Answer to Question #287926 in Differential Equations for Varun

Question

Find the extreme values of

x^4+y^4-2(x-y)^2

Answer 1

$f(x,y)=x^4+y^4-2(x-y)^2\\ \dfrac{\partial f}{\partial x}=4x^3-4(x-y)=4(x^3-x+y)\\ \dfrac{\partial f}{\partial y}=4(y^3-y+x)\\ \text{At extremum, }\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial y}=0\\ \therefore x^3-x+y=0~~~(i), y^3-y+x=0~~~~(ii)\\ \text{Adding (i) and (ii), }\\ x^3+y^3=0\Rightarrow x=-y\\ \text{Putting x=-y in (i) and (ii), }\\ (x,y)=(0,0), (\sqrt 2, -\sqrt 2), (-\sqrt 2, \sqrt 2)\\ \dfrac{\partial^2 f}{\partial x^2}=4(3x^2-1), \dfrac{\partial^2 f}{\partial x^2}=4(3y^2-1)\\ \dfrac{\partial^2 f}{\partial x \partial y}=4=\dfrac{\partial^2 f}{\partial y \partial x}\\ \therefore \text{Hassein Matrix,} \\ H(x,y)=\begin{bmatrix} 12x^2-4&4\\4&12y^2-4\end{bmatrix}\\$

$\text{Determinant, } D(x,y)=144x^2y^2-48(x^2+y^2)\\ \therefore D(\sqrt2, -\sqrt2)=D(-\sqrt2, \sqrt2)=384>0\\ \dfrac{\partial^2 f}{\partial x^2}|_{(\sqrt2, -\sqrt2)}=\dfrac{\partial^2 f}{\partial x^2}|_{(-\sqrt2, \sqrt2)} =20>0\\ \text{Hence, $f$ has two local minima at }\\ \text{$(\sqrt2, -\sqrt2)$ and $(-\sqrt2, \sqrt2)$}\\ ~\\ \text{Now, } D(0,0)=0\\ \text{So, we can't decide anything by second derivative test}\\ \text{Now if we approach at (0,0) by $y=x$,}\\ f(x,y)=f(x,x)=2x^4>0\\ \text{Now if we approach at (0,0) by $y=-x$,}\\ f(x,y)=f(x,x)=2x^4-8x^2<0~~ \text{ near zero.}\\~\\ \text{So there is no extremum at $(0,0)$.}$