Lagrange multipliers to find min/max with parabola

Question

Use Lagrange multipliers to find extreme values of $f(x, y) = x^2 + y^2$ on part of parabola $y = x^2 - 1, -2 \le x \le 2$

I got the equations

$$\left\{\begin{matrix} 2x = \lambda(-2x)\\ 2y = \lambda\\ y - x^2 = -1 \end{matrix}\right.$$

Using the equation $\nabla f = \lambda \nabla g$

From this I see (system of equations):

$y = \frac{\lambda}{2}$, $\{x = 0$, or, $\lambda = 0\}$, and $x = \pm \sqrt{\frac{\lambda}{2} + 1}$ (using last eq and solving for $x$)

Critical points of $f$ is $(0, 0)$

Using the lambda equations, we see possible points $(0, -1), (-1, 0), (1, 0)$

Using the edges of $g$, we get the points $(-2, 3), (2, 3)$

So the set finally is

$\{(0, 0), (-2, 3), (2, 3), (0, -1), (-1, 0), (1, 0)\}$

I get max = $13$ [correct]

I get min = 0 [wrong]

The minimum is supposed to be 3/4.

Can you help me out?

Answer 1

$$2x = \lambda (-2x)$$

$$2y=\lambda$$ $$y-x^2=-1$$

Case $1$: If $x=0$, $y=-1$.

Case $2$: If $x \neq 0$, $\lambda = -1$ from the first equation, hence $y=-\frac{1}{2}$, $x^2=\frac12$, hence $x=\pm\frac1{\sqrt2}$.

Can you find the minimum and maximum now?

Comment about your working:

From $$2x=\lambda(-2x)$$

We have $$2x(1+\lambda)=0$$

The right conclusion can be $x=0$ or $\lambda = -1$.

Answer 2

Alternatively (graphical method): Consider the contour curve: $$x^2+y^2=z,$$ which is a circle center $(0,0)$ and the radius $\sqrt{z}.$

Hence the largest circle intersecting the parabola is through the points $(-2,3), (2,3)$ and the smallest circle is through the vertex of the parabola $(0,-1)$ or the tangent to parabola and the circle (which is a little challenging).