦 ′′ = 𝑓 ′′ (𝑥) =
𝑑 𝑑𝑦
𝑑2𝑦
( ) = 2 = 6𝑥 − 36
𝑑𝑥 𝑑𝑥
𝑑𝑥
10
For comparison all three functions are plotted in Figure 8. Note that in Figure 8-a the primitive function 𝑓(𝑥) is rising
throughout the domain of the function. However, up to the inflection point 𝐹, 𝑦 rises at a decreasing rate. Up to 𝑥 = 6,
the slope of the function decreases. In Figure 8-b this is shown as the decreasing portion of the first derivative.
Correspondingly, in Figure 8-c, the second derivative function is negative—𝑓′′(𝑥) < 0.
At the inflection point 𝐹 the first derivative reaches the minimum, which implies the slope of 𝑦′ at point 𝑀 is zero,
which, in turn, means 𝑦′′ = 0.
Figure 8-a
𝑦 = 𝑓(𝑥) = 𝑥 3 − 18𝑥 2 + 141𝑥 + 100
y
F
6
x
Figure 8-b
𝑦 ′ = 𝑓 ′ (𝑥) = 3𝑥 2 − 36𝑥 + 141
y'
M
6
x
Figure 8-c
𝑦 ′′ = 𝑓 ′′ (𝑥) = 6𝑥 − 36
y''
6
x
11
6.1. Using the Second Derivative to Determine Relative Maximum and Relative
Minimum of the Primitive Function
Using the profit function in Figure 1 and changing the symbols, let’s use that function to show how the second
derivative can aid us in the determining whether an extremum of a function—where 𝑓′(𝑥) = 0—is a relative
maximum or a relative minimum.
𝑦 = −𝑥 3 + 13𝑥 2 + 9𝑥 − 100
The first and second derivatives of the function are:
𝑦′ =
𝑑𝑦
= −3𝑥 2 + 26𝑥 + 9
𝑑𝑥
𝑦 ′′ =
𝑑2𝑦
− 6𝑥 + 26
𝑑𝑥 2
The graph of these function are shown in Figure 9. To obtain the relative extremum of 𝑦, set the first derivative equal
to zero.
−3𝑥 2 + 26𝑥 + 9 = 0
𝑥=9
The y value for 𝑥 = 9 is: 𝑦 = −93 + 13(92 ) + 9(9) − 100 = 305. Without drawing the graph, what does the second
derivative evaluated at 𝑥 = 9 tell us regarding the value 𝑦 = 305?
At 𝑥 = 9, 𝑦′′ = −6(9) + 26 = −28 < 0.
The fact that at 𝑥 = 9, 𝑦′′ < 0 is the clue that 𝑦 = 305 is a relative maximum.
Figure 9-a
𝑦 = −𝑥 3 + 13𝑥 2 + 9𝑥 − 100
y
M
x
9
Figure 9-b
𝑦 ′ = 3𝑥 2 + 26𝑥 + 9
y'
9
x
12
Figure 9-c
𝑦 ′′ = −6 + 26
y
9
-28
x
Example
Use the “second-derivative test” to find the relative maximum and relative minimum of the following:
𝑦 = 𝑥 3 − 12𝑥 2 + 36𝑥 + 8
Set the first derivative equal to zero.
𝑦 ′ = 3𝑥 2 − 24𝑥 + 36 = 0
The solutions are
𝑥1 , 𝑥2 =
24 ± √242 − 4(3)(36) 24 ± 12
=
= 2,6
2(3)
9
At 𝑥1 = 2
𝑦1 = 23 − 12(22 ) + 36(2) + 8 = 40
At 𝑥2 = 6
𝑦2 = 63 − 12(62 ) + 36(6) + 8 = 8
Now find the second derivative.
𝑦 ′′ = 6𝑥 − 24
At 𝑥1 = 2
𝑦1′′ = −12 < 0
⇒
𝑦1 = 40 is a relative maximum
At 𝑥2 = 6
𝑦2′′ = 12 > 0
⇒
𝑦2 = 8 is a relative minimum
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Figure 10-a
𝑦 = 𝑥 3 − 12𝑥 2 + 36𝑥 + 8
y
40
8
2
6
x
Figure 10-b
𝑦 ′ = 3𝑥 2 − 24𝑥 + 36
y'
2
6
x
Figure 10-c
𝑦 ′′ = 6𝑥 − 24
y''
12
2
6
x
-12
Summary
Second-derivative test for relative extremum
If the value of first derivative of 𝑦 = 𝑓(𝑥) at 𝑥0 is 𝑓′(𝑥0 ) = 0, then the value of 𝑦 = 𝑓(𝑥0 ) will be
a) A relative maximum if the second-derivative 𝑓 ′ (𝑥0 ) < 0.
b) A relative minimum if the second-derivative 𝑓 ′ (𝑥0 ) > 0.
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