If the volume of a right circular cone is reduced by 15% by reducing its height by 5%, by what percent must the radius of the base reduced?

Printable View

- Sep 13th 2013, 05:49 PMalden35If the volume of a right circular cone is reduced by 15% by reducing its height by 5%
If the volume of a right circular cone is reduced by 15% by reducing its height by 5%, by what percent must the radius of the base reduced?

- Sep 13th 2013, 07:23 PMchiroRe: If the volume of a right circular cone is reduced by 15% by reducing its height b
Hey alden35.

Hint: Express the volumes and height of the two cones in terms of r, h, dr, and dh where dr is the change of radius and dh is the change of height. - Sep 14th 2013, 02:48 PMSorobanRe: If the volume of a right circular cone is reduced by 15% by reducing its height b
Hello, alden35!

Quote:

If the volume of a right circular cone is reduced by 15% by reducing its height by 5%,

by what percent must the radius of the base reduced?

$\displaystyle \text{We have: }\:V \:=\:\tfrac{\pi}{3}r^2h\;\;[1]$

$\displaystyle \text{Then: }\:0.85V \;=\;\tfrac{\pi}{3}(xr)^2(0.95h)\;\;[2]$

$\displaystyle \text{Divide }[2]\text{ by }[1]\!:\;\frac{0.85V}{V}\;=\;\frac{\frac{\pi}{3}(xr)^2 (0.95h)}{\frac{\pi}{3}r^2h} \quad\Rightarrow\quad 0.85 \:=\:x^2(0.95)$

. . $\displaystyle x^2 \:=\:\frac{0.85}{0.95} \quad\Rightarrow\quad x \:=\:\sqrt{\frac{17}{19}} \:=\: 0.945905303$

$\displaystyle \text{Hence: }\:x \:\approx\:0.946 \:=\:94.6\%$

$\displaystyle \text{Therefore, the radius must be reduced by }5.4\%.$

- Sep 15th 2013, 12:56 PMHallsofIvyRe: If the volume of a right circular cone is reduced by 15% by reducing its height b
- Sep 15th 2013, 07:55 PMchiroRe: If the volume of a right circular cone is reduced by 15% by reducing its height b
Yeah that could also work as well :)