Maximum volume of a cone.

**Defaultjoe** · Jun16-09, 12:02 PM

I'm machining a component as a means of testing one of our companies new machines. The objective is to manufacture a cone of maximum possible volume. The volume of a cone is given by: pir²h/3.

Given that h = 6 - r, how am i to calculate the maximum possible volume by means of integration.

Many thanks, Joe.

**HallsofIvy** · Jun16-09, 02:15 PM

You can't "by means of integration! You can "by means of differentiation". Putting h= 6- r in $LaTeX Code: \\pi r^2h/3$ gives $LaTeX Code: \\pi r^2(6-r)/3= \\pi/3(6r^2- r^3)$ . The derivative of that is $LaTeX Code: \\pi/3(12r- 3r^2)= \\pi(4r- r^2)$ . 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).