minimum value

**jacks** · January 7th 2012, 09:45 AM

If $x,y\in\mathbb{R}$ and $x^2+xy+y^2 = 1$ . Then find Minimum value of $x^3y+xy^3+4$

**alexmahone** · January 7th 2012, 10:13 AM

Originally Posted by jacks

If $x,y\in\mathbb{R}$ and $x^2+xy+y^2 = 1$ . Then find Minimum value of $x^3y+xy^3+4$

Using the method of Lagrange multipliers,

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

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

Adding, we get

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

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

Case 1: $x+y=0$

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

$x=\pm 1$

$y=\mp 1$

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

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

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

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

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

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

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

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

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

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

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

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

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

Case 2a: $x-y=0$

$3x^2=1$

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

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

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

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

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

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

$2xy=1$

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

No solutions.

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

So, the minimum value of $x^3y+xy^3+4$ is 2.

**CaptainBlack** · January 7th 2012, 10:47 AM

Originally Posted by jacks

If $x,y\in\mathbb{R}$ and $x^2+xy+y^2 = 1$ . Then find Minimum value of $x^3y+xy^3+4$

Switch to polars and substitute $r^4$ from the constraint into the objective then simplify to give a standard 1D unconstrained problem.

CB

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