Question 441698: The problem is as follows:
A portion of a wire 70 inches in length is bent to form a rectangle having the greatest possible area such that the length of the rectangle exceeds three times its width by 2 inches and the dimensions of the rectangle are whole numbers. Find the length of the wire that is NOT used to form the rectangle.
I noted the following:
* I understand that the perimeter of a rectangle is 2L + 2W = P
* I understand that the length of the triangle is L = 3W + 2
* Using U to represent the unused wire, I need to solve for U in either 2L + 2W + U = 70 or 2L + 2W = 70 - U.
* I start off inputting the linear equation for L by writing down 2(3W + 2) + 2W + U = 70. I get 6W + 4 + 2W + U = 70, which simplifies to 8W + U = 70.
* From there, I've tried inputting the linear equation value of W, which is W = L/3 - 2/3, but I hit a dead end.
* I've also attempted various inputs of various equations and all I get are the frustrating solutions that give you the same integer on both sides of the equal sign instead of the actual solution for the variable.
If someone can assist, I'd appreciate a thorough explanation of why I encounter problems where I input the substitution and the solution I get is the same integer on both sides of the equal sign instead of the actual solution for the variable. This would be greatly appreciated.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A portion of a wire 70 inches in length is bent to form a rectangle having the greatest possible area such that the length of the rectangle exceeds three times its width by 2 inches and the dimensions of the rectangle are whole numbers. Find the length of the wire that is NOT used to form the rectangle.
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Let width be "w".
Then length = "3w+2"
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Perimeter = 2(w+ 3w+2) = 70 inches
4w + 2 = 35
4w = 33
w is 8 if it has to be a whole number
3w+2 is 26 if it has to be a whole number.
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Length not used is 70-2(8+26) = 2 inches
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Cheers,
Stan H.
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