Let $f : X(\subseteq \mathbb{R}^2) \to \mathbb{R}$ by the function $f(x,y) = x^ny^m,$ where $D = \{ (x,y) \ \vert \ x^2+ y^2 <1 \}$ and $n \in \mathbb{Z}\_{\geq 1}$. I want to determine the supremum of $f$. 

We can compute the partials, $$\frac{\partial f}{\partial x} = ny^mx^{n-1}$$ and $$\frac{\partial f}{\partial y} = mx^ny^{m-1}$$ which gives a critical point at $(0,0)$. This looks like a minimum since $f(0,0)=0$. 

To determine the supremum, we would need to look on the bound of $D$. This gives us $$f(x) = x^n (1-x^2)^{m/2}.$$ Calculating the derivative of this, $$f'(x) = nx^{n-1}(1-x^2)^{m/2} - mx^{n+1}(1-x^2)^{(m-2)/2} =0 \\
= x^{n-1}(1-x^2)^{\frac{m}{2}-1}[n(1-x^2)-mx^2] =0 \ \ $$ which says that $x=0$, $1-x^2 =0$ or $$n(1-x^2)-mx^2=0 \implies n-(m+n)x^2=0 \implies x = \pm \sqrt{\frac{n}{n+m}}.$$ Can we then assert that the supremum occurs at $$\left(\pm \sqrt{\frac{n}{m+n}}, \pm \sqrt{\frac{m}{m+n}} \right)$$ and the supremum is $$f\left(\pm \sqrt{\frac{n}{m+n}}, \pm \sqrt{\frac{m}{m+n}} \right)=\frac{n^nm^m}{(m+n)^{m+n}}?$$