# Now form the lagrangian l 3 x 4 y ? x 1 3 y 1 3 25

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Now form the Lagrangian L = 3 x + 4 y - λ ( x 1 / 3 y 1 / 3 - 25). The first-order conditions are 3 = ( λ/ 3) x - 2 / 3 y 1 / 3 , 4 = ( λ/ 3) x 1 / 3 y - 2 / 3 . Dividing, we find 3 / 4 = y/x , or 3 x/ 4 = y . Substituting in the constraint yields x 2 / 3 (3 / 4) 1 / 3 = 25. Raising to the 3 / 2 power gives 3 x/ 2 = 125, or x * = 250 / 3, and y * = 125 3 / 2. The Hessian of the Lagrangian is: D 2 L = λ 9 2 x - 5 / 3 y 1 / 3 x - 2 / 3 y - 2 / 3 x - 2 / 3 y - 2 / 3 - 2 x 1 / 3 y - 5 / 3 . The Hessian itself is positive definite, with determinant λ 2 x - 2 / 3 y - 2 / 3 / 27 and a positive entry in the upper left. This insures we have a minimum (we don’t need to look at the bordered Hessian when the Hessian is positive definite). 3. Find all maxima of u ( x, y ) = x + y under the constraints x 0, y 0, x + py 10. Answer: The Lagrangian is L ( x, y, μ 0 , μ 1 , μ 2 ) = x + y - μ 0 ( x + py - 10) + μ 1 x + μ 2 y . The first-order conditions are 1 - μ 0 + μ 1 = 0 and 1 / 2 y - 0 + μ 2 = 0. The complementary slackness conditions are μ 0 ( x + py - 10) = 0, μ 1 x = 0, μ 2 y = 0. Of course, μ 0 , μ 1 , μ 2 0. Rewriting the first first-order condition, we find μ 0 = 1 + μ 1 1, whic h implies x + py = 10 by complementary slackness. The second first-order condition can only hold if y > 0, which implies μ 2 = 0 by complementary slackness. Thus 1 / 2 y = 0 > 0, so y = 1 / 4 p 2 μ 2 0 .

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