Function $f(x,y)=\frac{x^6+y^4}{x^3+xy^2}$ on the domain $\Omega=\{(x,y)\in\mathbb R^3; x>0, y\ge0\}$.
The answer for the problem is that the function doesn't have max or min. My first try was to calculate the gradient and see when it is equal to $0$. For me solving the system requires too much time, so is there a faster method? Can I use the domain given to answer the problem?
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Sometimes it's worth trying some typical parameters from the domain like $(x,0)$, $(0,y)$, $(x,x)$ or $(x,-x)$. Here you can calculate: $f(x,0) = x^3$. We see that this is unbounded from the above and arbitrarily close to zero. Since $f(x,y)>0$ for all $(x,y)$ we see that $f$ doesn't have minimum nor maximum.
It seems that solving the system isn't as hard as you think it is. $\frac{\partial f }{\partial y}=\frac{2y\left(-x^6+2x^2y^2+y^4\right)}{x\left(x^2+y^2\right)^2}$ (once we have simplified $\frac{4y^3\left(x^3+xy^2\right)-2xy\left(x^6+y^4\right)}{\left(x^3+xy^2\right)^2}$), so if we want to solve $\frac{\partial f }{\partial y}=0$ for $y$, then we can quite easily using the basic rules of algebra find that $y=0$. However, if we substitute this into $\frac{\partial f}{\partial \:x}$ (which is $\frac{3x^8+5x^6y^2-3x^2y^4-y^6}{\left(x^3+xy^2\right)^2}$), then we get $\frac{3x^8}{x^6}=0$, which means that $x=0$. This is not in the domain of $f(x,y)$. Therefore, a maximum, minimum or saddle point does not exist. I hope this helps.