Question 1183328
If the height of a right circular cone is decreased by 8 percent by what percent must the radius of the base be decreased so that the volume of the cone is decreased by 15 percent?
:
8% reduction = .92h
15% reduction = .85v
let a = reduced amt multiplier time the radius 

{{{1/3}}}{{{pi*(ar)^2*.92h}}} = .85({{{1/3}}}{{{pi*r^2*h}}})
multiply both sides by 3
{{{pi*(a^2r^2)*.92h}}} = .85({{{pi*r^2*h}}})
divide both side by pi*r^2*h
{{{a^2*.92 = .85}}}
{{{a^2 = .85/.92}}}
a = {{{sqrt(.924)}}}
a = .9612 times the radius, therefore
Radius has to be reduced by 3.88%