When I looked at this again I discovered that there were two solutions; one was between x=1 and 2, and the other between 8 and 9 (the one I found). It has helped having the original question, which has another part, I see. The other solution is point A at x=1.10107 approx. As for the gradient, we differentiate to get 12/x-3sqrt(x)/2, which is zero when 24=3x^3/2, i.e., x^3/2=8 and x^3=64, so x=4, making f(x)=8.64 to 2 dec places. This is a maximum. I'm no expert in judging levels, but I'd say because it includes logs and calculus it would be fairly advanced. Harking back to my youth, I guess I would be doing this sort of question at age 16 or 17, but I don't know what today's standards are. Good to hear from you, Peter. Please keep in touch.