Appendix to Chapter 4

1. Appendix to Chapter 4 Demand Theory: A Mathematical Treatment

2. 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

3. 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

4. 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

5. 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….

6. 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

7. 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

8. 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

9. 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

10. 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)

11. 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

12. 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

13. 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

14. 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

15. 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