College

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 [tex]$\frac{1}{3}$[/tex] foot wide, [tex]$\frac{1}{3}$[/tex] foot high, and 10 feet long can support 20 tons, find how much a similar beam can support if the beam is [tex]$\frac{2}{3}$[/tex] foot wide, [tex]$\frac{1}{2}$[/tex] foot high, and 18 feet long.

The maximum weight is [tex]$\square$[/tex] tons. (Round to one decimal place as needed)

Answer :

To solve this problem, we need to understand how the maximum weight a rectangular beam can support varies. The weight varies jointly as the width and the square of its height, and inversely as its length.

Here's how you can approach it:

1. Identify the formula:
The formula that describes the relationship is:
[tex]\[
W = k \cdot \frac{\text{width} \cdot (\text{height})^2}{\text{length}}
\][/tex]
where [tex]\( W \)[/tex] is the maximum weight the beam can support, and [tex]\( k \)[/tex] is a constant of proportionality.

2. Determine the constant [tex]\( k \)[/tex] using the first beam:
For the first beam, we have:
- Width = [tex]\( \frac{1}{3} \)[/tex] foot
- Height = [tex]\( \frac{1}{3} \)[/tex] foot
- Length = 10 feet
- Weight = 20 tons

Plug these values into the formula to solve for [tex]\( k \)[/tex]:
[tex]\[
20 = k \cdot \frac{\frac{1}{3} \cdot \left(\frac{1}{3}\right)^2}{10}
\][/tex]

Simplifying inside the fraction:
[tex]\[
20 = k \cdot \frac{\frac{1}{3} \cdot \frac{1}{9}}{10}
\][/tex]
[tex]\[
20 = k \cdot \frac{1}{27 \cdot 10}
\][/tex]
[tex]\[
20 = k \cdot \frac{1}{270}
\][/tex]

Solving for [tex]\( k \)[/tex]:
[tex]\[
k = 20 \times 270 = 5400
\][/tex]

3. Calculate the maximum weight for the second beam:
Now, use the same formula to find the weight the second beam can support with the given dimensions:
- Width = [tex]\( \frac{2}{3} \)[/tex] foot
- Height = [tex]\( \frac{1}{2} \)[/tex] foot
- Length = 18 feet

The formula becomes:
[tex]\[
W = 5400 \cdot \frac{\frac{2}{3} \cdot \left(\frac{1}{2}\right)^2}{18}
\][/tex]

Calculate the fraction:
[tex]\[
W = 5400 \cdot \frac{\frac{2}{3} \cdot \frac{1}{4}}{18}
\][/tex]
[tex]\[
W = 5400 \cdot \frac{\frac{2}{12}}{18}
\][/tex]
[tex]\[
W = 5400 \cdot \frac{1}{108}
\][/tex]

Simplifying:
[tex]\[
W = 50
\][/tex]

Thus, the maximum weight the second beam can support is 50.0 tons.