Maximum volume (right circular cone)...

burgerandcheese

burgerandcheese

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Q: A right circular cone has a base radius r and height h. As r and h vary, its curved surface area is kept constant. Show that its volume is a maximum when h=r√2

Please tell me if this is the right way to do it, because I still can't get h = r√2

I let the formula for the curved surface area = k (constant) and found the height, h in terms of the radius, r.
Then I substituted that h into the volume formula for a right circular cone. Now I have the volume, V in terms of its radius, r. I let dV/dr = 0 and solved for r.
After that I substitute that r into the first formula where I put h in terms of r.

Also, when the question said "r and h vary", does that mean r and h are not equal?

Formula I used:

Curved surface area = πr√(r² + h²)
Volume of right circular cone = πr² * h/3


 
The basic process you've outlined is a perfectly fine way of going about the problem. The only thing I'd do slightly differently is when you get to the step of setting \(\displaystyle \dfrac{dV}{dr} = 0\), solve that instead for k (the surface area). You'll end up having to consider two cases, when r < 0 and r > 0, but for obvious reasons the r < 0 case can be discarded. That way you can use the known formula for surface area and this new formula you found, to end up with k = (some function) and k = (some other function), from which you can conclude those two quantities must be equal.

That aside, the fact that I used your method and arrived at the correct answer and you did not tells me that any error(s) must be in the implementation. However, we cannot troubleshoot work we can't see. Please reply showing all of the workings you used, even the parts you know for sure are wrong, so that we can figure out where things are going off the rails for you.
 
ksdhart2 said:
The basic process you've outlined is a perfectly fine way of going about the problem. The only thing I'd do slightly differently is when you get to the step of setting \(\displaystyle \dfrac{dV}{dr} = 0\), solve that instead for k (the surface area). You'll end up having to consider two cases, when r < 0 and r > 0, but for obvious reasons the r < 0 case can be discarded. That way you can use the known formula for surface area and this new formula you found, to end up with k = (some function) and k = (some other function), from which you can conclude those two quantities must be equal.

That aside, the fact that I used your method and arrived at the correct answer and you did not tells me that any error(s) must be in the implementation. However, we cannot troubleshoot work we can't see. Please reply showing all of the workings you used, even the parts you know for sure are wrong, so that we can figure out where things are going off the rails for you.
Click to expand...

I did it :) instead of solving for r in dV/dr = 0 , I solved for k^2 and plugged that into the first equation.
Took some time differentiating pi/3 * r^2 * sqrt([{k^2 / pi^2} - r^4]/ r^2) though huhu

Thank you!!!
 
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