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?!
Let $4x+y^2=k$.

Thus, $$2x^2-4x+k-4=0,$$ which gives
$$4-2(k-4)\geq0$$ or
$$k\leq6.$$
The equality occurs for $x=\frac{4}{2\cdot2}=1$ and $y=\sqrt{2},$ which says that $6$ is a maximal value.

In another hand, since $-\sqrt2\leq x\leq\sqrt2,$ we obtain:
$$4x+y^2=4x+4-2x^2=2(3-(x-1)^2)\geq2(3-(-\sqrt2-1)^2)=-4\sqrt2.$$
The equality occurs for $x=-\sqrt2$ and $y=0,$ which says that we got a minimal value.
Hint:


Step 1. Re-write your equation as $y=f(x)$.
Step 2. Find the derivative of $f(x)$.
Step 3. Check where this derivative is $0$ and where it is positive and negative.


Can you take it from here?