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”. 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”? 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: What do you you want? That is, what is your Objective: To become Master of the Universe (and still have a life) To become Master of the Universe (and still have a life) Control Variables: Control Variables: Hours studying economics (since econ is the key to happiness and wisdom) Hours studying economics (since econ is the key to happiness and wisdom) Constraints: Constraints: Time, energy, tolerance of mind-numbing tedium Time, energy, tolerance of mind-numbing tedium

The magic word: “marginal” MARGINAL ____ : The change in ____ when something else changes. MARGINAL ____ : The change in ____ when something else changes. Approximate Formula: The marginal contribution of x to y=(change in y)/(change in x) 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. Exact Formally (calculus): If y=f(x), the marginal contribution of x to y is dy/dx.

Example: The Benefits Studying

Benefits and Costs Hours Studying per Week Total Benefit (Increase in Average Annual Wage) Approx. Marginal Benefit Total CostApprox. Marginal Cost 0 $ -0 1$ $1.00 2$20.00$9.004$3.00 3$27.00$7.009$5.00 4$32.00$5.0016$7.00 5$35.00$3.0025$9.00

Interesting Observation Marginal Benefits decrease and marginal benefits increase. Marginal Benefits decrease and marginal benefits increase. Questions for the class Questions for the class Is this sensible? Is this sensible? Is there a certain similarity between costs and benefits? 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 $1.00 $ $20.00$9.00$4.00$3.00$ $27.00$7.00$9.00$5.00$ $32.00$5.00$16.00$7.00$ $35.00$3.00$25.00$9.00$10.00

Net Benefits Hours Studying per Week Total Benefit Approx. Marginal Benefit Total CostApprox. Marginal Cost Net Benefit (Total Benefit -Total Cost) 0 $ - 1$11.00 $1.00 $ $20.00$9.00$4.00$3.00$ $27.00$7.00$9.00$5.00$ $32.00$5.00$16.00$7.00$ $35.00$3.00$25.00$9.00$ hours is the best

Think About Optimization as a Sequence of Steps Hours Studying per Week Total Benefit Approx. Marginal Benefit Total CostApprox. Marginal Cost Net Benefit (Total Benefit -Total Cost) 0 $ - 1$11.00 $1.00 $ $20.00$9.00$4.00$3.00$ $27.00$7.00$9.00$5.00$ $32.00$5.00$16.00$7.00$ $35.00$3.00$25.00$9.00$10.00 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 greater than marginal costs, then do more. If marginal benefits are less than marginal costs, then do less. If marginal benefits are less than marginal costs, then do less. To optimize, find the level of activity where marginal benefits with marginal costs 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 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 Objective: max profits Constraints are defined by the market demand and the costs. Let’s assume Constraints are defined by the market demand and the costs. Let’s assume Q=500(5-P) Q=500(5-P) Variable Production cost = $.5/unit Variable Production cost = $.5/unit Fixed production cost = $500 Fixed production cost = $500

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

Benefits of Selling Programs PriceQuanityRevenueMR $ 4.00 $ $ 2, $ 3.75 $ $ 2, $ 2.75 $ 3.50 $ $ 2, $ 2.25 $ 3.25 $ $ 2, $ 1.75 $ 3.00 $ 1, $ 3, $ 1.25 $ 2.75 $ 1, $ 3, $ 0.75 $ 2.50 $ 1, $ 3, $ 0.25 $ 2.25 $ 1, $ 3, $(0.25)

Costs are incurred in production and Sales QuanityCostMC 500 $ $ $ $ $ $ $ $ 1, $ $ 1, $ $ 1, $ $ 1, $ 0.50

Optimal Decision Making In the Firm: A Simple Example Explicitly describe the three elements of the optimization problem Goal: Profit Maximization Goal: Profit Maximization Decision Variables: Price or Quantity Decision Variables: Price or Quantity Constraints: Constraints: On Costs: It takes stuff to make stuff and stuff isn’t free.) 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. 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.) 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 Marginal Revenue (MR): The change in total revenue when output changes Approximated as: Change in TR/ Change in Q Approximated as: Change in TR/ Change in Q Calculus: dTR/dQ) Calculus: dTR/dQ)

Example QuantityPriceTotal RevenueApprox. Marginal Revenue

Putting it together PriceQuanityRevenueMRCostMCProfit $ $ 2, $ $ 1, $ $ 2, $ 2.75 $ $ 0.50 $ 1, $ $ 2, $ 2.25 $ $ 0.50 $ 1, $ $ 2, $ 1.75 $ $ 0.50 $ 1, $ 3.001,000 $ 3, $ 1.25 $ 1, $ 0.50 $ 2, $ 2.751,125 $ 3, $ 0.75 $ 1, $ 0.50 $ 2, $ 2.501,250 $ 3, $ 0.25 $ 1, $ 0.50 $ 2, $ 2.251,375 $ 3, $(0.25) $ 1, $ 0.50 $ 1,906.25

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

Assume Fixed Costs are $5000 PriceQuanityRevenueMRCostMCProfit $ $ 2, $ 5, $ (3,250.00) $ $ 2, $ 2.75 $ 5, $ 0.50 $ (2,968.75) $ $ 2, $ 2.25 $ 5, $ 0.50 $ (2,750.00) $ $ 2, $ 1.75 $ 5, $ 0.50 $ (2,593.75) $ 3.001,000 $ 3, $ 1.25 $ 5, $ 0.50 $ (2,500.00) $ 2.751,125 $ 3, $ 0.75 $ 5, $ 0.50 $ (2,468.75) $ 2.501,250 $ 3, $ 0.25 $ 5, $ 0.50 $ (2,500.00) $ 2.251,375 $ 3, $(0.25) $ 5, $ 0.50 $ (2,593.75)

Why Doesn’t the Change in Fixed Costs Change the Way the Firm Operates? Because there is change MR or MC Because there is change MR or MC But what if by shutting down (Q=0) fixed costs can be eliminated? But what if by shutting down (Q=0) fixed costs can be eliminated?

PriceQuanityRevenueMRCostMCProfit $ $ - $ $ 2, $ 5, $ (3,250.00) $ $ 2, $ 2.75 $ 5, $ 0.50 $ (2,968.75) $ $ 2, $ 2.25 $ 5, $ 0.50 $ (2,750.00) $ $ 2, $ 1.75 $ 5, $ 0.50 $ (2,593.75) $ 3.001,000 $ 3, $ 1.25 $ 5, $ 0.50 $ (2,500.00) $ 2.751,125 $ 3, $ 0.75 $ 5, $ 0.50 $ (2,468.75) $ 2.501,250 $ 3, $ 0.25 $ 5, $ 0.50 $ (2,500.00)

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. 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 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. 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) 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? 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 Consider the simple functiony=6x-x2 If we calculate the value of y for various values of x, we get If we calculate the value of y for various values of x, we get x12345 y58985

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. 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.) 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). 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. 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. 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. 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.) 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. The derivative of a constant is zero. If Y=C for all X, then dY/dX=0 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). (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) The derivative of a linear function is the coefficient (the thing multiplied by the variable) If Y=bX, then dY/dX=b If Y=bX, then dY/dX=b (Which makes sense, since the slope of this function is just the coefficient.) (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 The rule for a “power function” is as follows If Y=bX n, then dY/dX=nbX n-1 For example, if Y=3x 2, then dY/dX = 6X For example, if Y=3x 2, 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 x 0 =1. Notice, by the way, the rule for linear functions can be viewed as a special case of the power function rule—since x 0 =1. Notice also that the power function rule is good for evaluating functions involving quotients. For example the function 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/X 2

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 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), If Y=f(x)+g(x),then dY/dX=df(x)/dx+dg(x)/dx dY/dX=df(x)/dx+dg(x)/dx By combining these rules we can find the derivative of any polynomial. By combining these rules we can find the derivative of any polynomial. If Y=a + bx + cx2+…+dxn, If Y=a + bx + cx2+…+dxn, then then dY/dX=b+2cX+…+ndXn-1 dY/dX=b+2cX+…+ndXn-1

The derivative of the product of two functions is as follows The derivative of the product of two functions is as follows If Y = f(x )g(x), 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 Let X stand for the number of hours studying Benefits = 12X-X2 Benefits = 12X-X2 Marginal benefit = X Marginal benefit = X Cost = X2 Cost = X2 Marginal Cost = 2X Marginal Cost = 2X Setting marginal benefit = marginal cost implies 12-2X=2X or X=3 Setting marginal benefit = marginal cost implies 12-2X=2X or X=3