Quadratic Word Problem: Arena Problem

Question

Steps to take:
1. Draw diagrams
2. Decide what X will stand for
3. Write an equation that uses the variable and the info they give you
4. Set the equation equal to zero
5. Factor
6. Solve

 

Word Problem: You are building an arena that consists of a rectangular ice rink and a concrete walkway along one side. The entire arena (including the concrete walkway) has a total area of 5000 square feet. The length of the ice rink is 30 feet greater than the width. The dimensions of the concrete walkway are 20 ft. by w feet, where w is the width of the rink. Find the length and width of the entire arena.

Answer 1

Let's attempt to solve this problem.  Assign the following...

 

W_r=x=width of rink

L_r=length of rink

A_r=area of rink

A_w=area of walkway

 

From the problem wording, we can assign side lengths for the rink and walkway to our diagram below needed to calculate areas of each.  We cannot assume the walkway is the same length as the rink...

 

 

1.  Draw a diagram for the parameters described and label side lengths...

 

   L_w=W_r=x  (length of walkway equals width

                                     of rink)

  -----------------

|                    | W_w=20'       <----(Walkway)

  ----------------------------

|                                  |

|                                  |             <----(Rink)

|                                  | W_r=x 

|                                  |

  ----------------------------

   L_r=W_r+30'=x+30' (length of rink equals width

                                  of rink plus 30')

 

2.  

As stated above, we will assign "x" to the width of the rink "W_r".

 

3.

A_tot=5000 ft²...given, problem statement

A_tot=A_r+A_w......Eq1, add areas of rink and walkway

A_r=L_rxW_r.........Eq2, area of rink

A_w=L_wxW_w.....Eq3, area of walkway

 

So let's substitute Eqs2,3 into Eq1...

  A_tot=[L_rxW_r]+[L_wxW_w]

5000=[(x+30)(x)]+[20(x)]...substitute values

                                             assigned above in 

                                             diagram

5000=x²+30x+20x.....distribute

5000=x²+50x..............Eq4, combine like terms

 

4.

Let's rewrite Eq4 and set it equal to zero...

500=x²+50x................Eq4

    0=x²+50x-5000.......subtract 5000, both sides

x²+50x-5000=0...........Eq5, rewrite in reverse

                                      order, equate to zero

 

5.

Let's factor Eq4...

  x²+50x-5000=0.......Eq5

(x+100)(x-50)=0.........Eq5 factored

 

6.

Equate each factor to zero, solve for "x"...

x+100=0........first factor

        x=-100...subtract 100, both sides

 

x-50=0...........second factor

     x=50.........add 50, both sides

 

Disregard "x=-100" value as distance cannot be negative.

 

So x=W_r=L_w=50 feet,

L_r=W_r+30=x+30=50+30=80 feet

 

**********************

 

Let's check our solution...

A_r=L_rxW_r

    =80x50

    =4000 ft²

 

A_w=L_wxW_w

     =50x20

     =1000 ft²

 

  A_tot=A_r+A_w...............Eq1

5000=4000+1000......substitute

5000=5000.................true, √check, our solution 

                                     is correct

 

Always check your solutions and work!

 

Summary:

Rink dimensions are 80'x50'

Walkway dimensions are 20'x50'