Answer: 61
Solving Steps:
Solving Steps:
The area of a rectangular field is 1,710 square feet. The length of the field is 45 feet. What is the width of the field.
1 answer:
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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 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.
