If 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?

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.

Hello, alden35!

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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?

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

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

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

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

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


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

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Originally Posted by chiro

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.

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Or wait for Soroban to do the problem for you!

Yeah that could also work as well :)