So suppose I have some function$$A= \frac{1}{16}\left(2a^2+2b^2-2b(a+1)+1\right)$$ with $0\leq a \leq b \leq 1$, of which I know that it has a minimum at $(\frac13, \frac23)$. I also know the hessian to be a constant $H=\frac34$ and the gradient to be given by $$\nabla A = \frac{1}{16}\left(4a-2b,\;4b-2a-2\right).$$
This is an ever increasing function, but we wish to know the maximum in the region where $0\leq a\leq b \leq 1$. How do you do this? I suspect the corners to be equally valid maxima, but I wouldn't know how to prove it. (i.e. $(0,0),(0,1),(1,0),(1,1)$).
Edit: My solution based on lord shark's comment:
Suppose we set $b=0$, then $A = 1$, since $a\leq b$.
Suppose we set $a=0$, then $A = \frac{1}{16}(2b^2-2b+1)$, which is symmetric about $\frac12$.
Suppose we set $a=1$, then $A = 1$, since $a\leq b$.
suppose we set $b=1$, then $A = \frac{1}{16}(2a^2-2a-1)$, which is symmetric about $\frac12$.
Then from here we can see that $\max A = 1$.
Note that I've already proven the existence of only one critical point, so I don't have to consider local maxima.