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Math Help - Find max and min

  1. #1
    Member kezman's Avatar
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    Find max and min

    This problem I think it can be soved with Lagrange multipliers but I tried to solve this with out them.
    Find the maximum and minimum values of the variable x on the Equation:

     4y^{2}-2xy+x^{2}=3

    I dont know if Im having all the solutions when I solve for y:
     (2y - \frac{1}{2}x)^{2}+\frac{3}{4}x^2=3

     y = \dfrac{\sqrt{3 - \frac{3}{4}x^{2}} + \frac{1}{2}x}{2}

    For which I ask:
     \sqrt{3 - \frac{3}{4}x^{2}} > 0

    Solving it I have:

     |x| > 2
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  2. #2
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    Quote Originally Posted by kezman View Post
    This problem I think it can be soved with Lagrange multipliers but I tried to solve this with out them.
    Find the maximum and minimum values of the variable x on the Equation:

     4y^{2}-2xy+x^{2}=3

    I dont know if Im having all the solutions when I solve for y:
     (2y - \frac{1}{2}x)^{2}+\frac{3}{4}x^2=3

     y = \dfrac{\sqrt{3 - \frac{3}{4}x^{2}} + \frac{1}{2}x}{2}

    For which I ask:
     \sqrt{3 - \frac{3}{4}x^{2}} > 0

    Solving it I have:

     |x| > 2
    The problem is that there are 2 functions.
    When you take the square root (asuming it is positive) you get two different functions. One for positive and one for negative.
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  3. #3
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    Hello, kezman!

    If I read the problem correctly, your approach is wrong . . .


    Find the maximum and minimum values of x on the equation:

    . .  4y^{2}-2xy+x^2\:=\:3

    If we're supposed to maximize/minimize x, we must solve for x.

    We have the quadratic: . x^2 - 2yx + 4y^2 - 3 \:=\:0

    Quadratic Formula: . x \;=\;\frac{2y \pm\sqrt{4y^2 - 4(4y^2-3)}}{2}

    . . and we have: . x \;=\;y \pm \sqrt{3-3y^2}

    And that is the function we must maximize/minimize.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    With Lagrange multipliers, we want to maximize x
    . . with the constraint: 4y^2 - 2xy + x^2 \:= \:3

    So we have: . F(x,y,\lambda) \:=\:x + \lambda(4y^2 - 2xy + x^2 - 3)

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