Maximum volume of a cone.

  • Thread starter Defaultjoe
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  • #1
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.
 

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  • #2
HallsofIvy
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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).
 

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