# 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(57354)   (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. ------------------------ Let the width be "x" cm ; length is "2x+1" cm ----------------------- Changing dimensions: width is "x+5" cm ; length is "2x+6" cm ----------------------- EQUATION: (x+5)(2x+6) >= 150 Divide both sides by 2 to get: (x+5)(x+3) >= 75 x^2+8x-60 >= 0 -------- x <= [-8 +- sqrt(64-4*-60)]/2 x <= [-8 +- sqrt(304)]/2 x = [-8 +- 17.435596]/2 0 <= x <= 4.717798 --------------------- width is "x+5" cm ; length is "2x+6" cm 0 < width <=9.717798 15.435596 <= length < 150 ----------------------------------- 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 cm² greater." (x + 5)( 5 + 2x + 1) ≧ 150 + x + 2x² (x + 5)(6 + 2x)≧ 150 + x + 2x² (Use FOIL - first, outside, inside, and last to multiply) 30 + 6x + 10x + 2x²≧ 150 + x + 2x² 30 + 16x + 2x²≧ 150 + x + 2x² 16x - x ≧ 120 15x≧120 x≧8 Answer: Width: 8 Length: 17