# Finding the critical point of a function $f(x,y)$ with restrictions

I have the following problem. Find the critical points for:

$$f(x,y) = xy + 2x -ln(x^2y),$$

in the first quadrant $x>0$ and $y>0$, and show that $f(x,y)$ has a minimum in the first quadrant.

I found $\frac{\partial }{\partial x}$ and $\frac{\partial }{\partial y}$ and they are $y+2 +\frac{-2}{x}$ and $x - \frac {1}{y}$, respectively. However, I do not know where to go from here. Any help would be appreciated.

You have to find points $(x,y)$ such that

(1) $y+2 -\frac{2}{x}=0$

and

(2) $x - \frac {1}{y}=0$

From (2) we get $y= \frac {1}{x}$. Now proceed with (1).

• Okay, I got that $x= \frac {1}{2}$ and $y= 2$. Does that mean I'm done ? – A curious one Nov 28 '16 at 14:33
• I know $( \frac {1}{2},2)$ lie in the first quadrant but how do I show that this critical point is a minimum ? – A curious one Nov 28 '16 at 14:42
• Tipp: Hessian matrix – Fred Nov 28 '16 at 14:45
• okay I'll look into it. – A curious one Nov 28 '16 at 14:52