SOLUTION: The weight that can be safely supported by a 2-by 6-inch support beam varies inversely with its length. A builder finds that a support beam that is 10 feet long will support 800 po
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Question 1114099: The weight that can be safely supported by a 2-by 6-inch support beam varies inversely with its length. A builder finds that a support beam that is 10 feet long will support 800 pounds. Find the weight that can be safely supported by a beam that is 16 feet long.
The weight that can be safely supported by the beam is _ lb. Found 2 solutions by stanbon, greenestamps:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The weight that can be safely supported by a 2-by 6-inch support beam varies inversely with its length. A builder finds that a support beam that is 10 feet long will support 800 pounds. Find the weight that can be safely supported by a beam that is 16 feet long.
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w = k/L
800 = k/10
k = 8000
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w = 8000/L
w = 8000/16
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w = 500 lbs
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Cheers,
Stan H.
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The weight that can be safely supported by the beam is 500 lb.
The solution by tutor stanbon is fine. But let me suggest a couple of alternative solution processes that you might find easier for you to use.
The formal definition of inverse variation applied to this problem says
where W is the weight that can be supported, L is the length of the beam, and k is some constant.
In the solution by stanbon, you use the given information to find the constant k; then you use that value to find the weight that can be supported by the longer beam.
For me, it is easier to find the answer without specifically solving for the constant k.
To do that, I prefer to think of inverse variation for this problem as .
Since the product of weight and length is a constant, to solve the problem I just do
An even faster path to the solution is possible if you think of inverse variation as meaning that if one of the variables is multiplied by a factor of n, then the other number needs to get divided by that same factor.
Applied to this problem, I simply see that the new length is 8/5 times the original; that means the weight that can be supported is 5/8 as much:
(