Whats maximum and minimum of $4x + y^2$ without Lagrange 
Whats maximum and minimum of $4x + y^2$ without Lagrange
constraints = $2x^2 + y^2 =4$

Substitute $y^2= 4-2x^2$ into $4x+y^2$ we get $4x+4-2x^2$ manipulate it into $-2(x-1)^2 + 6$ the maximum is 6. Or using differentiation we get $x=1$ and only produces maximum value. What about the minimun? Also find the maximum in different way?

$f(x)=4x+4-2x^2=$
$-2(x-1)^2+6$;
1)$2x^2+y^2=4$; 
An ellipse: $x^2/(√2)^2+y^2/2^2=1$.
Semi minor axis : $√2$ ;
$y^2=0: x_{1,2}=\pm √2$;
Minimum: 
$x=-√2$: $f_{min}(-√2)=-2(-√2-1)^2 +6=-2(1+√2)^2 +6$.
See also comments.
