What are the maximum and minimum values of $4x + y^2$ subject to $2x^2 + y^2 = 4$?

Question

$$ 2x^2 + y^2 = 4 $$ $$ Y = \sqrt{4-2x^2} $$

$$4x + y = 2x^2 + \sqrt{4-2x^2}$$ 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?

Answer 1

you will get $$4x+y^2=4x+4-2x^2$$

Answer 2

I hope the image explains it all.

Answer 3

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.