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Finding max/min using Lagrange Multipliers

  1. Oct 14, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Find the maximum and minimum values of 2x2 + y2 on the curve x2 + y2 - 4x = 5 by the method of Lagrange Multipliers.

    2. Relevant equations

    I will express my Lagrange multipliers as λ.

    3. The attempt at a solution

    Okay so we want the max min of f(x,y) = 2x2 + y2 given the constraint that : x2 + y2 - 4x = 5.

    So the first thing I want to note is I can express my constraint in terms of a function g = x2 + y2 - 4x - 5. I believe that the max or min will occur somewhere on my constraint which happens to be a boundary. Since I don't have any interior to examine, I don't even need to know critical points of f. So the first thing I should do is form my Lagrange equation :

    F = f + λg = 2x2 + y2 + λ(x2 + y2 - 4x - 5)

    Now I should take the derivative of big F with respect to x, y and then λ and form a system of equations right?

    Is this good so far? ( First time trying one of these myself ).
     
  2. jcsd
  3. Oct 14, 2012 #2

    micromass

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    Yes, that is already good so far. Now you will need to calculate partial derivatives and find a system of equations.
     
  4. Oct 14, 2012 #3

    Zondrina

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    Um okay. So I take all my derivatives :

    Fx = 4x + 2λx - 4λ = 2x(λ + 2) - 4λ
    Fy = 2y + 2yλ = 2y(λ + 1)
    Fλ = x2 + y2 - 4x - 5

    Now I find my equations so I can find my values for λ afterwards.

    Fx = 0 [itex]\Rightarrow[/itex]
    0 = 2x(λ + 2) - 4λ
    0 = x(λ + 2) - 2λ

    Fy = 0 [itex]\Rightarrow[/itex]
    0 = 2y(λ + 1)
    0 = y(λ + 1)

    Fλ = 0 [itex]\Rightarrow[/itex]
    0 = x2 + y2 - 4x - 5
    x2 = 4x + 5 - y2

    Now here is where I'm not really sure what to do ^.
     
  5. Oct 14, 2012 #4

    Ray Vickson

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    Pretend, for the moment, that you already know λ. How would you find x and y? Do they satisfy the constraint?

    RGV
     
  6. Oct 14, 2012 #5

    micromass

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    So you got the following system of equations

    [tex]\left\{\begin{eqnarray}
    x(\lambda +2) & = & 2\lambda\\
    y(\lambda +1) & = & 0\\
    5+4x-x^2 & = & y
    \end{eqnarray}\right.[/tex]

    From the second equation we can already deduce [itex]y=0[/itex] or [itex]\lambda=-1[/itex]
    So we can split up in two cases:

    • Case I: y=0
      In this case, the equations reduce to

      [tex]\left\{\begin{eqnarray}
      x(\lambda +2) & = & 2\lambda\\
      5+4x-x^2 & = & 0
      \end{eqnarray}\right.[/tex]

      Solve this equation.

    • Case II: λ=-1
      In this case, we get

      [tex]\left\{\begin{eqnarray}
      -2 & = & x\\
      y & = & 5+4x-x^2
      \end{eqnarray}\right.[/tex]

      Solve this equation.

    Can you continue?
     
  7. Oct 14, 2012 #6

    Zondrina

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    Since you graciously murdered more than half the problem for me, lets see If I can finish it off.

    In case I where y=0, the first thing I would do is factor the second equation giving me either x = -1 or x = 5. So going back to the first equation, x = -1 yields λ = -2/3 and x = 5 yields λ = -10/3.

    So I would continue to the case II where λ = -1, yielding x = -2 which I can use to solve the equation right under it giving me y = -7.

    Is this what you meant?
     
  8. Oct 14, 2012 #7

    micromass

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    Yes, so you get the answers (-1,0), (5,0) and (-2,-7). These are the candidates for maximum and minimum. To see which one is the maximum/minimum, just plug them in in [itex]f(x,y)=2x^2+y^2[/itex] and see which point gives the highest/lowest values.
     
  9. Oct 14, 2012 #8

    Zondrina

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    Ahhh the way that works is amazing! So :

    f(x,y) = 2x2 + y2
    f(-1, 0) = 2
    f(5, 0) = 50
    f(-2, -7) = 8 + 49 = 56

    So my local maximum would occur at (-2, -7) and my local minimum would occur at (-1, 0).

    The only thing though... is that I notice that (-1, 0) satisfies my constraint while (-2, -7) surely does not. So would that imply I have only an absolute minimum at (-1, 0)?
     
  10. Oct 14, 2012 #9

    micromass

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    I don't get what you mean with (-1,0) satisfies the constraint while (-2,-7) doesn't. Which constraint?
     
  11. Oct 14, 2012 #10

    Zondrina

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    From my original problem statement, I'm given that I have to find the min/max values of f(x,y) along the path :

    x2 + y2 - 4x = 5 ( The constraining condition )

    Does it not matter my particular points do not satisfy this condition?

    If so, then I'm assuming the constraint only allows me to find my big F = f + λg which would then allow me to find my system of equations after I take the required partials.
     
  12. Oct 14, 2012 #11

    micromass

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    Oh no, I messed up. I wrote [itex]y=5+4x-x^2[/itex], while it should of course be [itex]y^2=5+4x-x^2[/itex]. I'm sorry, you'll need to solve the system of equations again.
     
  13. Oct 14, 2012 #12

    Zondrina

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    I knew something was up :(. Okay so lets re-do this.

    Case I where y = 0 is still perfect. No problems there.

    Now for Case II where λ=-1. We get x = -2 which we can use to solve the second equation to get : y2 = -7, but uh oh, we're working in ℝ2 so this means nothing to us!

    Thus the only points to consider would be (-1, 0) and (5, 0) which yield :

    f(-1, 0) = 2
    f(5, 0) = 50

    So a local min occurs at (-1, 0) and a local max occurs at (5, 0) which both satisfy the constraint as desired.

    So my suspicion was correct about my points having to satisfy my constraint equation?
     
  14. Oct 14, 2012 #13

    micromass

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    Seems right.

    Yes, you were correct to be suspicious :)
     
  15. Oct 14, 2012 #14

    Zondrina

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    Yay :) thanks so much for all your help. I think I really understand this now so I should be good to go for anything involving this.
     
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