We have a concrete forming tube with a V, where V is the product of its length l and the A of its circular base.
V=l* A_(base)
We are told that we can make 23ft^3 of cement per bag. We need to write a to find number of bags of cement needed to fill a tube with l=4 feet and with either r=4 inches, r=5 inches, or r=6 inches.
Writing a Rule for the Volume V of Any Tube
Since the volume V of a tube is the product of its length l and the area A of its circular base with radius r, we can write a rule as follows.
V=l* A_(base) ⇒ V=lπ r^2
This conversion came from the formula for the .
Determining What Operation to Use
Now, we will write a formula for the number of bags of cement. Then, we will substitute the given radii into that formula. Assume that y is the number of bags of cement. We know that one bag of cement can make V_b= 23ft^3 and length of the tube we want to fill is l=4ft. Let's use this information!
V_b * y=lπ r^2
2/3 * y= 4Ï€ r^2
y=4Ï€ r^2(3/2)
y=4Ï€ r^2/1(3/2)
y=12Ï€ r^2/2
y=6Ï€ r^2
The formula for the number of bags of cement is y=6 π r^2. Since the formula is in terms of feet, let's rewrite the given radii in terms of feet. To do so, we will multiply each by the of 1 ft12 in..
| r
|
Multiply by 1 ft/12 in.
|
r
|
| 4 in.
|
4 in. * 1 ft/12 in. = 4/12ft
|
≈ 0.33 ft
|
| 5 in.
|
5 in. * 1 ft/12 in. = 5/12ft
|
≈ 0.42 ft
|
| 6 in.
|
6 in. * 1 ft/12 in. = 6/12ft
|
0.50 ft
|
Let's start with the 4 inch radius calculation and substitute 0.33 feet for r in the formula.
y=6Ï€ r^2
y=6 π ( 0.33)^2
y=6Ï€(0.11)
y=2.05
We cannot use partial bags of cement, so 2.05 should be rounded up to 3 bags of cement. If we follow the same process for the other given radii, we have the following.
| Formula
|
r
|
Substitution
|
Results
|
Number of Bags y
|
| y=6 π r^2
|
0.33
|
6 π ( 0.33)^2
|
2.05
|
3
|
| 0.42
|
6 π ( 0.42)^2
|
3.33
|
4
|
| 0.50
|
6 π ( 0.50)^2
|
4.71
|
5
|
