Lecture 4

Remark note that the logic goes as follows first we

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Unformatted text preview: < α < 1 and x1 , x2 ≥ 0 are consumption of good 1 and 2 (such a function is called Cobb-Douglas ). Let p1 , p2 > 0 the price of each good and w > 0 be the wealth. Then the consumer’s utility maximization problem (UMP) is maximize subject to α log x1 + (1 − α) log x2 x1 ≥ 0, x2 ≥ 0, p1 x1 + p2 x2 ≤ w. The budget set {(x1 , x2 ) | x1 ≥ 0, x2 ≥ 0, p1 x1 + p2 x2 ≤ w} is nonempty and compact, and u(x) = α log x1 + (1 − α) log x2 is upper semicontinuous by defining log 0 = −∞. Hence UMP has a solution. Since the constraints are linear, the Guignard constraint qualification holds. Therefore we can apply the Karush-Kuhn-Tucker theorem. Let L(x, λ, µ) = α log x1 + (1 − α) log x2 + λ(w − p1 x1 − p2 x2 ) + µ1 x1 + µ2 x2 , where λ ≥ 0 is the Lagrange multiplier corresponding to the budget constraint p1 x1 + p2 x2 ≤ w ⇐⇒ w − p1 x1 − p2 x2 ≥ 0 and µn is the Lagrange multiplier corresponding to xn ≥ 0 for n = 1, 2. By the first-order condition, we get ∂ : ∂ x1 ∂ : ∂ x2 α − λp1 + µ1 = 0, x1 1−α − λp2 + µ2 = 0. x2 By the complementary slackness condition, we have λ(w − p1 x1 − p2 x2 ) = 0, µ1 x1 = 0, µ2 x2 = 0. 2014W Econ 172B Operations Research (B) Alexis Akira Toda Since log 0 = −∞, x1 = 0 or x2 = 0 cannot be an optimal solution. Hence x1 , x2 > 0, so by complementary slackness we get µ1 = µ2 = 0. Then by the α first order condition we get x1 = λα1 , x2 = 1−2 , so λ > 0. Substituting these p λp into the budget constraint p1 x1 + p2 x2 = w, we get α 1−α 1 + = w ⇐⇒ λ = , λ λ w so the solution is (x1 , x2 ) = αw (1 − α)w , p1 p2 . Remark. • Note that the logic goes as follows. First, we show by some means (e.g., Bolzano-Weierstrass) that an optimal solution exists. Second, we verify that the Karush-Kuhn-Tucker theorem applies and derive the necessary conditions for optimality. Third, since the necessary conditions led to a unique candidate, it must be the solution. • Since log 0 = −∞, we know that the optimal solution cannot be x1 = 0 or x2 = 0, so the constraints x1 ≥ 0, x2 ≥ 0 never binds at the solu...
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