Question

A wooden beam has a rectangular cross-section of height h and width w. The strength S of the beam is directly proportional to the width and the square of the height. What are the dimensions of the strongest beam that can be cut from a round log of diameter 20 inches? (Hint: S=kh^2w, where k is the proportionality constant)

Answer 1

The strongest beam that can be cut from a round log of diameter 20 inches has a height of 10 inches and a width of 20 inches.

Step-by-step explanation:

The strength of a beam is directly proportional to the width and the square of the height, as given by the equation S = kh^2w, where S is the strength, h is the height, w is the width, and k is the proportionality constant. To find the dimensions of the strongest beam that can be cut from a round log of diameter 20 inches, we need to maximize the strength.

Since the cross-section of the beam is rectangular, we can assume that the width is equal to the diameter of the log, 20 inches. Let's denote the height of the beam as h. We want to find the value of h that maximizes the strength S.

To maximize S, we need to find the maximum value of h^2. Since h is the height of the beam and the log has a diameter of 20 inches, the height cannot be greater than 20 inches. Therefore, the maximum value of h^2 is (20/2)^2 = 100 square inches.

Therefore, the dimensions of the strongest beam that can be cut from the log are a height of 10 inches and a width of 20 inches.

Answer 2

Answer: the dimensions of the strongest beam that can be cut from a round log with a diameter of 20 inches are h = 20 inches and w = 20 inches. (if not read explanation)

Step-by-step explanation:

To find the dimensions of the strongest beam that can be cut from a round log with a diameter of 20 inches, we can follow these steps:

Given that the strength of the beam, S, is directly proportional to the width, w, and the square of the height, h, we can express this relationship as S = kh^2w, where k is the proportionality constant.

Since we want to find the strongest beam, we need to maximize the value of S.

To maximize S, we can differentiate it with respect to both h and w, and then set the derivatives equal to zero to find the critical points.

Taking the derivative of S with respect to h, we get dS/dh = 2khw.

Taking the derivative of S with respect to w, we get dS/dw = kh^2.

Setting dS/dh = 0 and solving for h, we find that hw = 0, which means either h = 0 or w = 0. However, we know that the dimensions of the beam cannot be zero, so we can ignore this solution.

Setting dS/dw = 0 and solving for w, we find that w = 0, which again is not a valid solution.

Therefore, we can conclude that there are no critical points and the maximum value of S is achieved at the endpoints of the feasible region.

9. The feasible region in this case is a rectangle with height h and width w, where the width is constrained by the diameter of the log. Since the log has a diameter of 20 inches, the maximum value of w is 20 inches.

Plugging in the maximum value of w into the equation S = kh^2w, we have S = k(20)^2h = 400kh^2.

Since we want to find the dimensions of the strongest beam, we need to maximize S. Therefore, we want to maximize h.

Since h appears in the equation as h^2, we can see that the larger the value of h, the larger the value of S.

However, the height of the beam cannot exceed the diameter of the log. Therefore, the maximum value of h is 20 inches.

Therefore, the dimensions of the strongest beam that can be cut from a round log with a diameter of 20 inches are h = 20 inches and w = 20 inches.