What are the maximum and minimum values of $4x + y^2$ subject to $2x^2 + y^2 = 4$? $$ 2x^2 + y^2 = 4 $$
$$ Y = \sqrt{4-2x^2} $$
$$4x + y =  2x^2 +  \sqrt{4-2x^2}$$
Find the derivative of $$ 2x^2 +  \sqrt{4-2x^2} $$  set as = 0
$$X^2 = 64/33$$
$$ F(64/33) = 34\sqrt{33}/33 $$
How to solve it the right way?

Note that $y^2=-2x^2+4$
When performing substitution of $4x+y^2 \rightarrow 4x-2x^2+4$
Finding the vertex (maxima):
$$-2x^2+4x+4=-2(x^2-2x)+4=-2[(x-1)^2-1]+4=-2(x-1)^2+6$$
Which presents the relative(and only) maxima.
Note that there is no relative minima, as the parabola goes downwards.
