Modeling Word Problem -- Quadratic Functions #55

Question

One campus of Houston Community College has plans to construct a rectangular parking lot on land bordered on one side by a highway. There are 640 ft of fencing available to fence the other three sides. Let x represent the length of each of the two parallel sides of fencing.

 

a). express the length of the remaining side to be fenced in terms of x

b). what are the restrictions of x

c). determine a function for total area that represents the area of the parking lot in terms of x.

d). determine the values of x that will give an area between 30,000 and 40,000 ft²

e). what dimensions will give a maximum area, and what will this area be?

 

Answer 1

a)  The total fencing is 640ft which will be 2x + (side 3).

     Let y = length of side 3

     2x + y = 640

     y = 640-2x

 

b)  2x<640

     x< 320 ft.  The parallel sides must be less than 320 ft in length.

     The restrictions are 0<x<320

 

c)  The area of the parking lot is length*width

     A = x(640-2x)

     A = -2x² + 640x

 

d)  -2x² + 640x = 30,000

     -2x² + 640x - 30,000 = 0

      x² - 320x + 15,000 = 0

      x = [320 ±√((-320)²-4(15,000))]/2

      By quadratic equation x ≈ 57.04 ft or 262.96 ft

 

      If x = 57.04 ft, y = 640-2x = 525.92

      Dimensions 57.04 by 525.92 ft

 

     If x = 262.96 ft,  y = 640-2x = 114.08

     Dimensions 262.96 by 114.08

 

    Either of those dimensions will give approximately 30,000 ft²

 

    Now set up the problem the same way except with 40,000

    -2x² + 640x = 40,000

    -2x² + 640x - 40,000 = 0

     x² - 320x + 20,000 = 0

     Again use the quadratic equation to solve for x

     x = [320 ±√((-320)²-4(20,000))]/2

     x = (320 ± 149.67)/2

     x ≈ 234.84 or 85.17

 

     If x = 234.84, y = 640-2x = 170.32

     Dimensions 234.84 by 170.32

 

    If x = 85.17, y = 640-2x = 468.6

    Dimensions 85.17 ft by 468.6 ft

 

e) Maximum dimensions

    For y = -2x² + 640x

    The maximum area is at the vertex

    The vertex is at  (-b/2a, f(-b/2a))

   

    -b/2a = -640/(-4) = 160

     The maximum area will be achieved at x = 160 ft

     y = 640-2x = 640-2(160) = 320

     Dimensions 160 ft by 320 ft