SOLUTION: I have been struggling with this problem for a while now and I was wondering if anyone could help me? I would deeply appreciate it!! Please and Thank you!! Solving Problems In

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Question 126637This question is from textbook Algebra Structure and Method
: I have been struggling with this problem for a while now and I was wondering if anyone could help me? I would deeply appreciate it!! Please and Thank you!!
Solving Problems Involving Inequalities
Solve.
This question is from textbook Algebra Structure and Method

Found 2 solutions by stanbon, swooshfloop:
Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 1cm greater than twice the width. If each dimension were increased by 5cm, the area would be at least 150cm^2 greater. Find the least possible dimensions.
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Let the width be "x" cm ; length is "2x+1" cm
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Changing dimensions:
width is "x+5" cm ; length is "2x+6" cm
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EQUATION:
(x+5)(2x+6) >= 150
Divide both sides by 2 to get:
(x+5)(x+3) >= 75
x^2+8x-60 >= 0
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x <= [-8 +- sqrt(64-4*-60)]/2
x <= [-8 +- sqrt(304)]/2
x = [-8 +- 17.435596]/2
0 <= x <= 4.717798
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width is "x+5" cm ; length is "2x+6" cm
0 < width <=9.717798
15.435596 <= length < 150
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Cheers,
stan H.

Answer by swooshfloop(1) (Show Source):
You can put this solution on YOUR website!
Original Width: x
Original Length: 2x + 1
Original Area: (x)(2x + 1)
Increased Width: x + 5
Increased Length: 5 + 2x + 1
Increased Area: (x + 5)(5 + 2x + 1)
"If each dimension were increased by 5 cm, the area would be at least 150 cm� greater."
(x + 5)( 5 + 2x + 1) ≧ 150 + x + 2x�
(x + 5)(6 + 2x)≧ 150 + x + 2x�
(Use FOIL - first, outside, inside, and last to multiply)
30 + 6x + 10x + 2x�≧ 150 + x + 2x�
30 + 16x + 2x�≧ 150 + x + 2x�
16x - x ≧ 120
15x≧120
x≧8
Answer:
Width: 8
Length: 17