MATH SOLVE

3 months ago

Q:
# 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?

Accepted Solution

A:

Answer: (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 ft²Step-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__(b) The two parallel sides cannot be negative, and they cannot exceed half the length of the fence available, so ... 0 ≤ x ≤ 360__(c) Area is the product of the length (720-2x) and the width (x). The desired function is ... A = x(720 -2x)__(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} . . . feetFor an area of 40,000 ft², the values of x will be ... x = 180 ±20√31 ≈ {68.645, 291.355} . . . feetThe 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.__(e) The vertex of the area function is at the axis of symmetry: x = 180. The corresponding dimensions are ... 180 ft × 360 ftand the area of that is 64,800 ft².