# Find the stationary points of the curve $z(x,y)=xy(12-4x-3y)$

$$z(x,y)=xy(12-4x-3y)$$

First, I expanded the brackets:

$$z(x,y)=12xy-4x^2y-3xy^2$$

Then I found the partial derivatives with respect to $y$ and $x$: $$(\frac{\partial z}{\partial x})_y=12y-8xy-3y^2$$ $$(\frac{\partial z}{\partial y})_x=12x-4x^2-6xy^2$$

For there to be a stationary point, both of the above must equal zero: $$(\frac{\partial z}{\partial x})_y=12y-8xy-3y^2=0$$ $$(\frac{\partial z}{\partial y})_x=12x-4x^2-6xy^2=0$$

However, I can't seem to be able to get past this point (I can't solve either equation). How can the stationary points be found?

First equation $y(12-8x-3y)=0$ so either $y=0$ or $12-8x-3y=0$.
If $y=0$, the second equation is then $12x-4x^2=0$, this gives you $x=3,0$. These are two points $(3,0)$ and $(0,0)$.
If $12-8x-3y=0$, write y in terms of $x$, and plug into the second equation, you will get another three points.