The Use of Mathematics in Microeconomics

The subject matter of economics is social behavior. As economists, we sometimes use mathematics to inform us about that behavior. The purpose of this discussion is to equip you with the mathematical tools you will need to deal confidently with the topics and concepts we will cover in this course.

While our discussions will be limited to very elementary mathematical applications, they will serve to demonstrate the usefulness of math in economic model building. Our studies will take us through three levels of analysis: equilibrium analysis, comparative statics, and optimization. Each of these is examined in turn below.

Equilibrium (Static) Analysis

Equilibrium in economics has been defined in various ways. Fritz Machlup defines an equilibrium as "a constellation of selected interrelated variables so adjusted to one another that no inherent tendency to change prevails in the model which they constitute" (Equilibrium and disequilibrium: misplaced concreteness and disguised politics, Economic Journal, March 1958). The highlighted phrase captures the essence of an equilibrium in a mathematical model, no tendency to change. That is true as long as there are no changes in external forces.

In a static-equilibrium model the problem is usually defined in terms of finding the set of values for the endogenous variables that will satisfy the equilibrium conditions specified in the model. This can be illustrated with a simple model of price determination.

Consider a commodity that is bought and sold in an isolated market. The three endogenous variables of the model are quantity supplied (Qs), quantity demanded (Qd), and price (P). The normal assumptions hold: 1) quantity demanded is inversely related to price, 2) quantity supplied is directly related to price, and 3) equilibrium exists when excess quantity demanded equals zero and only at positive prices and quantities.

Qd = Qs

Qd = 20 - 2P

Qs = 2 + 3P

The model contains three endogenous variables to be determined and four parameters that summarize the external forces that affect demand and supply. These parameters show up in the above equations as slope and intercept values (-2 and 20, respectively, in the demand equation and 3 and 2 in the supply equation).

Recall that the slope of a function expresses the ratio of the changes implied by the function. When the two variables are plotted in the standard x-y plane, slope is a measure of the change y (plotted on the vertical axis) divided by the change in x (plotted on the horizontal axis), or in other words, the rise over the run. The intercept is the value of a function when the independent variable has a value of zero. For the function y=a+bx, the intercept is a, since y=a when x=0.

The solution to the above static-equilibrium problem can be determined as follows. According to the equilibrium condition, there is no tendency for change when excess demand equals zero, or when quantity demanded equals quantity supplied.

20 - 2P = 2 + 3P

5P = 18

P = 3.6

Substituting this price back into either the supply or the demand equation will yield the value for the equilibrium quantity.

Q = 20 - 2P

Q = 20 - 2(3.6)

Q = 20 - 7.2

Q = 12.8

When price is 3.6, quantity demanded will equal quantity supplied at 12.8 units.

Comparative Statics Analysis

Equilibrium analysis provides us with a means of determining the values of endogenous variables when a given set of external conditions exist. As is often the case, we are interested in how those values change when the external conditions vary. Since we are not able to hold other things constant for very long, we need a way of comparing different equilibrium states that are associated with different sets of values for parameters and exogenous variables. Comparative statics analysis is a means of making such comparisons.

In comparative statics we ignore the adjustment process entirely. Our focus is on comparing the initial (prechange) equilibrium values for the endogenous variables with the final (postchange) equilibrium values. The comparative statics approach is essentially one of finding a rate of change; specifically, the rate of change of the value of the endogenous variable with respect to the change in the particular parameter or exogenous variable. For this reason, the mathematical concept of the derivative takes on a special significance in comparative statics analysis.

The derivative of a function y=f(x) is dy/dx. This notation emphasizes that a derivative measures the rate of change. It is a measure of the slope of a function at any point and a way to get at the concept of the margin. This is important in economics since marginal analysis is central to the theory of choice behavior. All economic decisions are made at the margin. Seldom are we faced with "all-or-none" situations. The process of reaching a preferred position involves making tradeoffs; i.e., a little less of one item in return for a little more of another.

Approaching the static-equilibrium problem from the perspective of the rate of change, we can illustrate the difference equation and how we use calculus to solve equilibrium problems.

Qd = Qs

Qd = 8 - 2P

Qs = - 4 + 4P

The spread between the price buyers are willing to pay (Pd) and the price sellers are willing to accept (Ps) can be observed at each quantity. The quantity we are most interested in is the one where (Pd - Ps) = 0. To solve the difference equation, first write each equation in terms of the respective prices.

Pd = 4 - 0.5Q

Ps = 1 + 0.25Q

Pd - Ps = [4 - 0.5Q] - [1 + 0.25Q]

Setting the difference equation equal to zero and solving for Q identifies the output level where bid and ask prices are equal.

[4 - 0.5Q] - [1 + 0.25Q] = 0

0.75Q = 3

Q = 4

Substituting this quantity into any of the above equations identifies the price to which the market converges.

P = 4 - 0.5Q

P = 4 - 0.5(4)

P = 4 - 2

P = 2

The central problem in comparative statics analysis is finding the rate of change. This can be identified in terms of finding the derivative of some function y=f(x). Before going into comparative statics models, we must first familiarize ourselves with some of the rules of differentiation.

1. constant-function rule. The derivative of a constant (k) is always zero.

dk/dx = 0

Example: d(10) = 0

2. power-function rule. The derivative of a power function

y = xn

dy/dx = nxn-1

or more generally,

y = cxn

dy/dx = ncxn-1

Example: y = 2x9

dy/dx = 18x8

3. sum-difference rule. The derivative of the sum (difference) of two functions is the sum (difference) of the derivatives of the two functions.

y = f(x) + g(x)

dy/dx = d[f(x) + g(x)] = d[f(x)]/dx + d[g(x)]/dx

Example: y = x2 + 4x3

dy/dx = 2x + 12x2

4. product rule. The derivative of the product of two functions is equal to the first function times the derivative of the second, plus the second function times the derivative of the first.

y = f(x)g(x) = FG

dy/dx = F(dG/dx) + G(dF/dx)

Example: y = (2x + 3)(3x2)

dy/dx = (2x + 3)(6x) + (3x2)(2) = 18x2 +18x

5. quotient rule. The derivative of the quotient of two functions is equal to the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, divided by the denominator squared.

y = f(x)/g(x) = F/G

dy/dx =

Example: y = 6x2/4x3

4x3(12x) - 6x2(12x2)

dy/dx = --------------------------------


48x4 - 72x4

= -------------------



= - --------- = - 3/2x2


6. log function rule.

y = log f(x) = log F

dy/dx = 1/F(dF/dx)

Example: y = log x3

dy/dx = [1/x3](3x2) = 3/x

To this point we have considered only derivatives of functions of a single independent variable. However, in comparative statics analysis we often encounter situations where several parameters and exogenous variables appear in the same equation. In these circumstances the endogenous variables may be a function of more than one parameter or exogenous variable. The final step in applying the concept of derivatives to comparative statics is to learn how to find the derivative of a function of more than one variable. This is where the concept of the partial derivative comes in handy. The rules of partial differentiation are the same as those for basic differentiation. The thing to be aware of is that exogenous variables that are not changing are treated as constants. The symbol for a partial derivative is also different; "" rather than "d."

The notion of a total differential completes the list of derivatives we will use in this course. A total differential of a function q=q(a,b) measures the amount of the change in q resulting from small changes in a and b. The sum of all the changes is:

dy = (q/a)da + (q/b)db.

Example of comparative statics analysis

Suppose that the static-equilibrium model is given below (Y = income):

Qd = Qs

Qd = 150 - 4P + .01Y

Qs = -10 + 6P

Y = 4000

Setting Qd = Qs and solving for the values of the endogenous variables yields:

150 - 4P +.01(4000) = -10 + 6P

10P = 150 + 10 + 40

10P = 200

P = 20

Q = 110

Assume that income increases to 5000. The new equilibrium values for P and Q are 21 and 116, respectively. In other words, for every 1000 increase in income, quantity demanded increases 6 units.


We will use the concepts of minima and maxima quite extensively because of the focus on neoclassical price theory and its related assumptions of optimization: utility and profit maximization and cost minimization. Each may be identified by finding the value of the independent variable at which the slope of the function equals zero. This is referred to as the necessary (or first-order) condition. To determine whether it is a max or min, we must refer to the sufficient (or second-order) condition.

Example of profit maximization

Let us assume that

TR = 1000Q - 2Q2

TC = Q3 - 59Q2 + 1315Q + 2000

= TR - TC = - Q3 + 57 Q2 - 315Q - 2000

d/dQ = -3 Q2 + 114Q -315 = 0 when Q = 3 or 35

But since the second derivative is

d2/d Q2 = -6Q +114 > 0 when Q=3 and < 0 when Q=35

Thus, the profit maximizing output level is 35.