To solve the problem, we will use the formula for variation based on the information given. The strength \( S \) of a wooden beam varies jointly as the width \( w \) of the beam and the square of the thickness \( t \) of the beam, and inversely as the length \( l \) of the beam. This can be expressed mathematically as:
\[ S = k \frac{w t^2}{l} \]
where \( k \) is a constant of proportionality.
Step 1: Find the constant \( k \)
We know from the problem that a beam with the following specifications can support a load of 405 lb:
- Length \( l = 72 \) in
- Width \( w = 6 \) in
- Thickness \( t = 2 \) in
Substituting these values into the equation, we have:
\[ 405 = k \frac{6 \cdot 2^2}{72} \]
Calculating \( 2^2 \):
\[ 405 = k \frac{6 \cdot 4}{72} \]
This simplifies to:
\[ 405 = k \frac{24}{72} \]
Now, simplifying \( \frac{24}{72} = \frac{1}{3} \):
\[ 405 = k \cdot \frac{1}{3} \]
To find \( k \), multiply both sides by 3:
\[ k = 405 \cdot 3 = 1215 \]
Step 2: Find the maximum load for the new beam
Now we want to find the maximum load \( S \) that can be supported by a beam with the following specifications:
- Length \( l = 144 \) in
- Width \( w = 12 \) in
- Thickness \( t = 4 \) in
Using the formula for strength using the found value of \( k \):
\[ S = 1215 \frac{12 \cdot 4^2}{144} \]
Calculating \( 4^2 = 16 \):
\[ S = 1215 \frac{12 \cdot 16}{144} \]
Calculating \( 12 \cdot 16 = 192 \):
\[ S = 1215 \frac{192}{144} \]
Now, simplifying \( \frac{192}{144} \):
\[ \frac{192}{144} = \frac{4}{3} \]
So we have:
\[ S = 1215 \cdot \frac{4}{3} \]
Calculating \( 1215 \cdot \frac{4}{3} \):
\[ S = 1215 \cdot 1.3333 \approx 1620 \]
Thus, rounding our answer to the nearest whole number, the maximum load that can be safely supported by the new board is
\[ \boxed{1620} \text{ lb.} \]
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