Find the global extrema of $f(x,y)=4xy-3x-y+2$ on the triangular region with $(0,0),(5,0)$, and $(0,4)$

Question

I start with $f_x=4y-3$ and $f_y=4x-1$ so therefore $y=\frac{3}{4}$ and $x=\frac{1}{4}$. This gives the critical point $(\frac{1}{4},\frac{3}{4})$.

Next, $f(\frac{1}{4},\frac{3}{4})=4(\frac{1}{4})(\frac{3}{4})-3(\frac{1}{4})-(\frac{3}{4})+2$. So $f(\frac{1}{4},\frac{3}{4})=\frac{5}{4}$.

Now I want to find $x=0$, so $0\leq y \leq 4$: $f(x,y)=4y-y+2$... Right here I'm not sure if I am supposed to have the $4y$ or not. In relation to this I have to find the derivative of $f(x,y)$ which gives me numerical values with no variables, so I'm not sure how to get $(x,y)$ when $x=0$.

Answer 1

Along $x=0$, we have $g(y)=f(0,y)=-y+2$. Then $g'(y)=-1$ and $g''(y)=0$ and $g(0)=2$ and $g(4)=-2$. From $g'(y)=-1$, what we can say about $(0,2)$ and $(0,-2)$?

Along $y=0$, we have $h(y)=f(x,0)=-3x+2$. Then $h'(x)=-3$ and $h''(x)=0$ and $h(0)=2$ and $h(5)=-13$. From $h'(x)=-3$, what we can say about $(2,0)$ and $(-13,5)$?