Suppose that the maximum weight that a certain type of rectangular beam can support varies inversely as its length and jointly as its width and the square of its height. Suppose also that a beam 3 inches wide, 4 inches high, and 6 feet long can support a maximum of 28 tons. What is the maximum weight that could be supported by a beam that is 5 inches wide, 4 inches high, and 20 feet long?
Let x be the maximum weight that could be supported by the beam in tons. Then we can set up the following proportion using the given information:
(28 tons)/(3 in. x 4^2 in.^2) = (x tons)/(5 in. x 4^2 in.^2)
Simplifying, we have:
28/48 = x/80
Cross-multiplying, we get:
48x = 28 x 80
Solving for x, we get:
x = (28 x 80)/48 = 46.67 tons
Therefore, the maximum weight that could be supported by a beam that is 5 inches wide, 4 inches high, and 20 feet long is approximately 46.67 tons.