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: Finding max/min given contraint

  1. Dec 15, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the product of the maximum and minimum values of the function f(x, y) = xy
    on the ellipse (x^2)(1/9) + y^2 = 2


    3. The attempt at a solution

    I tried soving using lagrange multiplier and got:

    fx = y - (2/9)(x*lambda)
    fy = x - 2y*lambda
    flambda = (x^2)(1/9) + y^2 - 2

    then I set these = 0, but my answer came out wrong.. any suggestions for figuring out the min/max?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 15, 2008 #2
    For y I'm getting, [tex] y = \pm 1 [/tex]

    The way I set this up was as follows:

    Define:

    [tex] f(x,y) = xy; g(x,y) = \frac{x^{2}}{9} + y^{2} = 2 \Rightarrow h(x,y,\lambda) = f(x,y) + \lambda g(x,y) [/tex]

    (Note that the sign in front of [tex] \lambda [/tex] does not matter)

    So let's take our partials, we get:

    [tex] \frac{dh}{dx} = y + \frac{2 \lambda}{9}x, \frac{dh}{dy} = x + 2 \lambda y, \frac{dh}{d\lambda} = \frac{x^2}{9} +y^2 - 2 [/tex]

    We know that each of those partials vanish i.e. we can set each to 0.

    The first one gives us

    [tex] 9y = -2 \lambda x [/tex]

    and the second one gives us

    [tex] \frac{x}{y} = -2 \lambda[/tex]

    And by simple substitution we get:

    [tex] 9y^{2} = x^{2} [/tex]

    So let's substitute it into our 3rd equation to get:

    [tex] y^{2} + y^{2} = 2 [/tex]

    Which yields our desired result of [tex] y = \pm 1 [/tex]. Now we can plug this into our g(x,y) to get [tex] x = \pm 3 [/tex]

    Note that it doesn't matter which value for y we pick therefore our solution set will be:

    [tex] (1,3), (1,-3), (-1,3), (-1,-3) [/tex]

    Now if you don't want to do this using Lagrange Multipliers, we can just realize that we can rewrite our g(x,y) as

    [tex] y = \pm \sqrt{2 - \frac{x^2}{9}} [/tex]

    and now we can substitute this into our f(x,y) get an equation of one variable i.e.

    [tex] \bar{f}(x) = \pm x \cdot \sqrt{2 - \frac{x^2}{9}} [/tex]

    Now we can proceed using the techniques you learned in Calculus 1 (I'm going to use Maple because I'm lazy)

    > a:=x*sqrt(2-x^2/9);

    a := [tex] \frac{1}{3} x \sqrt{18 - x}[/tex]

    > solve(diff(a,x)=0,x);

    -3, 3

    Note that choosing the negative root produces the same results.
     
    Last edited: Dec 15, 2008
  4. Dec 15, 2008 #3
    what equations are you using to get y = to +/- 1?
     
  5. Dec 15, 2008 #4
    Refresh your page :)
     
  6. Dec 15, 2008 #5
    Thanks!!! This helps a lot
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook