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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}{4}[/tex] foot wide, [tex]\frac{1}{3}[/tex] foot high, and 13 feet long can support 30 tons, find how much a similar beam can support if the beam is [tex]\frac{2}{3}[/tex] foot wide, [tex]\frac{1}{4}[/tex] foot high, and 10 feet long.

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

Answer :

To solve the problem, we need to utilize the relationship given in the question about how the maximum weight a beam can support varies. Let's go through the process step-by-step.

### Step-by-Step Solution:

1. Understand the Relationship:

The maximum weight [tex]\( W \)[/tex] that a beam can support varies:
- Jointly as its width [tex]\( w \)[/tex], meaning [tex]\( W \propto w \)[/tex].
- Jointly as the square of its height [tex]\( h^2 \)[/tex], meaning [tex]\( W \propto h^2 \)[/tex].
- Inversely as its length [tex]\( l \)[/tex], meaning [tex]\( W \propto \frac{1}{l} \)[/tex].

Combining these, we can express the relationship as:
[tex]\[
W = k \cdot \frac{w \cdot h^2}{l}
\][/tex]
where [tex]\( k \)[/tex] is a constant.

2. Use the Given Data to Find [tex]\( k \)[/tex]:

For the first beam:
- Width [tex]\( w_1 = \frac{1}{4} \)[/tex] foot
- Height [tex]\( h_1 = \frac{1}{3} \)[/tex] foot
- Length [tex]\( l_1 = 13 \)[/tex] feet
- Maximum weight supported [tex]\( W_1 = 30 \)[/tex] tons

Substitute these values into the formula to find [tex]\( k \)[/tex]:
[tex]\[
30 = k \cdot \frac{\frac{1}{4} \cdot \left(\frac{1}{3}\right)^2}{13}
\][/tex]

Solving for [tex]\( k \)[/tex]:
[tex]\[
30 = k \cdot \frac{\frac{1}{4} \cdot \frac{1}{9}}{13}
\][/tex]
[tex]\[
30 = k \cdot \frac{1}{36 \cdot 13}
\][/tex]
[tex]\[
k = 30 \cdot 468 = 14040
\][/tex]

3. Calculate the Weight for the New Beam:

For the second beam:
- Width [tex]\( w_2 = \frac{2}{3} \)[/tex] foot
- Height [tex]\( h_2 = \frac{1}{4} \)[/tex] foot
- Length [tex]\( l_2 = 10 \)[/tex] feet

Use the formula with the new dimensions:
[tex]\[
W_2 = 14040 \cdot \frac{\frac{2}{3} \cdot \left(\frac{1}{4}\right)^2}{10}
\][/tex]
[tex]\[
W_2 = 14040 \cdot \frac{\frac{2}{3} \cdot \frac{1}{16}}{10}
\][/tex]
[tex]\[
W_2 = 14040 \cdot \frac{1}{240}
\][/tex]
[tex]\[
W_2 = \frac{14040}{240} = 58.5
\][/tex]

Thus, the maximum weight that the new beam can support is [tex]\( 58.5 \)[/tex] tons.

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Rewritten by : Barada