The safe load, $L$, of a wooden beam supported at both ends varies jointly as the width, $w$, and the square of the depth, $d$, and inversely as the length, $l$. A wooden beam $2 \mathrm{in}$. wide, $6 \mathrm{in}$. deep, and $8 \mathrm{ft}$ long holds up $20151 \mathrm{lb}$. What load would a beam 3 in. wide, 5 in. deep, and $11 \mathrm{ft}$. long, of the same material, support? Round your answer to the nearest integer if necessary. Answer Keypad Keyboard Shortcuts pounds

Step 1 ：Given that the safe load, $L$, of a wooden beam supported at both ends varies jointly as the width, $w$, and the square of the depth, $d$, and inversely as the length, $l$. This relationship can be expressed as $k = \frac{L}{wd^2l}$, where $k$ is the constant of variation.

Step 2 ：First, we need to find the constant of variation using the given values for the first beam. The first beam is $2 \mathrm{in}$ wide, $6 \mathrm{in}$ deep, and $8 \mathrm{ft}$ long and holds up $20151 \mathrm{lb}$. So, we substitute these values into the formula to get $k = \frac{20151}{2 \times 6^2 \times 96}$, which simplifies to $k = 2.9153645833333335$.

Step 3 ：Next, we use this constant to find the load that the second beam can support. The second beam is $3 \mathrm{in}$ wide, $5 \mathrm{in}$ deep, and $11 \mathrm{ft}$ long. We substitute these values and the constant of variation into the formula to get $L = k \times w \times d^2 \times l = 2.9153645833333335 \times 3 \times 5^2 \times 132$, which simplifies to $L = 28862$.

Step 4 ：Final Answer: The load that a beam $3 \mathrm{in}$ wide, $5 \mathrm{in}$ deep, and $11 \mathrm{ft}$ long, of the same material, can support is \(\boxed{28862}\) pounds.