Write an algebraic expression for the area of the pool’s border. Identify all variables

Question

Suppose you and your spouse have just put in a new 30’ by 35’ in-ground pool. While the pool itself was within budget, you and your spouse have a disagreement on how you want to border the four sides of the pool. You both agree the border should be a uniform width (3 feet); however, you disagree on what material should be used: 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 delivery charge, flagstone is $9.50 a square foot with free delivery, and grass is $5.50 a square foot plus $10 per linear foot 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

Answer 1

1) area of border = (30+2w)(35+2w)- 30(35)=

4w^2 +130w= 36+390= 426 ft^2

2) outside perimeter = 2(35+2w +30+2w)= 8w +130= 24+130 =154 ft

3) cement is cheapest = 6x426= $2,556

grass= 5.5x426= 2343 +10(154)= $3,883 > $2,556

brick = 8x426= $3408+85= $3,493 is 2nd cheapest

flagstone = 9.5(426)= $4047 is most expensive

4) 5000 budget limits width to

5000= 6(4w^2+130w)

24w^2+780w-5000=0

8w^2+195w -1250=0

w=-195/16+(1/16)sqr(78025)

w=279.33-195)/16=about 5.27 feet width for cement

for grass

33w^2+715w +80w+1300 -5000=0

33w^2+795w- 3700 =0

w= -795/66 +sqr(795^2+132(3700))/66

=(-795+sqr1,120,425)/66

= about 3.99 feet max