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**HallsofIvy** $\displaystyle f(x,y)=2x^4-4x^2y^2+5y^4$ so $\displaystyle \nabla f= (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}$. At any max or min in the interior of the set we must have $\displaystyle 8x^2- 8xy^2= 8x(x^2- y^2)= 8x(x- y)(x+ y)= 0$ so one of x= 0, x=y, x=-y must be true. Also $\displaystyle -8x^2y+ 20y^3= 4y(5y^2- 2x^2)= 4y(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y+ \sqrt{2}x)= 0$ so one of y= 0, $\displaystyle \sqrt{5}y= \sqrt{2}x$, or $\displaystyle \sqrt{5}y= -\sqrt{2}x$ must be true.

If x= 0, only y= 0 satisfies the other equations. If y= x again we have only x= y= 0. I don't see how x= -1/2, y= -1/5 satisfy those equations. I find only x= y= 0.

On the boundary, $\displaystyle g(x, y)= x^4+ y^4= 80$, $\displaystyle \nabla g= 4x^3\vec{i}+ 4y^3\vec{j}$ so we must have $\displaystyle (8x^3- 8xy^2)\vec{i}+ (-8x^2y+ 20y^3)\vec{j}= \lambda(4x^3\vec{i}+ 4y^3\vec{j}$ or $\displaystyle 8x^3- 8xy^2= \lambda(4x^3)$ and $\displaystyle -8x^2y+ 20y^3= \lambda(4y^3)$.

It's never necessary to find $\displaystyle \lambda$ since that is not relevant to the solution. Instead, eliminate $\displaystyle \lambda$ by **dividing** one equation by the other:

$\displaystyle \frac{8x^3- 8xy^2}{-8x^2y+ 20y^3}= \frac{4x^3}{4y^3}$

$\displaystyle \frac{2(x- y)(x+ y)}{(\sqrt{5}y- \sqrt{2}x)(\sqrt{5}y- \sqrt{2}x)}= \frac{x^2}{y^2}$.

That, together with $\displaystyle x^4+ y^4= 80$ gives two equations to solve for x and y.