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

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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 .
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern