Given the curve $ 2x^3+3x^2y+2y^3=30 $, motivate that there is a point closest to the origin, and show if there is a max distance.
I set up the distance function as $d(x,y) = \sqrt{x^2+y^2}$ but use $f(x,y) = x^2+y^2$ for simplicity.
I have then used Lagrange multipliers to find the local max and min for the distance function on the curve, let $$\mathcal{L}(x,y,\lambda) = x^2+y^2 + \lambda(2x^3+3x^2y+2y^3-30).$$ And I get the equation system as: $$\nabla \mathcal{L} = 0 \Leftrightarrow \begin{cases} \mathcal{L}'_x = 2x+6x^2 \lambda + 6xy \lambda = 0 & (1)\\ \mathcal{L}'_y = 2y+3x^2 \lambda + 6y^2 \lambda = 0 & (2)\\ \mathcal{L}'_\lambda = 2x^3+3x^2y+2y^3-30=0 & (3) \end{cases}$$
Which I solve and find the (real) solutions to be $(0,\sqrt[3]{15})$ and $(2,1)$.
Now that I know that these are local min and max, how would I motivate that the min is indeed a global min? And how do I motivate that the max is not a global max?