SOLUTION: Jack used 54 ft of forming and 24 cubic ft of concrete to pour a sidewalk 4 inches thick. Assuming that the length is the larger of the dimensions, what was the length of the side

Question 1124204: Jack used 54 ft of forming and 24 cubic ft of concrete to pour a sidewalk 4 inches thick. Assuming that the length is the larger of the dimensions, what was the length of the sidewalk?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the 54 feet of framing formed the perimeter of the sidewalk.
let L equal the length and W equal the width to get 2L + 2W = 54
divide both sides of this equation by 2 to get L + W = 27.

the volume of the sidewalk is equal to 24 cubic feet.
that would be L * W * D = 24
D is the depth, which is 4 inches which is 1/3 feet deep.
therefore L * W * D = 24 becomes L * W * 1/3 = 24
multiply both sides of this equation by 3/2 to get L * W = 72

you have 2 equations that need to be solved simultaneously.

they are L + W = 27 and L * W = 72

solve for L in both equations to get L = 27 - W and L = 72/W

since they're both equal to L, then 27 - W = 72 / W

multiply both sides of this equation by W to get 27W - W^2 = 72

subtract 27W from both sides of this equation and add W^2 to both sides of this equation and order the terms by descending order of degree to get:

0 = W^2 - 27W + 72 which is the same as W^2 - 27W + 72 = 0

factor this quadratic equation to get (W - 3) * (W - 24) = 0
solve for W to get W = 3 or W = 24.

since W is the smaller dimension, then W = 3.
since L + W = 27, then L must be equal to 24 because 3 + 24 = 27.

you have L = 24 and W = 3

L + W = 24 + 3 = 27
L * W = 24 * 3 = 72
L * W * D = 24 * 3 * 1/3 = 24

the solution looks good.
the solution is that the length of the sidewalk is 24 feet.

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