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}{4}[/tex] foot high, and 12 feet long can support 13 tons, find how much a similar beam can support if the beam is [tex]\frac{1}{3}[/tex] foot wide, [tex]\frac{1}{2}[/tex] foot high, and 15 feet long.

The maximum weight that the rectangular beam can support is [tex]\square[/tex].

(Simplify your answer. Round to one decimal place as needed.)

Answer :

To solve this problem, we need to understand the relationship described for the beam's maximum weight: it varies jointly with the width and the square of the height, and inversely with the length. This means we can use a formula like:

[tex]\[ W = k \times \frac{w \times h^2}{l} \][/tex]

where:
- [tex]\( W \)[/tex] is the maximum weight the beam can support.
- [tex]\( w \)[/tex] is the width of the beam.
- [tex]\( h \)[/tex] is the height of the beam.
- [tex]\( l \)[/tex] is the length of the beam.
- [tex]\( k \)[/tex] is the constant of variation.

Step 1: Calculate the constant of variation [tex]\( k \)[/tex]

We start with the information from the first beam:
- Width [tex]\( w_1 = \frac{1}{3} \)[/tex] foot
- Height [tex]\( h_1 = \frac{1}{4} \)[/tex] foot
- Length [tex]\( l_1 = 12 \)[/tex] feet
- Supported weight [tex]\( W_1 = 13 \)[/tex] tons

Substitute these values into the formula to solve for [tex]\( k \)[/tex]:

[tex]\[ 13 = k \times \frac{\left(\frac{1}{3}\right) \times \left(\frac{1}{4}\right)^2}{12} \][/tex]

Rearranging and solving for [tex]\( k \)[/tex]:

[tex]\[ k = \frac{13 \times 12}{\left(\frac{1}{3}\right) \times \left(\frac{1}{16}\right)} \][/tex]
[tex]\[ k = \frac{13 \times 12 \times 3 \times 16}{1} \][/tex]

Step 2: Use [tex]\( k \)[/tex] to find the supported weight of the new beam

Now, we have the new beam's specifications:
- Width [tex]\( w_2 = \frac{1}{3} \)[/tex] foot
- Height [tex]\( h_2 = \frac{1}{2} \)[/tex] foot
- Length [tex]\( l_2 = 15 \)[/tex] feet

We substitute these into the formula with the calculated [tex]\( k \)[/tex] to find [tex]\( W_2 \)[/tex]:

[tex]\[ W_2 = k \times \frac{\left(\frac{1}{3}\right) \times \left(\frac{1}{2}\right)^2}{15} \][/tex]

After solving this, we find that the new beam can support approximately 41.6 tons.

Therefore, the maximum weight that the new beam can support is 41.6 tons.