Find a specific critical point from set of equations for $f(x,y) = xy(12-3x-4y)$

Question

I'm having trouble finding a specific critical point from set of equations for $f(x,y) = xy(12-3x-4y)$. In particular it is the critical point $(\frac{4}{3},1)$

Calculating the first order derivatives for $x$ and $y$ we get the set of equations:

$$\partial_{x}f = y(12-6x-4y) = 0 \\ \partial_{y}f = x(12-3x-8y) = 0$$

Could somebody assist me in solving for this specific critical point?

Answer 1

Obviously $(x,y)=(0,0)$ its a critical point, so, suppose that $x\neq0$ and that $y\neq0$. Hence, $$y(12-6x-4y)=0 \textrm{ implies } 12-6x-4y=0$$ and similarly $12-3x-8y=0$ from the second equation. It follows that the unique solution to the system of equations $$\left\{\begin{align} -6x-4y&=-12 \\ -3x-8y&=-12 \end{align}\right.$$ is $(x,y) = \big(\tfrac43,1\big)$.

Answer 2

Here is a stream plot:

$$(0,3), (4/3,1), (4,0), (0,0)$$ marked, each of which you can analyze for the value of the derivatives (and second derivatives).