Answer
$x^2+2y^2 \leq 4 $; Region inside the ellipse $x^2+2y^2=4$
Work Step by Step
We have $\iint_{R} (4-x^2-2y^2) dA$
The integral must be a maximum on the region when $4-x^2-2y^2\geq 0$
In order to find the direction of inequality multiply with $-1$, a negative number .
So, $\iint_{R} (4-x^2-2y^2) dA=4-x^2-2y^2\leq 0$
$\implies x^2+2y^2 \leq 4$
This is the region inside the ellipse $x^2+2y^2=4$
