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## PowerPoint Slideshow about 'Lecture 1c Marginal Analysis and Optimization' - gaerwn

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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:
- To become Master of the Universe (and still have a life)
- Control Variables:
- 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.

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

3 hours is the best

Common Sense Conclusion

- 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

Applying the Principle to Equibase

- Our spreadsheets let us work with any number of different assumptions about demand and costs, and so let’s assume that Equibase merged with the tracks
- Objective: max profits
- Constraints are defined by the market demand and the costs. Let’s assume
- Q=500(5-P)
- Variable Production cost = $.5/unit
- Fixed production cost = $500

Distinguishing costs and benefits for the firm

- Selling programs generates revenue, the “benefit” of the activity.
- Note R = PxQ (a very pure definition)
- MR = Change in R when Q changes
- More formally (change in R)/(change in Q)

Optimal Decision Making In the Firm: A Simple Example

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

- Total Revenue (TR): PxQ, nothing more-nothing 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)

Why Is the Optimal Q=1,125

- For all Q less than this, MR<MC, meaning an increase in output would raise revenue by more than costs.
- For all Q greater than this MR<MC, meaning an increase in output would raise revenue by less than the increase in costs.
- 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.

Thinking about optimization this way helps understand two important (related) principles

- Fixed Costs Don’t matter (at least to the optimal solution)
- Anything the “unnaturally” distorts marginal costs leads to a reduction in the optimal amount that can be achieved

Why Doesn’t the Change in Fixed Costs Change the Way the Firm Operates?

- Because there is change MR or MC
- But what if by shutting down (Q=0) fixed costs can be eliminated?

Anything that “unnaturally” distorts marginal costs leads to a reduction in the optimal amount that can be achieved

- Remember we saw that charging a per unit fee reduced the combined profits. Now we know why.
- The tracks had control of the retail price and hence the sales volume.
- The per unit fee was reflected in the sales price, but it didn’t represent a “real” cost. It appeared real enough to the tracks but it didn’t represent any actual expenditure, it was just a transfer from one division to the other. Thus it caused the tracks to raise the price of the programs beyond the optimal amount.
- In other words, the tracks were led to believe that there was a variable cost (beyond the printing costs)
- Can you think of other examples where firms make this mistake?

Everything you ever needed to know about calculus to solve optimization problems in economics

- Consider the simple function

y=6x-x2

- If we calculate the value of y for various values of x, we get

The graph of the function would look like this

As you can see both from the table and the graph, if x=3, y is at its maximum value.

How Calculus Helps

- 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).

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

- 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).
- The derivative of a linear function is the coefficient (the thing multiplied by the variable)
- 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

- 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

- By combining these rules we can find the derivative of any polynomial.
- 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]

Formal Analysis (Calculus)

- 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

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