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