Answer:
tank #1: [tex]r=2\ ft, h=9\ ft,V=113.04\ ft^3[/tex]
tank #2: [tex]r=3\ ft, h=4\ ft,V=113.04\ ft^3[/tex]
tank #3: [tex]r=4\ ft, h=2\ ft,V=100.48\ ft^3[/tex]
Step-by-step explanation:
we know that
The volume of a right cylindrical storage is given by the formula
[tex]V=\pi r^{2}h[/tex]
we have that
Juan needs a right cylindrical storage tank that holds between
110 and 115 cubic feet of water
so
For the maximum volume
[tex]115=(3.14)r^{2}h\\36.62=r^{2}h[/tex] ----> equation A
For the minimum volume
[tex]110=(3.14)r^{2}h\\35.03=r^{2}h[/tex] ____> equation B
Tank # 1
Assume a value for r and then solve for h
For r=2 ft
using equation A
substitute
[tex]36.62=(2)^{2}h[/tex]
solve for h
[tex]h=36.62/4\\h=9.2\ ft[/tex]
Remember that are whole numbers
so
[tex]h=9\ ft\\r=2\ ft[/tex]
Verify the volume
[tex]V=(3.14)(2)^{2}(9)=113.04\ ft^3[/tex]
[tex]110 \leq 113.04 \leq 115[/tex] ----> is ok
Tank # 2
Assume a value for r and then solve for h
For r=3 ft
using equation B
substitute
[tex]35.03=(3)^{2}h[/tex]
solve for h
[tex]h=35.03/9\\h=3.9\ ft[/tex]
Remember that are whole numbers
so
[tex]h=4\ ft\\r=3\ ft[/tex]
Verify the volume
[tex]V=(3.14)(3)^{2}(4)=113.04\ ft^3[/tex]
[tex]110 \leq 113.04 \leq 115[/tex] ----> is ok
Tank # 3
Assume a value for r and then solve for h
For r=4 ft
using equation A
substitute
[tex]36.62=(4)^{2}h[/tex]
solve for h
[tex]h=36.62/16\\h=2.3\ ft[/tex]
Remember that are whole numbers
so
[tex]h=2\ ft\\r=4\ ft[/tex]
Verify the volume
[tex]V=(3.14)(4)^{2}(2)=100.48\ ft^3[/tex]
[tex]110 \leq 100.48 \leq 115[/tex] ----> is ok
