Max and min of $f(x,y)=|2x-y|(x^2+y^2-20)$ The function is continuous in $R^2$.If I consider$ f(x,x)=|x|(2x^2-20)$ is not limited up for $x\rightarrow \infty$.But when i calculate partial derivatives of f(x,y) i have to study different cases about sign of (2x-y)?
 A: If a function has an absolute form, it might become less-abstract if you convert the form to remove absolute form. 
In your case :
$$ f(x,y) =  \begin{cases}   (2x-y)(x^{2} + y^{2} - 20), \: \text{ for} \:\:\:  2x-y \ge 0 \\ -(2x-y)(x^{2} + y^{2} - 20), \: \text{ for}\:\:\: 2x-y < 0
\end{cases} $$
Then investigate the min or max for each region. 

*If you found an optimum value using partial derivative of the first function, at $(a, b)$ with $a < \frac{b}{2}$. Then this would not be
  valid, since that point $(a,b)$ is not in domain of the first
  function.

Hope this helps.
A: the function is even for $(x,y)$ thus define $y'=-y_1$ and we re-write the equation as $|2x+y_1|(x^2+y_1^2-20)$ now to maximize this take $(x,y_1)$ as positive. $$f(x,y_1)=(2x+y_1)(x^2+y_1^2-20)$$
We can maximize this by creating two partial differential equations and our solution will be $(x,-y_1)$ and $(-x,y_1)$
A: You could try this with polar coordinates. The function then becomes
$$f(r,\vartheta)=(r^3-20r)|2\cos\vartheta-\sin\vartheta|$$
which is much simpler to find extreme values. As you noticed, function is unbounded from above. For minimum of $f$, notice that it is achieved at global minimum of $r^3-20r$ on $[0,+\infty)$ and global maximum of $|2\cos\vartheta-\sin\vartheta|$ on $[0,2\pi)$.
Since $(r^3-20r)' = 3r^2-20$, global minimum of $r^3-20r$ on $[0,+\infty)$ is achieved either at $0$, $+\infty$ or $\frac{2\sqrt 5}{\sqrt 3}$. By inspection, it is at $r = \frac{2\sqrt 5}{\sqrt 3}$.
For the other one, we have that $(2\cos\vartheta-\sin\vartheta)' = -2\sin\vartheta-\cos\vartheta$, so extremes are achieved either at $0$, $\arctan(-\frac 12)$ or $\arctan(-\frac 12)+\pi$. Because of absolute value, $|2\cos\vartheta-\sin\vartheta|$ has same value at $\arctan(-\frac 12)$ and $\arctan(-\frac 12)+\pi$ and from formulas $$\cos\arctan x = \frac{1}{\sqrt{1+x^2}},\ \sin\arctan x = \frac{x}{\sqrt{1+x^2}}$$
it turns out that the value is $\sqrt 5$.
All put together, minimum of $f$ is $-\frac{400}{3\sqrt 3}$.
