# Cylinder Volume Question??? (Picture attached)?

### 2 Answers

- PinkgreenLv 71 year agoFavorite Answer
Let H cm be the height of the can when R=4.2 cm, then

piHR^2=725

=>

H=725/(17.64pi)=13.082 cm

approximately.

Now, pi[725/(17.64pi)-h](4.2+r)^2=725

=>

(725/17.64-pi*h)(17.64+8.4r+r^2)=725

=>

h=725(r^2+8.4r)/[17.64pi(r+4.2)^2]

=>

(a) h(r)=(725/pi)[1/17.64-1/(r+4.2)^2]

(b) When the radius increases by 1 cm,

h(1)=(725/pi)[1/17.64-1/(5.2)^2]=4.547 cm

h is decreased by 4.547 cm approximately.

Check:

the volume of the new can=

pi(13.082-4.547)(5.2^2)=725.04 ml, valid.

(c) When the height is reduced by 1 cm,

(r+4.2)^2=1/(1/17.64-pi/725)

=>

r=1/sqr(1/17.64-pi/725)-4.2=0.17 cm

the radius increases by 0.17 cm approximately.

Check:

The volume of the new can=

pi(13.082-1)(4.2+0.17)^2=724.86 ml, valid.

- Mike GLv 71 year ago
For the original can 725 = πy(4.2)^2 where y = original height in cm

y = 13.0825 cm

a) For the new can

725 = π(13.0825-h)(4.2+r)^2

13.0825-h = 725/[π(4.2+r)^2]

h(r) = 13.0825 -725/[π(4.2+r)^2]

c) (i) h(0.5) = 2.635 cm

ii) h(1.3) = 5.454 cm

iii) h(1.5) = 5.980 cm

d) For this write r as a function of h and then plug in the three values h