SOLUTION: The length of a rectangle is a = 2 cm less than twice the width. Express as an integer the maximum width of the rectangle when the perimeter is less than 59 cm.
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Question 500098: The length of a rectangle is a = 2 cm less than twice the width. Express as an integer the maximum width of the rectangle when the perimeter is less than 59 cm. Answer by oberobic(2304) (Show Source):
You can put this solution on YOUR website! L = length
W = width
L = 2W -2 cm
P = perimeter
P < 59
.
Solve the equality first.
.
P = 59
.
P = 2(L+W)
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Substitute for L = 2W -2
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2( 2W -2 + W) = 59
2(3W -2) = 59
6W -4 = 59
6W = 63
W = 63/6
W = 21/2
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L = 2(21/2)-2
L = 21-2
L = 19
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So, the suggested dimensions are:
L = 19
W = 10 1/2
which makes
P = 2 (19 + 10 1/2)
P = 2 (29 1/2)
P = 59
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But we are told to ensure the perimeter < 59.
And we are told to make the width and integer.
So,
W = 10
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L = 2*W -2
L = 18
.
P = 2(18+10)
P = 56
.
Answer:
Width = 10 cm
Length = 18 cm
.
Done.
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