Global maxima of $f(x,y)=x^2y$ restricted to D Let $f(x,y) = x^2y$ and $D = \{(x,y): y\geq0 \land 2x^2+y^2 \leq a\}$ with $a>0$.
I need to find $a$ such that the global maxima of $f$ restricted to $D$ is $\frac{1}{8}$.
I found, using Lagrange multipliers, that $a$ = $\frac{3}{4}$. With that value, $f(\frac{1}{2}, \frac{1}{2}) = f(-\frac{1}{2}, \frac{1}{2}) = \frac{1}{8}$, and those points are local maxima.
Tha part where I am struggling with is to prove that those are global maxima. 
The restriction now states that $2x^2+y^2 \leq \frac{3}{4}$, but from there I can't get a close enough bound for $x^2y$. What else can I try?
Thank you.

The function is monotonic in $y$, so that by definition, for each constant $x$, the maximum will be found on the upper half of the ellipse's boundary. 
Thus, you can search for local maxima of:
$$f(x,y) = x^2\sqrt{a-2x^2}$$ 
Which will then be global maxima of $f(x,y)$ on $D$, and also happen to coincide with the local minima you already found.
