Appendix to Chapter 4

Demand Theory: A Mathematical Treatment

Consumer Maximization

Maximize U(X,Y) subject to the constraint that all income is spent on the two goods

PxX + PyY = Income (I) Use technique of constrained optimization:

– Describes the conditions of utility maximization

Lagrangian Method

Used to maximize or minimize a function subject to a constraint

Lagrangian is the function to be maximized or minimized

λ = lagrangian multiplier Take the utility function to be maximized and

subtract the lagrangain multiplier multiplied by the constraint as a sum equal to zero

Lagrangian Method

U(X, Y) – λ (PxX + PyY – I) If we choose values of X that satisfy the

budget constraint, the sum of the last term will be zero

Differentiate this function three times with respect to X, Y and λ and equate them to zero

This will give us the three necessary conditions for maximization

Lagrangian Method

We will end up with the following three conditions:– MUx – λPx = 0– MUy – λPy = 0– PxX + PyY – I = 0

What do these mean? – MUx = λPx: Marginal Utility from consuming one

more X = a multiple (λ) of its price– MUy = λPy: Marginal Utility….

Lagrangian Method

If we combine the first two equations (the third is the budget constraint), we get:– λ = MUx/Px = MUy/Py – This is the equal marginal principal from chapter

three– To optimize (maximize utility subject to a budget

constraint), the consumer MUST GET THE SAME UTILITY FROM THE LAST DOLLAR SPENT ON BOTH X AND Y

Marginal Utility of Income

λ = MU of income, or marginal utility of adding one dollar to the budget

We will see in an example how this works, but for now:– If λ = 1/100– Then if Income increases by $1, Utility will

increase by 1/100

Example: Cobb-Douglas Utility Function

U(X, Y) = XaY1-a

We can express this function as linear in logs: alog(X) + (1-a)log(Y)

These two are equivalent in that they yield identical demand functions for X and Y

Lagrangian Set-up

alog(X) + (1-a)logY – λ(PxX +PyY – I) Differentiating with respect to X, Y and λ, and

setting equal to zero gives three necessary conditions for a maximum

X: a/X – λPx = 0 Y: (1-a)/Y – λPy = 0 λ: PxX + PyY – I = 0 Solve for PxX and PyY and substitute into the

third equation

Lagrangian Set-up

Solving for PxX and PyY gives:– PxX = a/λ– PyY = (1-a)/λ

Now: substituting these back into the budget constraint gives:– a/λ + (1-a)/λ – I = 0– And solving for λ gives: λ = 1/Income (I)

Lagrangian

If λ = 1/I then we can use λ as a function of Income to solve for X and Y using the two original conditions

Recall:– PxX = a/λ and PyY = (1-a)/λ– Now: PxX = a/(1/I) = Ia– And: PyY = (1-a)I– So: X = Ia/Px and Y = I(1-a)/Py

Lagrangian

Notice that the demand for X is dependent on Income and the price of X, while the demand for Y is dependent on Income and the price of Y

Demand for X, Y, NOT dependent on the price of the other good

Cross-price elasticity is equal to zero

Meaning of Lagrangian Multiplier

λ = Marginal Utility of an additional dollar of Income

If λ = 1/100, then if income increases by $1, utility should increase by 1/100

Duality

Optimization decision is either a maximization decision OR a minimization decision

We can use a Lagrangian to:– Maximize utility subject to the budget constraint,

OR– Minimize the budget constraint subject to a given

level of utility

Duality and Minimization

Lagrangian problem would be:– Minimize PxX + PyY subject to U(X,Y)=U*

Formal set up would look like this:– PxX + PyY – μ(U(X,Y) – U*)– Where U* = a fixed, given level of utility just the

same as Income was fixed in the maximization problem

This method will yield the same demand functions as the maximization approach