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Thread: 3 eqns w/ 3 unknowns: 4y2-2x = 2*lambda*x, 8*xy = 2*lambda*y, x2+y2=1

  1. #1
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    3 eqns w/ 3 unknowns: 4y2-2x = 2*lambda*x, 8*xy = 2*lambda*y, x2+y2=1

    Hi everyone!

    I've been stuck on this one for some hours now (about 5 hours to be honest) and I just can't solve it. It is actually a "find min. and max. points"-question but I've come so far that I just have to solve for the Lagrange multiplier and the three equations that I have are:
    4y2-2x = 2*lambda*x
    8*xy = 2*lambda*y
    x2+y2=1

    I need to solve for x and y and x has to be greater than zero.
    I have to do this by hand and I really hope that some of you smart people can figure this out! the result should be three points.
    Thanks in advance, I appreciate every help I can get!

    EDIT: the function that I've initially been given is 4xy2-x2 and it is in the domain of R2 with the restriction: x2+y<=1, x>=0
    Last edited by danishkid; 11-05-2017 at 05:44 PM.

  2. #2
    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by danishkid View Post
    I've been stuck on this one for some hours now (about 5 hours to be honest) and I just can't solve it. It is actually a "find min. and max. points"-question but I've come so far that I just have to solve for the Lagrange multiplier and the three equations that I have are:

    4y2-2x = 2*lambda*x
    8*xy = 2*lambda*y
    x2+y2=1

    I need to solve for x and y and x has to be greater than zero.
    Try using what you learned back in algebra! I'll replace "lambda" with "z", for easy of typesetting. This means we have:

    . . . . .[tex]4y^2\, -\, 2x\, =\, 2xz[/tex]

    . . . . .[tex]8xy\, =\, 2zy[/tex]

    . . . . .[tex]x^2\, +\, y^2\, =\, 1[/tex]

    Solving the first equation for "z=", we get:

    . . . . .[tex]\dfrac{2y^2}{x}\, -\, 1\, =\, z[/tex]

    Solve the second equation for "z=", we get:

    . . . . .[tex]4x\, =\, z[/tex]

    Putting these together, we get:

    . . . . .[tex]4x\, =\, \dfrac{2y^2}{x}\, -\, 1[/tex]

    . . . . .[tex]4x\, +\, 1\, =\, \dfrac{2y^2}{x}[/tex]

    . . . . .[tex]2x^2\, +\, \dfrac{x}{2}\, =\, y^2[/tex]

    Solving the third of the original equations, we get:

    . . . . .[tex]y^2\, =\, 1\, -\, x^2[/tex]

    Putting these together, we get:

    . . . . .[tex]2x^2\, +\, \dfrac{x}{2}\, =\, 1\, -\, x^2[/tex]

    What did you get when you solved this quadratic equation? Where did this lead?

  3. #3
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    Quote Originally Posted by danishkid View Post
    Hi everyone!

    I've been stuck on this one for some hours now (about 5 hours to be honest) and I just can't solve it. It is actually a "find min. and max. points"-question but I've come so far that I just have to solve for the Lagrange multiplier and the three equations that I have are:
    4y2-2x = 2*lambda*x
    8*xy = 2*lambda*y
    x2+y2=1 HOW DO YOU GET THIS?

    I need to solve for x and y and x has to be greater than zero.
    I have to do this by hand and I really hope that some of you smart people can figure this out! the result should be three points.
    Thanks in advance, I appreciate every help I can get!

    EDIT: the function that I've initially been given is 4xy2-x2 and it is in the domain of R2 with the restriction: x2+y<=1, x>=0
    I want to start from the original problem.

    First, did you give us the actual constraint or is it [tex]x^2 + y^2 \le 1.[/tex]

    Second, why did you ignore the second constraint of [tex]x \ge 0[/tex]?

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