Find the point P on the parabola y = x² closest to the point (3, 0).
$L=d^2 = (x-3)^2 + (x^2-0)^2 = (x-3)^2 + x^4$
$\dfrac{dL}{dx} = 2(x-3) + 4x^3$
$\dfrac{dL}{dx} = 0$
$2(x-3)+4x^3=0$
$2(3-x) = 4x^3$
$3-x=2x^3$
This has 1 real solution
$x=1$
$\dfrac{d^2L}{dL^2} = \left . 2+12x^2\right |_{x=1} = 14>0$
So this is a minimum.
The point is $(1,1)$