Under condition $2x^2 + y^2= 4$ for real numbers $x, y$, find the maximum and minimum value of $4x + y^2$. How can I solve this problem. I can find absolute maximum and minimum value of equation with given interval. But here, I don't know where should I start. Could you explain step by step?!

$4x+y^2=4x +4-2x^2=$
$4+ 2x(2-x);$
Constraint:
$2x^2=4-y^2$; 
$x_{1,2} =\pm \sqrt{2-y^2/2}$
$4-y^2 \ge 0$; or $|y| \le 2$; and
$-√2 \le x \le √2$.
Max & Min of 
$Y:=4+4x-2x^2=$
$ -2(x^2-2x)+4= -2(x-1)^2+6$;
$\max (Y _x)= 6$; (cf. lab bhatthachargee).
$\min (Y_x)$ :
The parabola $Y=-2(x-1)^2 +6$ has vertex at $(1,6)$, and is opening downward.
Since  $-√2 \le x \le √2$, we find 
the minimum at the boundary point:
$x=-√2$:
$\min Y_x=-2(-√2-1)^2+6$.
