Question 1186499: The safe load LF a wooden beams to put it up both and varies jointly as the wet W in the square of the deputy and inversely as the length , l . a wooden beams 7 inches wide 9 inches deep and 7 feet long holds up 39701 pounds what load will they be 9 inches wide 6 inches deep and 18 foot long of the same material support round your answer to the nearest integer if necessary
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52780)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
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In this problem, the maximum load varies directly as the width w, directly as the square of the depth d, and inversely as the length l. The given information is that the maximum load is 39701 pounds when the width is 7 inches, the depth is 9 inches, and the length is 7 feet. We are asked to find the maximum load when the width is 9 inches, the depth is 6 inches, and the length is 18 feet.
One way to solve the problem is to use the variation formula with the given set of data to find the constant of proportionality using the given information, then use that constant with the new set of data to find the new safe load.
But since we are only asked to find the safe load for one other set of data, we don't need to do that much work. Instead, we can just determine how the maximum load changes for the change in each of the parameters.
The width changes from 7 to 9 inches. Since the safe load varies directly with the width, this changes the safe load by a factor of (9/7).
The depth changes from 9 to 6 inches. Since the safe load varies directly as the square of the depth, this changes the safe load by a factor of (6/9)^2.
The length changes from 7 feet to 18 feet. Since the safe load varies inversely as the length, this changes the safe load by a factor of (7/18).
So the safe load with the new parameters is
 = 8822.4444...
Rounded to the nearest integer, the new safe load is
ANSWER: 8822 pounds
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