If the height of a right circular cone is decreased by 8%, by what percent must the radius of the base be decreased so that the volume of the cone is decreased by 15%?

Question

amistre64 · Answer

Vcone = (1/3)(base area)(h)

amistre64 · Answer

kinda depends on the base area to begin with I think

anonymous · Answer

Problem doesn't provide any measurements at all :P

amistre64 · Answer

$$V=\frac{\pi. r^2.h}{3}$$

amistre64 · Answer

$$V(1-.15)=\frac{\pi.r^2.h(1-.08)}{3}$$it looks like

amistre64 · Answer

$$r^2*(\frac{1}{.85})$$looks to be what they expet you to work with then

anonymous · Answer

The given answer is 4%.

amistre64 · Answer

how to work that in i dunno

amistre64 · Answer

sqrt(r^2sqrt(1/.85)) maybe

amistre64 · Answer

r(1.04) is what the comes to; but that seems to increase r

amistre64 · Answer

.85/.92 maybe?  nope... lol

anonymous · Answer

x^2=.85/.92

anonymous · Answer

that is what you multiply the radius by to decrease volume 15 %

you get .96 which is a 4 % decrease

amistre64 · Answer

i was thiiiisssss lose lol

anonymous · Answer

$$\frac{1}{3}.92h \pi(xr)^2=.85 \frac{1}{3}\pi r^2h$$

anonymous · Answer

the x gets squared, most of the other stuff drops out