Maximum value of $f(x,y)=x^2+2y^2$ subject to constraint : $y-x^2+1=0$?

Question

If $f(y)=2y^2+y+1,$

$f'(y)=4y+1=0\Rightarrow y=-\frac14,\; x=\pm\frac{\sqrt3}2$

$f''(y)=4>0,$ so I can't obtain a point of maxima.

What does this mean? Do I necessarily need to use Lagrange's Multiplier method for this?

Answer 1

You must remember: $$x^2=y+1\ge 0 \Rightarrow y\ge -1.$$ So, it has the additional constraint on $y$. According to the Extreme value theorem, you must check the border and critical points: $$\text{critical:} \ \ f(\pm \frac{\sqrt{3}}{2},-\frac14)=\frac78 \ \text{(local min)}\\ \text{border:} \ \ f(0,-1)=2 \ \text{(local max)}$$

Note: There is no global max, because for $x\to\infty$, $f(x,y)\to \infty$.

Answer 2

$$x^2=y+1$$

$$x^2+2y^2=y+1+2y^2=2\left(y+\dfrac14\right)^2+1-2\left(\dfrac14\right)^2\ge?$$