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.

i am wondering if you can do this. we need to find the maximum of $k$ so that $$y = \frac k{x^2} \tag 1$$ touches the ellipse $$2x^2 + y^2 = a \tag 2$$ in the first quadrant.
for $(1)$ an $(2)$ to touch, $$2x^2 + \frac{k^2}{x^4} = a $$ must have a double root. that is, with $x^2 = u$ the function $g(u) = 2u^3 - au^2 + k^2$  and its derivative $g'(u) = 6u^2 - 2au = 0$ must have a positive common root $u = a/3.$ so that 
$$0 = g(a/3) = \frac 2{27}a^3 - \frac 19a^3 + k^2 = 0 \to k = \left(\frac a 3\right)^{3/2} \text{ or } a = 3k^{2/3}$$ which agrees with the max pair found 
$a = \frac 34, k = \frac1 8.$
