Question 361533
An absolute minimum exists at x = 6.
{{{y = x^2 + 432/x}}},
{{{dy/dx= 2x-432/x^2}}}.  Letting this derivative equal 0, and solving, 
we get  {{{2x = 432/x^2}}}, or {{{x^3 = 216}}}, or x = 6.
Now {{{d^2y/dx^2 = 2+864/x^3}}}.  Substitute x = 6 into the 2nd derivative to find out that {{{d^2y/dx^2>0}}}.  Therefore there is an absolute minimum at x = 6.