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 4 inches wide, 3 inches high, and 18 feet long can support a maximum of 3 tons. What is the maximum weight that could be supported by a beam that is 7 inches wide, 4 inches high, and 6 feet long? [Itons $\times 5$

Step 1 ：Given 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. This can be represented by the formula \(k = \frac{y \cdot w \cdot h^2}{l}\), where \(k\) is the constant of variation, \(y\) is the weight, \(w\) is the width, \(h\) is the height, and \(l\) is the length.

Step 2 ：First, we need to find the constant of variation (\(k\)) using the given values. A beam 4 inches wide, 3 inches high, and 18 feet long can support a maximum of 3 tons. Substituting these values into the formula, we get \(k = \frac{3 \cdot 4 \cdot 3^2}{18}\).

Step 3 ：Next, we use this constant to find the maximum weight that the other beam can support. The other beam is 7 inches wide, 4 inches high, and 6 feet long. Substituting these values into the formula, we get \(y = \frac{k \cdot l}{w \cdot h^2}\).

Step 4 ：The result is in tons and is a decimal. To make it more understandable, we should convert it to pounds (since 1 ton is equal to 2000 pounds) and round it to the nearest whole number.

Step 5 ：Final Answer: The maximum weight that could be supported by a beam that is 7 inches wide, 4 inches high, and 6 feet long is \(\boxed{643}\) pounds.