Can anyone PLEASE help with this algebra question?

Question

A cylindrical aluminum can is being manufactured so that its height is 8 cm more than it's radius. Estimate values for the radius to the nearest hundredth that result in the can having a volume between 1000 and 1500 cubic cm inclusively.
V=Pie*r^2h.

Answer 1

The height, h, is r + 8

Substituting into the formula for volume is

πr²(r + 8)

We are interested where πr²(r + 8) = 1000

Try r=5
π(5)²((5) + 8) = 1021.0176

Try r=4
π(4)²((4) + 8) = 603.1858

so the value lies between r=4 and r=5.

Continue in this way and you might try

r=4.95
π(4.95)²((4.95) + 8) = 996.8505

So r is between 4.95 and 5

r=4.96: 1001.6551
try r=4.955 and if it is less than 1000 and the closest so far then r=4.96 (2 dp) since any other number would round up to 4.96 anyway

r=4.955: 999.2510373

so r = 4.96 to the nearest hundredth.

This can similarly done for volume = 1500
You'll find r=5.87 to the nearest hundred.

Answer 2

1/2 of 1500 is your radius

Answer 3

V=pi*r^2*(r+8) = pi( r^3+8r^2)
The smallest r results from V=1000cm^3
r^3+8r^2-1000/pi=0`
r=4,96 cm
r^3+8r^2-1500/pi=0
r=5,87cm

Answer 4

1000 < pi r^2(r+8)< 1500
Dividing by pi
318.3098862 < r^3+8 r^2< 477.4648293
Graph the function y = x^3 + 8 x^2
* Draw two horizontal lines at y= 318.3098862
& y= 477.4648293 to intersect the graph
* Draw two vertical lines at the two intersecting point to cut the x axis, to get the limits of the value of x(r)

Answer 5

For V = 1000 cm^3, r ~= 10.75 cm.
For V = 1500 cm^3, r ~= 11.57 cm.