I need to locate the optimal point(s) of the function $$f(x,y) = \frac{1}{xy} + x^2y$$ subject to the conditions $x \gt 0, y \gt 0$.
I have tried the following approach $$\frac{\partial f}{\partial x} = \frac{-1}{x^2y} + 2xy = 0 \rightarrow 1$$ $$\frac{\partial f}{\partial y} = \frac{-1}{xy^2} + x^2 = 0 \rightarrow 2$$ Now, $x \ne 0, y \ne 0,$ hence, we solve the two equations to get the optimal point, $(x^*, y^*)$. But the above system of equations has no solution.
Physically, when $x \rightarrow 0$ or $y \rightarrow 0$, the function blows up. Also, when either $x \rightarrow \infty$ or $y \rightarrow \infty$, the function also blows up. Since it takes finite values in between and is continuous and differentiable, it should have some minimum value(s), and hence an optimal point.
Where am I going wrong ?
SOLVED
Thanks to Arthur, I have realised the error in the above statements. I have assumed that whenever x or y becomes large, the function will blow up. But if I were to go along the path $y = \sqrt{\frac{1}{x^3}}$, and take $x$ smaller and smaller, $f = 2\sqrt x$ will go to 0. Hence, the infimum is $0$
$C=\{(x,y)|x>0,y>0\}$ is not compact.
Thus, we can get the situation for which a minimal value does not exist.
It's exactly which happens in our case.
By AM-GM $$\frac{1}{xy}+x^2y\geq2\sqrt{\frac{1}{xy}\cdot x^2y}=2\sqrt{x}.$$ The equality occurs for $\frac{1}{xy}=x^2y$ or $x^3y^2=1$.
Since, for $x\rightarrow0^+$ we get $\sqrt{x}\rightarrow0^+$, we obtain that $$\inf\left(\frac{1}{xy}+x^2y\right)=0$$