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The maximum weight that a rectangular beam can support varies jointly as its width, \( W \), and the square of its height, \( H^2 \), and inversely as its length, \( L \).

If a beam \( \frac{1}{2} \) foot wide, \( \frac{1}{4} \) foot high, and 16 feet long can support 30 tons, find how much a similar beam can support if the beam is \( \frac{1}{3} \) foot wide, \( \frac{1}{3} \) foot high, and 12 feet long.

A. 45 tons
B. 20 tons
C. 60 tons
D. 15 tons

Answer :

Final answer:

The ability of a rectangular beam to support weight is dictated by direct proportion to its width and the square of its height, and inverse proportion to its length. With the constant of proportionality calculated by the given example, a beam that's 1/3 foot wide, 1/3 foot high, and 12 feet long can support 45 tons. Hence the correct option is (a).

Explanation:

The question relates to a basic principle in mathematics, specifically the concept of direct and inverse proportionality. We are given that the maximum weight a rectangular beam can support varies directly as its width (W) and the square of its height (H^2), and inversely as its length (L). This leads to the formula for the maximum weight a beam can support: Weight = k * W * H^2 / L, where k is a constant of proportionality.

Using the provided dimensions, we can first find the constant of proportionality using the initial given beam dimensions: 1/2 foot wide, 1/4 foot high, and 16 feet long supporting 30 tons.

Solving for k, we find k = Weight * L / (W * H^2), which gives k = 30 tons * 16 feet / (0.5 feet * (0.25 feet)^2) = 15360.

Next, apply the calculated k to find the weight supported by the beam with dimensions: 1/3 foot wide, 1/3 foot high, and 12 feet long.

Substituting these values into the formula: Weight = k * W * H^2 / L = 15360 * (1/3 feet) * (1/3 feet)^2 / 12 feet = 45 tons. So the answer is a) 45 tons.

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