Find the maximum volume of the cylinder

danago
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A cylinder is placed into a right cone which has a radius of 1m and a height of 2m. Find the maximum volume of the cylinder.

I think i am able to do it, but I am not 100% sure if I am correct. I first used similar triangles to write the radius of the cone at any height from its apex. i came up with h=2r.

Now, if i let the radius of the cylinder be R, then the height that it rests at is 2R. Since the cone is 2m high, the height of the cylinder will be 2-2R.

The volume is given by V = \pi R^2 (2 - 2R), which i used calculus to maximize and came up with R=2/3, therefore V=~0.93. Does that look right?

Thanks in advance,
Dan.
 
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Looks fine. Kind of a confusing double use of 'height' though. I'd use another word for the apex distance if I were describing it.
 
Alright thanks for confirming that :smile:
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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