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Max and Min of function via Lagrange multipliers

  1. Oct 13, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    If n is a fixed positive integer, compute the max and min values of the function [tex] (x-y)^n = f(x,y), [/tex] under the constraint [itex] x^2 + 3y^2 = 1 [/itex]

    3. The attempt at a solution
    I got the 4 critical points [itex] (±\frac{\sqrt{3}}{2}, ±\frac{1}{2\sqrt{3}})\,\,\text{and}\,\, (±\frac{1}{2},±\frac{1}{2}) [/itex]
    Correct? My question is : to find the max and min of this function under this constraint, I split n into even and odd cases and found the values from there. Is that the right way to go about the question?
     
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  3. Oct 14, 2012 #2

    CAF123

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    Can anyone offer any advice?
    I alo noted that I got 4 values of x but I only had 3 equations. Does this make sense?
     
  4. Oct 14, 2012 #3

    micromass

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    I think you would get more feedback if you actually show your calculations instead of just your answer.
     
  5. Oct 14, 2012 #4

    Ray Vickson

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    Sure: you can get more than one value from a single equation. For example, the single equation x^2 = 4 has two roots: x = 2 and x = -2.

    RGV
     
  6. Oct 14, 2012 #5

    CAF123

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    I computed [itex] f_{x}, f_{y}, g_{x}, g_{y} [/itex] and solved the system of equations: [tex] n(x-y)^{n-1} = 2\lambda x, -n(x-y)^{n-1} = 6\lambda y\,\,\text{and}\,\, x^2 + 3y^2 =1 [/tex] Adding the first of these equations gives λ=0 or x=-3y. Sub x=-3y into the constraint to give y= ±1/sqrt(12) which implies x= ±(sqrt(3))/2. Now using λ=0 gives x=y , which again by the constraint gives x=±1/2 and y=±1/2. Correct?

    To find the max and min I split n into even and odd. I am pretty confident this is correct because I have suitable answers, but I was wondering if the question wanted me to do something else(purely because I had to prove the cases for any n=2k, n= 2k+1, etc.. which that sort of style of working I had not done before in calculus)
     
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