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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}{3}[/tex] foot wide, [tex]\frac{1}{4}[/tex] foot high, and 18 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}{3}[/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 the relationship given for the maximum weight a beam can support.

The maximum weight [tex]\( W \)[/tex] that a beam can support varies:

1. Jointly as its width [tex]\( w \)[/tex] and the square of its height [tex]\( h^2 \)[/tex].
2. Inversely as its length [tex]\( l \)[/tex].

This relationship can be expressed by the formula:
[tex]\[ W = k \times \frac{w \times h^2}{l} \][/tex]
where [tex]\( k \)[/tex] is a constant of proportionality.

Now, let's solve the problem step by step:

1. Find the constant [tex]\( k \)[/tex]:

We have data for the first beam:
- Width ([tex]\( w_1 \)[/tex]) = [tex]\( \frac{1}{3} \)[/tex] foot
- Height ([tex]\( h_1 \)[/tex]) = [tex]\( \frac{1}{4} \)[/tex] foot
- Length ([tex]\( l_1 \)[/tex]) = 18 feet
- Maximum weight ([tex]\( W_1 \)[/tex]) = 30 tons

Substitute these values into the formula to find [tex]\( k \)[/tex]:
[tex]\[
30 = k \times \frac{\left(\frac{1}{3}\right) \times \left(\frac{1}{4}\right)^2}{18}
\][/tex]

Simplify inside the fraction:
[tex]\[
30 = k \times \frac{\frac{1}{3} \times \frac{1}{16}}{18}
\][/tex]

Simplify further:
[tex]\[
30 = k \times \frac{1}{144}
\][/tex]

Solve for [tex]\( k \)[/tex]:
[tex]\[
k = 30 \times 144 = 4320
\][/tex]

2. Find the maximum weight for the second beam:

For the second beam:
- Width ([tex]\( w_2 \)[/tex]) = [tex]\( \frac{2}{3} \)[/tex] foot
- Height ([tex]\( h_2 \)[/tex]) = [tex]\( \frac{1}{3} \)[/tex] foot
- Length ([tex]\( l_2 \)[/tex]) = 18 feet

We use the same formula to calculate the new maximum weight ([tex]\( W_2 \)[/tex]):
[tex]\[
W_2 = 4320 \times \frac{\left(\frac{2}{3}\right) \times \left(\frac{1}{3}\right)^2}{18}
\][/tex]

Simplify inside the fraction:
[tex]\[
W_2 = 4320 \times \frac{\frac{2}{3} \times \frac{1}{9}}{18}
\][/tex]

Simplify further:
[tex]\[
W_2 = 4320 \times \frac{2}{243}
\][/tex]

Calculate [tex]\( W_2 \)[/tex]:
[tex]\[
W_2 \approx 35.6
\][/tex]

Therefore, the maximum weight that the second beam can support is approximately 106.7 tons when rounded to one decimal place.

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Rewritten by : Barada