Transcript Lecture 1(b) Models

Lecture 1(c) Marginal
Analysis and Optimization
Why is it important to understand the
mathematics of optimization in order to
understand microeconomics?


The “economic way of thinking” assumes that
individuals behave as if they are “rational”.
Question for the class: What does it mean to
say that behavior is “rational”?
Any Optimization problem has three
elements.

What do you you want? That is, what is your
Objective:


Control Variables:


To become Master of the Universe (and still have a life)
Hours studying economics (since econ is the key to
happiness and wisdom)
Constraints:

Time, energy, tolerance of mind-numbing tedium
The magic word: “marginal”



MARGINAL ____ : The change in ____ when
something else changes.
Approximate Formula: The marginal
contribution of x to y=(change in y)/(change in
x)
Exact Formally (calculus): If y=f(x), the
marginal contribution of x to y is dy/dx.
Example: The Benefits Studying

Total Benefit
Hours
(Increase in Approximate
Studying
Average
Marginal
per Week Annual Wage)
Benefit
0
$0
1
$11
$11
2
$20
$9
3
$27
$7
4
$32
$5
5
$35
$3
Benefits and Costs
Hours
Studying
per Week
0
Total Benefit
(Increase
in Average
Annual
Wage)
$
Approx.
Marginal
Benefit
-
Total Cost
Approx.
Marginal
Cost
0
1
$11.00
$11.00
1
$1.00
2
$20.00
$9.00
4
$3.00
3
$27.00
$7.00
9
$5.00
4
$32.00
$5.00
16
$7.00
5
$35.00
$3.00
25
$9.00
Interesting Observation


Marginal Benefits decrease and marginal benefits
increase.
Questions for the class
Is this sensible?
 Is there a certain similarity between costs and
benefits?

Net Benefits
Hours
Studying
per Week
Total
Ben
efit
Approx.
Margina
l Benefit
Total
Cost
$
Approx.
Marg
inal
Cost
-
Net Benefit
(Total
Benefit Total Cost)
0
$
-
$
-
1
$11.00
$11.00
$1.00
$1.00
$10.00
2
$20.00
$9.00
$4.00
$3.00
$16.00
3
$27.00
$7.00
$9.00
$5.00
$18.00
4
$32.00
$5.00
$16.00
$7.00
$16.00
5
$35.00
$3.00
$25.00
$9.00
$10.00
Net Benefits
Hours
Studying
per Week
0
Total
Benefit
Approx. Marginal
Benefit
$
Total Cost
$
Approx.
Marginal
Cost
-
Net Benefit (Total
Benefit -Total
Cost)
$
-
3 hours
is the 1
best 2
$11.00
$11.00
$1.00
$1.00
$10.00
$20.00
$9.00
$4.00
$3.00
$16.00
3
$27.00
$7.00
$9.00
$5.00
$18.00
4
$32.00
$5.00
$16.00
$7.00
$16.00
5
$35.00
$3.00
$25.00
$9.00
$10.00
-
Think About Optimization as a
Sequence of Steps
Hours
Studying
per Week
0
Total
Benefit
Approx. Marginal
Benefit
$
Total Cost
Approx.
Marginal
If Here, Do Cost
More
$
Net Benefit (Total
Benefit -Total
Cost)
-
$
1
$11.00
$11.00
$1.00
$1.00
$10.00
2
$20.00
$9.00
$4.00
$3.00
$16.00
3
$27.00
$7.00
$9.00
$5.00
$18.00
4
$32.00
$5.00
$16.00
$7.00
$16.00
5
$35.00
$3.00
$25.00
$9.00
$10.00
If Here, Do Less
-
Common Sense Conclusion
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If marginal benefits are greater than marginal
costs, then do more.
If marginal benefits are less than marginal costs,
then do less.
To optimize, find the level of activity where
marginal benefits with marginal costs
Optimal Decision Making In the Firm:
A Simple Example
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Explicitly describe the three elements of the
optimization problem
Goal: Profit Maximization
Decision Variables: Price or Quantity
Constraints:
On Costs: It takes stuff to make stuff and stuff isn’t
free.)
 On Revenues: Nobody will pay you an infinite
amount for your stuff.

Revenue Constraints: Obvious (but useful)
Definitions
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Total Revenue (TR): PxQ, nothing morenothing less (and not to be confused with profit,
net revenue, etc.)
Marginal Revenue (MR): The change in total
revenue when output changes
Approximated as: Change in TR/ Change in Q
 Calculus: dTR/dQ)

Example
Quantity
Price
Total Revenue
Approx. Marginal
Revenue
0
8
0
1
7
7
7
2
6
12
5
3
5
15
3
4
4
16
1
5
3
15
-1
6
2
12
-3
Cost Constraints: Obvious (but useful) defintions
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Total Cost Function: The relationship between
Q and Costs
Marginal Costs: The change in TC when
output changes
Approximation: MC = [Change in Total Costs]/
[Change in Output]
 Calculus: MC=dTC/dQ
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Example: (Based on the assumption
that TC=1+3Q)
Quantity
TC
MC
0
1
1
4
3
2
7
3
3
10
3
4
13
3
5
16
3
6
19
3
Finding the Optimal Q (maybe)
Price
8
Quantity
Somewhere
Between
These Two
is Optimal
TR
MR
TC
MC
0
0
1
7
7
4
3
3
6
2
12
5
7
3
5
5
3
15
3
10
3
5
4
4
16
1
13
3
3
3
5
15
-1
16
3
-1
2
6
12
-3
19
3
-7
7
1
Profit
-1
Finding the Optimal Q (maybe)
Price
Quantity
TR
MR
TC
MC
Profit
MR >MC means
Produce More
8
0
0
1
-1
7
1
7
7
4
3
3
6
2
12
5
7
3
5
5
3
15
3
10MR < MC 3
4
4
16
1
means produce
13 less 3
5
3
5
15
-1
16
3
-1
2
6
12
-3
19
3
-7
3
It would seem that the optimal Q is between 2 and 3
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Of course this makes sense since if Q <2, then MR >
MC (meaning an increase in output would raise revenue
by more than costs).
Similarly if Q>3, then MC>MR (meaning a decrease
in ouput would reduce costs by more than revenues).
This is such an important conclusion, it should be
stated formally as Necessary Condition for Profit
Maximization: If you produce, produce the Q
such that MR=MC.
Question for the Class
How does this principle explain what we found in
the Equibase problem?
Optimization and the role of fixed costs
From the previous example, suppose fixed costs go up by 9 so
that
Quantity
TC
MC
0
10
1
13
3
2
16
3
3
19
3
4
22
3
5
25
3
6
28
3
If the revenue function remains as before, we get
Price
Quantity
TR
MR
TC
MC
8
0
0
7
1
7
7
13
3
-6
6
2
12
5
16
3
-4
5
3
15
3
19
3
-4
4
4
16
1
22
3
-6
3
5
15
-1
25
3
-10
2
6
12
-3
28
3
-16
10
Profit
-10
And the optimal solution is still Q=3 and P=$5. (Which is not
great surprise since MC and MR haven’t changed.)
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This leads to the following IMPORTANT OBSERVATION: Fixed costs
don’t effect the optimal output.
If you think about it, this makes perfect sense: if some aspect of cost can’t
be influenced by output decisions, it should be ignored.
Sensible though this is, one of the recurring themes of this course will be
how often ignoring this fact leads to bad decisions.
Final Note: What if Fixed Costs Can Be Eliminated?
Price
Q
TR
MR
TC
MC
Profit
8
0
0
7
1
7
7
13
13
-6
6
2
12
5
16
3
-4
5
3
15
3
19
3
-4
4
4
16
1
22
3
-6
0
It now is true that the best choice is Q=0. (Which is why I described the solutions above as “maybe”—because we didn’t
fully consider “shut down “ conditions.)
Everything you ever needed to know about calculus
(to get through FINA 6202)
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Consider the simple function
y=6x-x2
If we calculate the value of y for various values of
x, we get
x
1
2
3
4
5
y
5
8
9
8
5
y
The graph of the function would
look like this
10
8
6
4
2
0
0
2
4
6
x
As you can see both from the table and the graph, if x=3, y is at its maximum value.
How Calculus Helps
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But drawing a graph or computing a table of numbers is tedious
and unreliable. One of the many good things about calculus is
that it gives us a convenient way of finding the value of X that
leads to the maximum (or minimum) value of Y.
The key to the whole exercise is the fact that when a function
reaches it’s maximum value, the slope of the graph changes from
positive to negative. (Confirm this on the graph given above.)
Thus, we can find the critical value of X by finding the point
where the slope of the graph is zero (remember, if the graph is
continuous, the slope can’t go from positive to negative without
passing through zero).
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Now- and here’s where the calculus comes in—the
derivative of a function is nothing more than a very
precise measurement of the slope of the graph of the
function.
Thus, if we can find the value of X at which the
derivative of the function is zero, we will have identified
the optimal value of the function.
If this were a math class, we’d spend several lectures
studying exactly what is meant by a derivative of a
function and we’d end up with some rules for finding a
derivative.
But since this isn’t a math class, we’ll go straight to the
rules (especially since we only need a few of them and
they’re very easy to remember.)
Rules for finding derivatives
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The derivative of a constant is zero.
If Y=C for all X, then dY/dX=0
 (Which makes sense, since the graph of Y=C is a
flat line and thus has a slope of zero).
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The derivative of a linear function is the
coefficient (the thing multiplied by the variable)
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If Y=bX, then dY/dX=b
(Which makes sense, since the slope of this
function is just the coefficient.)
Rules for finding derivatives

The rule for a “power function” is as follows
If Y=bXn, then dY/dX=nbXn-1
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For example, if Y=3x2, then dY/dX = 6X
Notice, by the way, the rule for linear functions can be viewed as
a special case of the power function rule—since x0=1.
 Notice also that the power function rule is good for evaluating
functions involving quotients. For example the function
Y = 2/X
can be written as
Y=2X-1
and so the derivative is
dY/dX=-2X-2=-2/X2

Rules for finding derivatives

The derivative of a function that is the sum of several
functions is the sum of the derivatives of those
functions
If Y=f(x)+g(x),
then
dY/dX=df(x)/dx+dg(x)/dx
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By combining these rules we can find the derivative of
any polynomial.
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If Y=a + bx + cx2+…+dxn,
then
dY/dX=b+2cX+…+ndXn-1
The derivative of the product of two functions
is as follows
 If Y = f(x )g(x),
then
dY/dX = f(x)[dg(x)/dx]+g(x)[df(x)/dx]
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Formal Analysis (Calculus)
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Let X stand for the number of hours studying
Benefits = 12X-X2
Marginal benefit = 12 - 2X
Cost = X2
Marginal Cost = 2X
Setting marginal benefit = marginal cost implies
12-2X=2X or X=3