



 
Hi Jordan, I approach was essentially the same as yours. The area of the side of a drum is $2 \times \pi \times r \times h$ where $r$ is the radius of the top and $h$ is the base. That gave me \[2 \times \pi \times 27 \times 54 = 14250.26 \mbox{ cm}^2\] The area of the bottom is \[\pi \times r^2 = \pi \times 27^2 = 2290.22 \mbox{ cm}^2\] Thus the total surface area is \[14250.26 + 2290.22 = 16540.49 \mbox{ cm}^2\] $8.8 m^2 = 8.8 \times 100 \times 100 = 88000 \mbox{ cm}^2$ and hence each drum requires \[\frac{16540.49}{88000} = 0.18796 \mbox{ liters of paint.}\] Thus you need $0.18796 \times 400 = 75.2$ liters. Is that what you got?  


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