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Lagrangian_Optimization_Example

Course: ECON 11, Winter 2007
School: UCLA
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Word Count: 808

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Example Lagrangian Constrained Optimization with the Lagrangian Multiplier The Lagrangian multiplier method allows us to solve for the utility maximizing bundle for consumers with a wide variety of utility functions. The utility functions may include several 1 goods, goods which are independent, complements, substitutes, as well as goods consumed in a certain proportion, or goods which have declining marginal...

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Example Lagrangian Constrained Optimization with the Lagrangian Multiplier The Lagrangian multiplier method allows us to solve for the utility maximizing bundle for consumers with a wide variety of utility functions. The utility functions may include several 1 goods, goods which are independent, complements, substitutes, as well as goods consumed in a certain proportion, or goods which have declining marginal utility, goods which become bads if the quantity reaches a certain point, as well as arguments which are not goods at all, such as labor which gives disutility but which is necessary to earn income. The general method we will use to solve these problems is the following: 1. We construct the Lagrangian problem. We wish to maximize total utility, subject to the constraint that the quantity of all of the goods, times their prices, cannot exceed the budget: Max L = (Utility Function) (PxQx + PyQy + PzQz Budget) Here we show three goods, X, Y and Z, but this applies to utility functions with any number of goods. 2. We then take the first order conditions. This means that we take the derivative of the utility function with respect to a small change in the quantity of the first good (giving us the marginal utility of the first good), and the derivative of the budget constraint with respect to a small change in the quantity of first good (giving us the price of the first good, times what the whole budget constraint is being multiplied by, , pronounced lambda). We continue doing this for the second good, the third good, however many arguments there are in the utility function, and then we take the derivative of the budget constraint itself. For a utility function with N arguments, we will have N + 1 first order conditions, because of the budget constraint. Lagrangian Example 3. We then want to find the relationship between the numerical quantities of one good and another. We find this using algebra and canceling terms. A typical way is to find two first order equations that have a term in common, such as a third good, and 2 then canceling the term, then canceling the term for the third good. This will be shown in the examples below. 4. Once we are able to describe each of the arguments in terms of one of the goods, we can plug these relationships back into the budget constraint and solve for that term. We then solve for the numerical quantities of all of the other arguments. It is a good idea to check that the budget constraint is satisfied. 5. Finally, we check to see that the marginal ratio utility/price is the same across all of the goods. This proves that the bundle maximizes utility. Lagrangian Example 3 EX-A A. A consumer's utility function increases with the product of the quantities of three goods X, Y and Z: U(X, Y, Z) = XYZ. The consumer faces a budget constraint. Here income is \$90 and the prices of X, Y and Z are \$1, \$2 and \$3 respectively. The utility maximizing bundle is: a. X = 10, Y = 10, Z = 10 b. X = 10, Y = 15, Z = 20 c. X = 30, Y = 15, Z = 10 d. X = 20, Y = 25, Z = 10 e. X = 15, Y = 20, Z = 15 The Lagrangian problem is thus: Max L = ___________ (__X + __Y + __Z ____) The first order conditions are: (A.1) dL dX (A.2) dL dY (A.3) dL dZ (A.4) dL d A good general rule to proceed is to solve for one variable in terms of another, and then plug that ratio into one of the first order equations. We begin by solving for X in terms of Z. We start by comparing the first order equations that each have the other good, Y, included, so that we may cancel it out, leaving = d(_____) ___ = ______ __ = 0 dX = d(_____) ___ = ______ __ = 0 dY = d(_____) ___ = _______ __ = 0 dZ = ____ ___X ___Y ___Z = 0 Lagrangian Example only X and Z. First we eliminate the term: From (A.1) and (A.3), (A.1) ____ ___ = 0 ____ = ___ (A.3) ____ ___ = 0 ____ = ___ Multiply (A.1) above by ___: 3____ = 3__ 4 Compare the left hand sides of (A.1) and (A.3), which are equal. We may now eliminate the term, and express X in terms of Z, by dividing each side by Y: ____ = 3 = _____ ____ = _____ X = ___Z We can similarly describe Y in terms of Z by comparing (A.2) and (A.3): (A.2) ____ ___ = 0 (A.3) ____ ___ = 0 Multiply (A.2) by 3 on both sides, and (A.3) by 2 on both sides, so that we can eliminate the terms: (A.2) ___ 2 = 0 3___ 6 = 0 3___ = 6 (A.3) ____ 3 = 0 2____ 6 = 0 2____ = 6 2____ = 3____ Divide both sides by X: 2___ = 3____ Lagrangian Example Y = _____Z We can now describe X and Y in terms of Z. We next solve for Z by plugging in these relationships to the budget constraint: (A.4) ____ ___X ___Y ___Z = 0 ____ ___Z ___Z ___Z = 0 ____ = ____Z Z* = ____ Y* = ____ X* = ____ Finally we check to see that the ratio of MU/P is equal across the three goods. The marginal utility is the derivative of the utility function with respect to a change in each of the goods: MUx = ____ = ____ * ____ = ____ MUy = ____ = ____ * ____ = ____ MUz = ____ = ____ * ____ = ____ MUx = ____ = MUy = ____ = MUz = ____ Px Py Pz 5 The benefit per dollar is the same across the three goods.
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