find the possible dimensions of each corral?

Question

You have 240 feet of wooden fencing to form two adjacent rectangular corrals.You want each corral to have an area of 1000 square feet.Find the possible dimensions of each corral.
Please explain step-by-step, and thanks.

Answer 1

Being adjacent they share a common side. Assuming the corrals are identical, then if we let x be the length of the common side and y be the length of the other side, the amount of fencing required is 3x+4y = 240ft. The area of each corral is xy =1000sq ft. So y = 1000/x. Substitute this value in the first equation and you have:

3x+4(1000/x)=240

3x^2 +4000 =240x

3x^2 - 240x+4000=0

From the quadratic formula, x= 56.33ft. or x= 23.67ft.

In the first case the corral would have dimensions:

56.33ft and 1000/56.33 = 17.75ft (approximately)

In the 2nd case, the dimensions are:

23.67ft. and 1000/23.67 = 42.25ft (apprx)

In both cases 240ft of fencing would be used. Check:

3x+4y=240ft

Case 1: 3(56.33)+4(17.75)= 239.99 = 240

Case 2: 3(23.67)+4(42.25) = 240.01 = 240

If you wanted the corrals to be closer to squares you would choose the 2nd dimensions.

Answer 2

The corrals are adjacent so they share one side. Call that side W; the side opposite W must also be W because the corrals are rectangular. t\This means the area of corral 1 is W*L1 where L1 is the length of corral 1. Similarly for corral 2, the area is w*L2. The total length of fencing used is 3*W+2*L1+2*L2.

This gives the following formulas:

W*L1 = 1000
W*L2 = 1000
3*W+2*L1+2*L2 = 240

Subsititute L1 = 1000/W and L2 = 1000/W to get

2000/W+2000/W+3*W = 240
4000/W + 3*W = 240
4000 + 3*W^2 - 240*W = 0
3*W^2 - 240*W + 4000 = 0

You will get two answers for W. For each you can compute L1 and L2 (which are equal to each other).

Answer 3

First off, do you have to use all of the fencing? And do the dimensions of each corral have to be integers?

Answer 4

Do your own homework. But, I got two corrals short end to each other at about 16' 8.15" by 59' 11.46". Or so. And no rancher would bother putting together that precise of a corral; it's impractical.

Answer 5

let the length and width of each corral be x ft and the width y ft

Then they either share their length or their width

Case 1
Share length:

3x + 4y = 240
So y = (240 - 3x)/4

A = xy = 1000
So x(240 - 3x)/4 = 1000
ie 3x² -240x + 4000 = 0
x = (240 ± √[(-240)² - 4*3*4000])/6
= (240 ± √(57600 - 48000)/6
= (240 ± √(9600)/6)
= (240 ± 40√(6))/6
= (120 ± 20√(6))/3
≈ 28.16 ft, 11.84 ft
When x = 28.16, y = 38.88
When x = 11.84, y = 51.12

Case 1
Share width:

3x + 4y = 240
y = (240 - 4x)/3

Area = xy = 1000
x(240 - 4x)/3 = 1000
4x² - 240x + 3000 = 0
ie x² - 60x + 750 = 0
x = (60 ± √((-60)² - 4*1*750))/2
= (60 ± √(3600 - 3000))/2
= 30 ± 5√6
≈ 42.25, 17.75

When x = 42.25, y = 23.67
When x = 17.75, y = 56.33

So the possible dimensions of the corals are
____________
|...........................|
| .shared side . |
| _____↓____ |
|...........................|
|...........................|
| ___________ |

length .. width
. (ft) ....... (ft) ..
11.84 ... 51.12 --- shared length
28.16 ... 38.88 --- shared length
17.75 ... 56.33 --- shared width
42.25 ... 23.67 --- shared width