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Thread: minimum value

  1. #1
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    Red face minimum value

    If $\displaystyle x,y\in\mathbb{R}$ and $\displaystyle x^2+xy+y^2 = 1$. Then find Minimum value of $\displaystyle x^3y+xy^3+4$
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Re: minimum value

    Quote Originally Posted by jacks View Post
    If $\displaystyle x,y\in\mathbb{R}$ and $\displaystyle x^2+xy+y^2 = 1$. Then find Minimum value of $\displaystyle x^3y+xy^3+4$
    Using the method of Lagrange multipliers,

    $\displaystyle 3x^2y+y^3=\lambda(2x+y)$

    $\displaystyle x^3+3xy^2=\lambda(x+2y)$

    Adding, we get

    $\displaystyle (x+y)^3=3\lambda(x+y)$

    $\displaystyle x+y=0$ or $\displaystyle (x+y)^2=3\lambda$

    Case 1: $\displaystyle x+y=0$

    $\displaystyle x^2-x^2+x^2=1$

    $\displaystyle x=\pm 1$

    $\displaystyle y=\mp 1$

    $\displaystyle x^3y+xy^3+4=-1-1+4=2$

    Case 2: $\displaystyle (x+y)^2=3\lambda$

    $\displaystyle 9x^2y+3y^3=(x+y)^2(2x+y)$

    $\displaystyle =(x^2+y^2+2xy)(2x+y)$

    $\displaystyle =2x^3+2xy^2+4x^2y+x^2y+y^3+2xy^2$

    $\displaystyle =2x^3+y^3+5x^2y+4xy^2$

    $\displaystyle 4x^2y-4xy^2-2x^3+2y^3=0$

    $\displaystyle 4xy(x-y)-2(x^3-y^3)=0$

    $\displaystyle 4xy(x-y)-2(x-y)(x^2+xy+y^2)=0$

    $\displaystyle (x-y)(4xy-2x^2-2xy-2y^2)=0$

    $\displaystyle (x-y)(2xy-2x^2-2y^2)=0$

    $\displaystyle (x-y)(x^2+y^2-xy)=0$

    $\displaystyle x-y=0$ or $\displaystyle x^2+y^2-xy=0$

    Case 2a: $\displaystyle x-y=0$

    $\displaystyle 3x^2=1$

    $\displaystyle x=\pm\frac{1}{\sqrt{3}}$

    $\displaystyle y=\pm\frac{1}{\sqrt{3}}$

    $\displaystyle x^3y+xy^3+4=\frac{2}{9}+4=\frac{38}{9}$

    Case 2b: $\displaystyle x^2+y^2-xy=0$

    $\displaystyle 2(x^2+y^2)=1$

    $\displaystyle x^2+y^2=\frac{1}{2}$

    $\displaystyle 2xy=1$

    $\displaystyle (x-y)^2=-\frac{1}{2}$

    No solutions.

    --------------------------------------------------

    So, the minimum value of $\displaystyle x^3y+xy^3+4$ is 2.
    Last edited by alexmahone; Jan 7th 2012 at 10:46 AM.
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  3. #3
    Grand Panjandrum
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    Re: minimum value

    Quote Originally Posted by jacks View Post
    If $\displaystyle x,y\in\mathbb{R}$ and $\displaystyle x^2+xy+y^2 = 1$. Then find Minimum value of $\displaystyle x^3y+xy^3+4$
    Switch to polars and substitute $\displaystyle r^4$ from the constraint into the objective then simplify to give a standard 1D unconstrained problem.

    CB
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