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.

so then what are you asking?

step-by-step explanation:

2 & 7

step-by-step explanation:

he climbs 300 ft/hr at a constant rate. the peak is at 1,500 ft.

when going up, he reaches 600 ft in 2 hours. remember, that he is going in a constant rate. divide 600 with 300:

600/300 = 2

assuming that he takes the same route back (meaning that it's a 1500 ft route back, you can solve.

note also that you are solving when he gets back to 600 ft after he reaches the peak.

first, find the amount of hours it takes to get up the peak. divide the total amount (1500), with the ft/her (300) to get the amount of hours (x):

1500/300 = x

x = 5

it takes jamie 5 hours to get up. now, he is returning down, and reaches 600 (his current destination). remember that he climbs at a constant 300 ft/hr. divide 600 with 300:

600/300 = x

x = 2

now add the two numbers together.

x = 5 + 2 = 7

jamie takes 7 hours to go up the mountain and reach his current position when descending.

(total amount/constant rate) = (time).

remember that it is a constant rate, so in total, when he gets back down the mountain, it will take the same amount of time as when he goes up. the total amount for the round trip is: 5 + 5 = 10.

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