He campus of a college has plans to construct a rectangular parking lot on land bordered on one side by a highway. there are 720 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 rectangle has width x. x x (a) express the length of the remaining side to be fenced in terms of x. (b) what are the restrictions on x? (c) determine a function a that represents the area of the parking lot in terms of x. (d) determine the values of x that will give an area between 20 comma 000 and 40 comma 000 ftsquared. (e) what dimensions will give a maximum area, and what will this area be?

  (a) 720 -2x

  (b) 0 ≤ x ≤ 360

  (c) A = x(720 -2x)

  (d) (30.334, 68.645) ∪ (291.355, 329.666) (two disjoint intervals)

  (e) x = 180 ft, the other side = 360 ft; total area 64,800 ft²

Step-by-step explanation:

(a) The two parallel sides of the fenced area are each x feet, so the remaining amount of fence available for the third side is (720 -2x) ft. Then ...

  length = 720 -2x

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(b) The two parallel sides cannot be negative, and they cannot exceed half the length of the fence available, so ...

  0 ≤ x ≤ 360

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(c) Area is the product of the length (720-2x) and the width (x). The desired function is ...

  A = x(720 -2x)

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(d) For an area of 20,000 ft², the values of x will be ...

  20000 = x(720 -2x)

  2x² -720x +20000 = 0

  x = (-(-720) ±√((-720)² -4(2)(20000)))/(2(2)) = (720±√358400)/4

  x = 180 ±40√14 = {30.334, 329.666} . . . feet

For an area of 40,000 ft², the values of x will be ...

  x = 180 ±20√31 ≈ {68.645, 291.355} . . . feet

The values of x producing areas between 20,000 and 40,000 ft² will be values of x in the intervals (30.334, 68.645) or (291.355, 329.666) feet.

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(e) The vertex of the area function is at the axis of symmetry: x = 180. The corresponding dimensions are ...

  180 ft × 360 ft

and the area of that is 64,800 ft².

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