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Explore the significance of optimization in microeconomics through rational behavior, marginal analysis, and profit maximization examples in this informative lecture.
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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
Think About Optimization as a Sequence of Steps If Here, Do More If Here, Do Less
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