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?!
 A: 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. 
A: $$4x+y^2=4x+4-2x^2=4-2(x-1)^2+2\le4+2$$
A: You want to optimize the expression $$4x+y^2$$ subject to the condition that $2x^2+y^2=4.$ In this case it is easy to just substitute for $y^2$ in the objective function to have $$4x+(4-2x^2).$$ This is a quadratic in one real variable which I'm sure you know how to optimize by completing the square. This will give you a maximum value of $6.$
A: $4x+y^2=4x +4-2x^2=$
$4+ 2x(2-x);$
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:
$x=-√2$:
$\min Y_x=-2(-√2-1)^2+6$.
A: Hint:
Write the condition as 
$$\dfrac{x^2}{2}+\dfrac{y^2}{4}=1$$
with $x=\sqrt{2}\cos t$ and $y=2\sin t$ we have
$$4x+y^2=6-4\left(\cos t-\dfrac{\sqrt{2}}{2}\right)^2$$
A: 
Here's a graphical approach, for a different perpsective.
A: 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?
