Yahoo Answers 5-4-14

Topic:
Algebra, Area
Suppose you and your spouse have just put in a new 30 foot by 35 foot in ground pool. While the pool itself was within budget, you and your spouse have a disagreement on how you want to border the pool. You both agree the surrounding material should be a uniform size (3 feet); however, you disagree on what that material should be: cement, brick, flagstone, or grass. Each of the materials has a different cost associated with it. Cement is $6 per square foot, brick is $8 a square foot with a $85 charge, flagstone is $9.50 a square foot with free delivery, and grass is $5.50 a square foot plus $10 to edge the outside perimeter. 1. Write an algebraic expression for the area of the pool’s border. Identify all variables. 2. Write an algebraic expression for the outside perimeter of the pool’s border. Identify all variables. 3. Which option is the cheapest? Support your answer mathematically. 4. If you are limited by $5,000, how wide can your chosen border be? Support your answer mathematically. L=Length of Pool W=Width of Pool B=Width of Border The area of the pool is just WB. The area of the border gives you a few options. Maybe the easiest to see is a bunch of rectangles around the pool. There are 2 long rectangles along the length, 2 along the width and 4 squares in the corners. Add the areas for the total: Border Area=2LB+2WB+4B^2 The perimeter has to include the border. If it was just the pool it would be 2(L+W). But the length and width need twice the width of the border added: Total perimeter=2(L+W+4B) Take your area and multiply it by the price (and add any needed charges) Cement=6*BA Brick=8*BA+85 Flag=9.5*BA Grass=5.5*BA+10 Number 4... is a bit tricky and since I got you pretty far I'll let you play with this one. Maybe a table/graph would help... border width vs. cost.