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?
Hello, alden35!
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?
$\displaystyle \text{We have: }\:V \:=\:\tfrac{\pi}{3}r^2h\;\;[1]$
$\displaystyle \text{Then: }\:0.85V \;=\;\tfrac{\pi}{3}(xr)^2(0.95h)\;\;[2]$
$\displaystyle \text{Divide }[2]\text{ by }[1]\!:\;\frac{0.85V}{V}\;=\;\frac{\frac{\pi}{3}(xr)^2 (0.95h)}{\frac{\pi}{3}r^2h} \quad\Rightarrow\quad 0.85 \:=\:x^2(0.95)$
. . $\displaystyle x^2 \:=\:\frac{0.85}{0.95} \quad\Rightarrow\quad x \:=\:\sqrt{\frac{17}{19}} \:=\: 0.945905303$
$\displaystyle \text{Hence: }\:x \:\approx\:0.946 \:=\:94.6\%$
$\displaystyle \text{Therefore, the radius must be reduced by }5.4\%.$