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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”.
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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
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)
Cost Constraints: Obvious (but useful) defintions • 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
Finding the Optimal Q (maybe) Somewhere Between These Two is Optimal
Finding the Optimal Q (maybe) MR >MC means Produce More MR < MC means produce less
It would seem that the optimal Q is between 2 and 3 • 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 costsFrom the previous example, suppose fixed costs go up by 9 so that
And the optimal solution is still Q=3 and P=$5. (Which is not great surprise since MC and MR haven’t changed.) • 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? 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) • 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