Maximum volume of a cone.

You can't "by means of integration! You can "by means of differentiation". Putting h= 6- r in [itex]\pi r^2h/3[/itex] gives [itex]\pi r^2(6-r)/3= \pi/3(6r^2- r^3)[/itex]. The derivative of that is [itex]\pi/3(12r- 3r^2)= \pi(4r- r^2)[/itex]. The maximum possible volume is given by the (non-zero) r value that makes that 0.

(neither r nor h can be negative which mean r cannot be less than 0 or greater than 6. Technically we should consider the endpoints 0 and 6 but those obviously give minimum volume (a flat disk or a straight line).