The question is:

A cone is to have a slant height of 22 cm. Find the vertical height h if its volume is to be a maximum.

Volume(cone) = ⅓πr²h

Can anyone help me? I have a little bit of idea as to how to approach this but my execution fails.



By pythagorus theorem,

$ V=π[(22)^2-h^2]h$

(for extreme points)
$\frac{dV}{dh}= π(22)^2-3h^2π=0$


$$\frac{d^2 V}{dx^2}=-6πh$$

$$\frac{d^2 V}{dx^2}<0$$


Volume is maximum at $$h=\frac{22}{√3}$$



Draw yourself a picture of the cross-section. Notice that you have a right triangle with the radius and the height as legs, and the slant as the hypotenuse. By the Pythagorean theorem, we have $h^2 + r^2 = 22^2$.

  • $\begingroup$ Yeah I did that, so h = √(22^2 - r^2). Is this able to be simplified even more? $\endgroup$ – rolatis Nov 16 '14 at 9:28
  • $\begingroup$ Well, you'll want to solve for $r$, since we wish to minimize $\frac{dV}{dh}$, but it'll be the same with the letters switched. At any rate, you won't be able to simplify that any more, so you'll have to use the chain rule when taking the derivative. $\endgroup$ – Kaj Hansen Nov 16 '14 at 9:30
  • $\begingroup$ Well can you just do it for me? I'm really confused. (I know this is bad, to just cop out, but I am just lost). $\endgroup$ – rolatis Nov 16 '14 at 9:43
  • $\begingroup$ Ok, first get your volume function in terms of just $h$ by eliminating $r$ using the equation in my post. Then just take a derivative with respect to $h$ and set it equal to zero. This will maximize your volume. $\endgroup$ – Kaj Hansen Nov 16 '14 at 9:46
  • $\begingroup$ Yeah I managed to do it. Thanks for the comments. $\endgroup$ – rolatis Nov 17 '14 at 3:15

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