Write algebraic expressions for the problem

Question

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 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.

 

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

 

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

 

Which option is the cheapest? Support your answer mathematically.

 

If you are limited by $5,000 for the cost of the border alone, how wide can your chosen border be? Support your answer mathematically.

Answer 1

Hi Jacob,

This problem is a detailed word(story) problem. The key to answering the questions is to translate the word phrases into concise, precise math expressions (variables and operations). Then, the math is easy. At first, you may need to read and re-read the problem. Then, after finding answers and verifying them, you might want to read the problem again (you’d be surprised at how many teachers on this WyzAnt Answers Forum have posted incorrect answers – I have).

O.K., let’s assign variables (that we must find to answer the questions):
     X = length of one side of pool
     Y = length of other side of pool
     A = area of pool’s border
     P = outside perimeter of pool’s border
     W = width of border alone; initially 3 ft.
     C = total cost for cement boarder
     B = total cost for brick boarder
     F = total cost for flagstone border
     G = total cost of grass border

Now, let’s translate:
     “30 x 35 feet” are the width and length of the pool alone
     “30 x 35 feet” and “border uniform 3 ft” mean
            The outside dimensions are (3+30+3) by (3+35+3) feet
     “Cement is $6 per square foot”    means    C = $6*A
     “Brick is $8 a square foot with a $85 delivery charge” means
            B = $8*A + $85
     “flagstone is $9.50 a square foot with free delivery” means
            F = $9.50*A + $0
     “grass is $5.50 a square foot plus $10 per linear foot to edge the outside perimeter”

              means G = $5.50*A + $10*P
     “Write an algebraic expression for the area of the pool’s border. Identify all variables.”

              Means write formula for A
     “Write an algebraic expression for the outside perimeter of the pool’s border.”

              Means write formula for P
     “Which option is the cheapest? Support your answer mathematically.”

              Means determine which of C,B,F,G is the smallest; support
      “If you are limited by $5,000 for the cost of the border alone, how wide can your chosen border be?”

              means find W to have cost of $5,000. Support.

We’ve done the difficult work. Now the easy math:

Let’s write formula for A (use outside dimensions mentioned above – pool’s dimensions):
       A = (W+X+W)*(W+Y+W) – X*Y (remember to identify W, X, Y)
       A = (3+30+3)*(3+35+3) – 30*35 = 36*41 – 30*35
       A = 426 sq ft.
Let’s write an expression for P:
       P = (W+X+W) + (W+Y+W) + (W+X+W) + (W+Y+Y) (identify W,X,Y)
       P = 36+41+35+41
       P = 152 sq ft.
Let’s compare costs:
      C = 6*A = 6*426 = $2556
      B = 8*A + 85 = 8*426 + 85 = $4273
      F = 9.5*A = 9.5*426 = $4047
      G = 5.5*A + 10*P = 5.5*426 + 10*152 = $3863
  Concrete has the lowest price.

Now, let’s consider how wide the border can possibly be, by using W instead of its initial value of 3 ft. Note: the problem specifically says, “your chosen border,” so we don’t have to re-compute costs for all materials, just the width using C.
      C = 6*A =6*(W+30+W)(W+35+W)
      C = 6*(2W + 30)(2W +35)
      C = 24W^2 + 420W + 360W + 1050 (using FOIL)
      24W^2 + 780W + 1050 <= 5000 (the limit for $)
      24W^2 + 780W - 3950 <= 0
           W = 4.4 ft or -37.0 ft (using quadratic formula; but W must be positive)
      W = 4.4 ft

Checking (always smart to do):
Is … ? (verify that the values of A,P,C,B,F,G fit the equations by using actual values,

        for example: C for W=4.4 is $4947)

Note: I may have made calculation errors, but you can easily find them by checking my math!

Answer 2

I'll give you the algebraic expressions, you determine the costs.

 

w = width of border

A = area of border

P = outside perimeter of border

 

P = 2(30 + 2·3) + 2(35 + 2·3)

 

A = (30 + 2·3)(35 + 2·3) - (30)(35)

A = (36)(41) - (30)(35)

A = 1476 - 1050

A = 426