SOLUTION: The safe load a beam can support varies jointly as the width and the square of the depth and inversely as the length. If a 2 x 8 inch beam 16 feet long is turned so that the width

Algebra ->  Surface-area -> SOLUTION: The safe load a beam can support varies jointly as the width and the square of the depth and inversely as the length. If a 2 x 8 inch beam 16 feet long is turned so that the width       Log On


   

Question 507544: The safe load a beam can support varies jointly as the width and the square of the depth and inversely as the length. If a 2 x 8 inch beam 16 feet long is turned so that the width is 2 inches, it can support 2,000 pounds. How much weight can the same bea, support if it is turned so that the width is 8 inches?
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
First write the equation of proportionality:
L+=+k%2Aw%2Ad%5E2%2Fl To find the value of k, the constant of proportionality, substitute the given values of: L = 2000, w = 2, d = 8, and l = 16
2000+=+k%2A2%2A8%5E2%2F16 Solve for k. Multiply both sides by 16.
32000+=+k%2A128 Divide by 128.
k+=+250 so we can rewrite the first equation as:
L+=+250%2Aw%2Ad%5E2%2Fl6 Now we can find the safe load, L, for the given parameters: w = 8, d = 2, and l = 16.
L+=+250%2A8%2A2%5E2%2F16
L+=+8000%2F16
L+=+500pounds.