Econ 520

Dr. Helen Roberts

Mathematics of Microeconomics

 

I. The Algebra of Supply and Demand

 

A. If the demand and supply curves are both linear, it is easy to get solutions for the conditions of equilibrium.

 

Demand: P = A - BQ (A and B are positive, so -B is a negative slope.)

Supply: P = C + DQ (A and C are vertical intercepts, D is positive slope.)

 

So these are 2 simultaneous equations. The solution values are:

 

Q* = (A - C)/(B+D) P* = (AD + BC)/(B+D)

 

B. Now consider shifts in the determinants of equilibrium (comparative statics). In the algebra, this means shifts in A, B, C, or D.

 

1. Shift Demand Rightward (increase)

A. Increase intercept A (shift curve upward) or decrease B (tilt about vertical intercept which makes the curve more horizontal).

 

Q** = (A’ - C)/(B+D) where A’ = A + D A, and DA = change in A

 

Since B, D, A and DA are all positive, Q** > Q*

 

B. Also, P** = (A’D + BC)/(B+D) > P*

 

2. Shift Supply Rightward (increase)

A. Decrease intercept (C), downward movement of curve, or decrease D (tilt curve, more horizontal). C’ = C + DC, where DC is a negative increment.

 

Q*** = (A - C’)/(B+D) > Q*

 

B. P*** = (AD + BC’)/(B+D) < P*

 

II. The Relationships Among Totals, Averages and Marginals

 

This holds for Total, Average and Marginal Utility, Total, Average and Marginal Cost, and more. This example is for a straight line demand curve with vertical and horizontal intercepts at 10.

 

Total: Revenue (R) = Price (P) times Quantity (Q)

 

Average Revenue (AR) = P = R/Q (P is $ per unit, R is $)

 

Marginal Revenue (MR) = limit (DR/ DQ ) (As DQ -> 0)

Or, the Marginal is the slope of the Total curve.

 

Q P, or AR R MR

 

 

0

 

10

 

0

 

--

 

1

 

9

 

9

 

9

 

2

 

8

 

16

 

7

 

3

 

7

 

21

 

5

 

4

 

6

 

24

 

3

 

5

 

5

 

25

 

1

 

6

 

4

 

24

 

-1

 

7

 

3

 

21

 

-3

 

8

 

2

 

16

 

-5

 

9

 

1

 

9

 

-7

 

10

 

0

 

0

 

-9

6 Propositions on Total, Average, and Marginal Relationships

 

1. When a Total is rising, the corresponding Marginal is positive.

2. When a Total is falling, the corresponding Marginal is negative.

3. When a Total reaches a maximum, or a minimum, the Marginal is 0.

4. When the Average is falling, the Marginal must lie below it.

5. When the Average is rising, the Marginal must be above it.

6. When an Average is neither rising nor falling (at a min or max or other stationary point), the Marginal must equal the Average.

A. From an Average to a Marginal (Demand curve is an Average relationship, relating average cost to buyers (price) to quantity.)

 

1. If this is the demand relationship: AR = P = 15 - Q

 

2. Multiply by Q to get the total paid (R): R = PQ = 15Q - Q2

 

3. Differentiate R: MR= (dR/dQ) = 15 - 2Q

 

4. Note that if AR if f(Q), then R is f(Q)Q, the product of 2 functions, which we can differentiate to get the relationship:

 

MR = (dR/dQ) = f(Q)(1) + (Q)f’(Q) , and since f(Q) is AR,

 

MR - AR = MR - f(Q) = f(Q) + Qf’(Q) - f(Q) = Qf’(Q)

 

This is furnishes the relationship in propositions 4, 5, and 6 above.

 

III. Ceteris Paribus (Other Things Equal)

 

A. Many applications: Utility -- CP gives indifference curves, changing goods amounts holding utility level equal; Budget line -- holding prices and income constant, changing goods amounts; and other functions with more than one variable.

 

B. Partial Differentiation

 

1. If y = f(x1,x2,. . . xn), and the x variables are all independent of each other, so if x1 changes it does not affect the others, then given a change in x1, Dx1, with x2, ...,xn all remaining fixed, then the change in y will be Dy and the limit of Dy/Dx1 as x1 ® 0 will be the partial derivative of y with respect to x1. Partial derivative indicates that all the other independent variables in the function are held constant, and treated as constants in the process of differentiation. A partial derivative is has a special symbol (not dy/dx anymore), a variant of the Greek d, written y/xi , and read "the partial derivative of y with respect to xi ." Also, the partial derivative function is no longer f’(x). Now the symbol would be f1 , where the subscript indicates that x1 alone is being allowed to vary.

2. If y = f(x1,x2) = 3x12 + x1x2 + 4x22 then to find y/x1 we treat x2 as a constant during differentiation. So the additive 4x22 will drop out, but terms remain if they are multiplicative constants, like x1x2.

 

y = 6x1 + x2 And y = x1 + 8 x2

x1 x2

 

Like the original function, the partial derivatives are functions of both x1 and x2. If (x1,x2) = (1,3) then f1 = 9 and f2 = 25

 

3. The total derivative is the rate of change (slope) of a function of more than one x variable , when the x variables are not independent of each other. For example, y = f(x,w) and x = g(w), so that w affects y directly, through the function f, and indirectly, through the function g which alters x, and indirectly, through the effects of x on y. So fw expresses only the direct effect, but a total derivative expresses both effects jointly.(dy/dw) = fx (dx/dw) + fw (dw/dw) Note that (dw/dw=1).

 

4. For example, the total derivative, dy/dw, of y = 3x - w2 where x = g(w) = 2w2 + w + 4 is (dy/dw) = 3(4w+1) + (-2w) = 10w + 3. An economic use is a utility function U = U(c,s) where c is the amount of coffee consumed and s is the amount of sugar consumed. If sugar and coffee are complements and the function s = g(c) indicates the relationship, then U = U[c,g(c)] . Then (dU/dc) = U/c + [U/g(c)]g’(c)

 

IV. Using Derivatives to Identify Maximum versus Minimum

 

A. Maximum: First Order Condition, 1st derivative = 0; Second Order Condition, 2nd derivative < 0

 

B. Minimum: FOC, 1st derivative = 0; SOC, 2nd derivative > 0

 

V. Lagrange Multipliers: Optimization Under Constraint

 

A. Given the objective function z = f(x,y) subject to the constraint that g(x,y) = c where c is a constant, we can write the Lagrangian function as

 

Z = f(x,y) + l[c - g(x,y)]

 

Z is considered a function of 3 variables: x, y, and l. The Lagrangian multiplier, l, is some unknown number. But if the constraint is satisfied, then the last term will vanish, whater the value of l. Then Z would be identical with z. To make l vanish, we treat it as an additional variable. The FOC necessary conditions for a maximum are

 

Zl = c - g(x,y) = 0

 

Zx = fx - lgx = 0

 

Zy = fy - lgy = 0

 

The first condition guarantees that the constraint is satisfied, lambda vanishes, and the constrained maximum has been reached by treating it as an unconstrained maximization problem in 3 choice variables. These 3 conditions can be solved for the values of x and y that maximize z, given the constraint.

 

B. A consumer has the utility function: U = x1x2 + 2x1. The marginal utilities are U1 and U2 (the partial derivatives) and are positive for all positive levels of x1 and x2. So for maximum utility, this consumer would buy infinite amounts of both goods, which is not a particularly useful result. If the consumer has $60, and the price of x1 is $4 while the price of x2 is $2, then the consumer’s purchasing power, measured by the budget constraint, is 4x1 + 2x2 = 60. This lowers the range of x1,x2 values, and thus the range of the utility function, which is the "objective" function.

 

The Lagrangian function: Z = x1x2 + 2x1 + l(60 - 4x1 - 2x2)

 

The FOC are Z/¶l = 60 - 4x1 - 2x2 = 0

 

Z/x1 = x2 + 2 - 4l = 0

 

Z/x2 = x1 - 2l = 0

 

This can be solved by substitution and elimination of variables to get x1=8 and x2=14. The utility of an income of 60 is 128.

 

C. If the income amount and prices are not known, the first constraint from the Lagrangian above becomes y - p1x1 - p2x2 = 0, where y is the income and p1 and p2 are the prices. The second FOC becomes x2 + 2 - p1l = 0. The third FOC becomes x1 - p2l = 0. These FOC can be solved for demand functions for goods x1 and x2 as functions of the prices and income by eliminating l, then solving for one of the x’s and substituting that value in the first FOC.

 

Demand for x1: x1 = (y + 2p2)/2p1

 

Demand for x2: x2 = (y - 2p2)/2p2

 

D. These functions can be differentiated and solved for the price and income elasticities.

 

1. For the own price elasticity of x1, e = (x1/p1)(p1/x1). Substitute the demand function for x1 after differentiating and cancel terms. So e11 (the price elasticity of x1 with respect to p1) = [-(y + 2p2) p1 2p1]/[2p12 (Y+2p1)] = -1.

 

2. For the cross price elasticity of x1, e = (x1/p2)(p2/x1). Same process, just differentiate x1 with respect to p2 and substitute in the demand function for x1.

 

3. For the income elasticity of x1, h = (x1/y)(y/x1). Substitute the demand function in for x1 after differentiating and cancel terms.

 

 

I. Present Values: Time Compounding and Discounting

 

The price of the right to a stream of payments, or the price today of an asset to be delivered in the future, is the present value.

 

For a payment in one year,

 

P = R1 / (1+i), where P is the present value, R1 is the principal plus yield coming all at once one year from now, and i is the interest rate that could be earned on the sum P used to buy the stream. So (1+i)P = R1 , or if you invest P you get back P itself plus iP at the end of the year.

 

For a stream of annual returns lasting forever,

 

P = R/i where R (no subscript) is the annual return repeated into the future.

 

The general discounting formula for any stream of income R1 in the first year, R2 in the second, and so on for N years is

 

P = [R1 /(1+i)] + [R2 /(1+i)2 ] + [R3 /(1+i)3 ]+ . . . + [RN /(1+i)N]

 

If all the R’s are equal, the general discounting formula becomes the annuity formula

 

P = (R/i){1 - [1/(1+i)N]}

 

II. Algebra of Growth Rates

 

The relationships of growth in component parts to the growth of the whole is easily calculated.

 

For multiplicative relationships, such as Cobb-Douglas functions, Y = AKa Lb, the growth rate of the whole is the sum of the growth rates of the parts, multiplied by the exponents (minus signs for division):

 

Y%D = A%D + aK%D + bL%D

 

If the relationship is additive, Y = K + L, the growth rate of the whole is the growth rates of the parts, each weighted by that part’s share of the total:

 

Y%D = [K/(K+L)]K%D + [L/(K+L)]L%D