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Question 929784: Please someone help I have Been lost on this problem for 3 days now!!! :-(
The inside of a cylindrical tank used to hold resin must be coated with a corrosion preventative. The tank measures 7 ft. in diameter and 5 ft. in height. One gallon of corrosion preventative will cover 10 ft^2. How many gallons will be needed to coat the tank? Assume that the sides, top, and bottom of the tank will be treated.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! You need to find the surface area of the tank (if the shell is thin enough, then the outside & inside surface areas are roughly the same)
We will use the formula SA = 2pi*r^2 + 2*pi*r*h, where,
SA = surface area
r = radius
h = height
In this case,
r = 7/2 (cut the diameter in half)
h = 5
SA = 2pi*r^2 + 2*pi*r*h
SA = 2pi*(7/2)^2 + 2*pi*(7/2)*5
SA = (119/2)*pi
SA = 186.924762888592 (this is approximate)
The surface area of the inner walls of the tank is approximately 186.924762888592 square feet.
We are told that "One gallon of corrosion preventative will cover 10 ft^2"
so we have this ratio: (1 gallon/10 ft^2)
and we use this to determine how many gallons we'll need
Multiply the surface area by the ratio shown above
(186.924762888592 ft^2) * (1 gallon/10 ft^2) = 186.924762888592/10 = 18.6924762888592
Round up to the nearest whole gallon to get 19
You will need 19 gallons of the corrosion preventative. You'll have leftover amount of corrosion preventative, but at least you won't come up short (as compared to getting 18 gallons).
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