Answer :
The maximum weight a beam can support varies jointly as its width and the square of its height and inversely as its length. Using the given proportionality, we find that a similar beam with dimensions [tex]\(w' = \frac{1}{2}\)[/tex] foot, [tex]\(h' = \frac{1}{4}\)[/tex] foot, and [tex]\(l' = 19\)[/tex] feet can support approximately 5.8 tons.
To solve this problem, we can use the concept of joint and inverse variation. Let's denote the maximum weight that the beam can support as (W) (in tons), the width as (w) (in feet), the height as (h) (in feet), and the length as (l) (in feet). The joint and inverse variation can be expressed as follows:
[tex]\[ W \propto w \cdot h^2 \cdot \frac{1}{l} \][/tex]
Given that a beam of dimensions [tex]\(w = \frac{1}{2}\) foot, \(h = \frac{1}{2}\)[/tex] foot, and [tex]\(l = 14\)[/tex] feet can support 10 tons, we can set up the equation:
[tex]\[ 10 \, \text{tons} \propto \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^2 \cdot \frac{1}{14} \][/tex]
Now, we can use this proportionality to find the constant of proportionality (k):
[tex]\[ 10 = k \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^2 \cdot \frac{1}{14} \][/tex]
Solving for (k), we get (k = 560).
Now that we have the constant of proportionality, we can use it to find the maximum weight (W') that a similar beam with dimensions [tex]\(w' = \frac{1}{2}\)[/tex] foot, [tex]\(h' = \frac{1}{4}\)[/tex] foot, and [tex]\(l' = 19\)[/tex] feet can support:
[tex]\[ W' = k \cdot w' \cdot (h')^2 \cdot \frac{1}{l'} \][/tex]
Substituting the values, we get:
[tex]\[ W' = 560 \cdot \left(\frac{1}{2}\right) \cdot \left(\frac{1}{4}\right)^2 \cdot \frac{1}{19} \][/tex]
Solving for (W'), we find [tex]\(W' \approx 5.8\)[/tex] tons.
The question probable maybe :-
The maximum weight that a rectangular beam can support varies jointly as its width and the square of its height and inversely as its length. If a beam 1/2 foot wide, 1/2 foot high, and 14 feet long can support 10 tons, find how much a similar beam can support if the beam is 1/2 foot wide, 1/4 foot high, and 19 feet long. The maximum weight is tons. (Round to one decimal place as needed.)
