1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Find the maximum and minimum values of the function (Answers included).

  1. Mar 26, 2012 #1


    User Avatar

    1. The problem statement, all variables and given/known data
    Find the maximum and minimum values of the function
    f(x,y) = 2x^2 - 28xy + 2y^2 - 1 on the disk x^2 + y^2 <= 1.

    (a) Find the maximum.
    (b) Find the minimum.

    2. Relevant equations
    Lagrange Multiplier method and partial differentiation.

    3. The attempt at a solution
    My work is attached as MyWork.jpg. If anything is unclear, tell me and I will restate it (or whatever it is you request). In the image, I forgot to plug in x and y in f(x,y) to get f(sqrt(1/2), sqrt(1/2). This yields the correct answer for the minimum: 2*(sqrt(1/2))^2 - 28*sqrt(1/2)*sqrt(1/2) + 2(sqrt(1/2))^2 - 1 whereas the correct answer for the maximum is: 2*(sqrt(1/2))^2 + 28*sqrt(1/2)*sqrt(1/2) + 2(sqrt(1/2))^2 - 1. So, my ultimate question is, what are the reasons for choosing positive or negative square roots for x and/or y?

    Attached Files:

  2. jcsd
  3. Mar 26, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    In the problem max f(x) s.t. g(x) <= 0, if you want the Max you should use L = f - λg, with λ ≥ 0; if you want the Min you should use L = f + λg, with λ >= 0 (or use f - λg with λ ≤ 0).

    How can you remember this? The easiest way is to form L so that it is more favourable than f for feasible points; that is, for feasible points, L should be ≥ f in a max problem and should be ≤ f in a min problem. So, if the constraint is g ≤ 0 we get something ≥ f in a max problem by subtracting g, that is, by using L = f - λg with λ ≥ 0.

    Note that for an equality constrained problem, with constraint g = 0, the sign of λ is not determined, and it does not matter whether you write L = f + λg or L = f - λg.

    So, in your analysis you need to look at the sign of your λ.

  4. Mar 26, 2012 #3


    User Avatar
    Science Advisor

    The Lagrange multiplier method requires that you have "=" in the constraint. You can use that to determine any max or min on the boundary. Just set the partial derivatives equal to 0 in the interior of the disc.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook