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Appendix to Chapter 4 Demand Theory: A Mathematical Treatment

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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

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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

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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

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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….

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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

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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

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Example: Cobb-Douglas Utility Function U(X, Y) = X a Y 1-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

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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

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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)

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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

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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

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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

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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

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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

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Economics 214 Lecture 37 Constrained Optimization.

Economics 214 Lecture 37 Constrained Optimization.

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