I'm struggling with an excise that I can't find a way to solve. I need to find the absolute minimum and maximum on a function: $$f(x,y)=x-y$$
on the curve defined as: $$x^4+y^4+xy=0$$
I tried with the Langrage multiplier method but the system does not have solution (or it's too complex for me). I don't think I can use polar coordinates because it is not a circumference.
Solutions should be $(-\sqrt2/2;\sqrt2/2)$ - minimum and $(\sqrt2/2;-\sqrt2/2)$ - maximum
Thanks.
Define $g(x,y)=x^4+y^4-xy$. Then $\nabla f=\lambda \nabla g$ gives $$(1,-1)=\lambda (4x^3+y,4y^3+x),$$ so $1=\lambda (4x^3+y)$ and $-1=\lambda(4y^3+x)$. Note $\lambda\neq 0$ since then we would have $1=0$.
Hence $\lambda (4x^3+y)=-\lambda(4y^3+x)$. Dividing by $\lambda$ gives $4x^3+y=-4y^3-x$, which rearranges to $$4x^3+4y^3+x+y=0,$$ or $$(x+y)(4(x^2+y^2-xy)+1)=0.$$ Hence $x=-y$ or $4(x^2+y^2-xy)+1=0$, but we cannot have $4(x^2+y^2-xy)+1=0$ since $$4(x^2+y^2-xy)+1=4\left(x-\frac{y}{2}\right)^2+3y^2+1\geq1.$$ Now we have $x=-y$, this can be substituted into $g(x,y)=0$ to find the values of $x$ and $y$ at the critical points.