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LaGrange multipliers!

  1. Aug 28, 2008 #1
    1. The problem statement, all variables and given/known data

    Using the method of lagrange multipliers, find the points on the curve 3x² - 4xy + 6y² = 140 which are closest and furthermost from the ORIGIN and the corresponding distances between them

    3. The attempt at a solution
    I have done roughly half the question but appear to be stuck!

    Firstly I find distance from a point (x,y) to the (0,0) using distance equation.
    This yields:

    d² = x² + y² = f(x,y)

    My constraint equation is given by the equation of the curve given, i.e.

    g(x,y) = 3x² - 4xy + 6y² = 140

    Now I want to use grad(f) = λgrad(g)

    2xi + 2yj = λ[(6x-4y)i + (-4x+12y)j]
    And by equating coefficients,
    2x = λ(6x-4y) (1)
    2y = λ(-4x+12y) (2)

    This is where I get stuck, I have never encountered a problem like this whereby one of the variables depends on itself. What is the best way to tackle this? Should I sub (1) into (2)?
  2. jcsd
  3. Aug 29, 2008 #2
    Well it appears to me, you now have three equation for your three unknowns x, y, lambda (one of which is not linear). It further appears to me that subbing these equation into one another might be helpful, give it a try.

    I would expect you to find four solutions, two corresponding to the max. distance and two corresponding to the min. distance.
    Last edited: Aug 29, 2008
  4. Aug 29, 2008 #3
    If I sub (2) into (1), I get:

    x = 3λx + 4λ²x - 12²λy

    Doesn't help! :confused:
  5. Aug 29, 2008 #4
    You should try and eliminate one unknown after the other, for example you could solve your equation (1) for x, giving


    If you plug this into equation (2) as well as into g(x,y)=140 you will be left with two equations involving only y and lambda. then you continue by solving this simplified system of equations.

    Of course you can also start by eliminating x or lambda...

    EDIT: It is probably easier to eliminate lambda first..but i didn't carry out the calculations, so this is only a guess:smile:
  6. Aug 29, 2008 #5
    hmmm, if I sub the x as shown in the above post into equation 2, I return y=0

    seems fishy..
  7. Aug 29, 2008 #6
    Yes, one "solution" would be y=0, resulting in x=0 as well, but g(0,0) is not equal to 140 is it, so (0,0) is not a solution. If you plug


    in (2) yu get a quadratic equation in y with two solutions --- one of which is zero... what about the other one.... I suggest you play around a little with these equation, try what happens if you eliminate differnent variables first.
  8. Aug 29, 2008 #7


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    One variable depends on itself? If you mean there is an x on both sides of one equation and a y on both sides of the other, problems like that are common in elementary algebra. Surely you have done problems like that before. Basically you have 3 equations in the three unknown values x, y, and [itex]\lambda[/itex].
    Since [itex]\lambda[/itex] is not part of the information you need to get, one step that may help in a problem like this is to divide one equation by the other: x/y= (3x- 2y)/(-2x+ 6y). That gives x(-2x+ 6y)= y(3x- 2y) or -2x2+ 6xy= 3xy- 2y2 and finally 2x2+ 2y2+ 9xy= 0 Now the problem is to find where that intersects 3x² - 4xy + 6y² = 140.
    Last edited: Aug 30, 2008
  9. Aug 29, 2008 #8
    Subbing [​IMG] into (2)

    returns y(18λ - 2 - 28λ²) = 0
    therefore y=0 or (18λ - 2 - 28λ²) = 0

    if y=0, x=0, but this does not satisfy g(x,y)=140 hence y=0 is not a solution

    if (18λ - 2 - 28λ²) = 0 there are 2 cases (by factorising)
    i) λ = 1/2
    ii) λ = 1/7

    CASE (i)
    sub λ = 1/2 into [​IMG]
    this returns x=2y

    subbing this into g(x,y) yields y(-2y+6)=140
    thus y=140 or -67
    subbing these two values into [​IMG] seperately gives 2 points, P1=(280,140) and P2=(-67/2,67)

    before proceeding to case ii) where λ = 1/7, is my method correct?
  10. Aug 30, 2008 #9
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