How to prove there's no global maxima?

Question

Given $f(x,y)=x^2+y^2$ subject to $y-x^2=2$, By Lagrange multipliers, I only found the point $(0,2)$.

Wolfram says this point is the global minimum and there's no global maxima.

How can I prove that?

Answer 1

If $y=x^2+2$, then $x^2+y^2=x^2+(x^2+2)^2=x^4+5x^2+4$. But then, if$$\varphi(x)=f(x,x^2+2)=x^4+5x^2+4,$$then $\varphi|_{[0,\infty)}$ is a strictly increasing function, whereas $\varphi|_{(-\infty,0]}$ is strictly decreasing. So, the minimum of $\varphi$ is attained when $x=0$ (and then $y=2$), and it has no maximum.

Answer 2

A bit of geometry:

$1)f_r(x,y)=x^2+y^2=r^2$ is a family of concentric circles with centres at $(0,0)$ and radii $r.$

$2) y=x^2+2$ is a parabola opening upwards with vertex at $(0,2)$.

Intersection of the family $f_r(x,y)$ and the parabola.

$0)r<2:$ no intersection

$1)r=2:$ intersection at $(0,2)$

$2)r>2:$ $2$ points of intersection for each $r$

Is there a maximal $r$?

Answer 3

Geometrically, $f(x,y)=x^2+y^2$ is the squared distance of a point from the origin.

$y-x^2=2 \implies y=x^2+2$ is a parabola with vertex at $(0,2)$. Since the coefficient of the quadratic term is positive, the parabola rises on either side of the vertex, meaning the vertex is the closest point and everything else is further away by the pytagorean theorem.