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Math Help - I think my algebra died

  1. #1
    Member billym's Avatar
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    Red face I think my algebra died

    Why am I getting stuck on these questions?

    Find and classify the stationary points:

    f(x,y)=(x-y)^3+x^2y^2

    f_x=3(x-y)^2+2xy^2

    f_y=-3(x-y)^2+2x^2y

    I forget how to find the roots. What to I need to revise? Any links?
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  2. #2
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    Quote Originally Posted by billym View Post
    Why am I getting stuck on these questions?

    Find and classify the stationary points:

    f(x,y)=(x-y)^3+x^2y^2

    f_x=3(x-y)^2+2xy^2

    f_y=-3(x-y)^2+2x^2y

    I forget how to find the roots. What to I need to revise? Any links?
    It is not trivial and requires quite some tricky algebra:

    f_x=0=f_y \Longrightarrow 3(x-y)^2+2xy^2=-3(x-y)^2+2x^2y  \Longrightarrow 6(x-y)^2-2xy(x-y)=0 \Longrightarrow

    \Longrightarrow 3(x-y)\left[2(x-y)-xy\right]=0 \Longrightarrow x=y\,,\,\,or \,\,2(x-y)=xy\Longleftrightarrow x-y=\frac{xy}{2}

    If x = y then plugging into f_x we get x=y=0, else we plug x-y=\frac{xy}{2}\,\,in\,\,f_x\,,\,\,f_y and:

    f_x=3\left(\frac{xy}{2}\right)^2+2xy^2 \Longrightarrow xy^2 \left(\frac{3}{4}x+2\right)=0, and again x = y = 0 or else x=-\frac{8}{3},
    and doing the same with the other partial derivative we get that also y=-\frac{8}{3}

    Nervertheless these last two values for x,y don't work, as you can check at once, so we're left with only...

    Tonio
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  3. #3
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    <br />
\Longrightarrow 6(x-y)^2-2xy(x-y)=0 \Longrightarrow<br />
    <br />
\Longrightarrow 3(x-y)\left[2(x-y)-xy\right]=0<br />
    There is a mistake there, it is supposed to be


    <br />
\Longrightarrow 2(x-y)\left[3(x-y)-xy\right]=0<br />
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