SOLUTION: Mang Lino is planning to construct a rectangular coop with the width 2 meter shorter that thrice the length. He wants to use at least 4 meters of chicken wire but at most 20 meters

Algebra ->  Inequalities -> SOLUTION: Mang Lino is planning to construct a rectangular coop with the width 2 meter shorter that thrice the length. He wants to use at least 4 meters of chicken wire but at most 20 meters      Log On


   

Question 1182496: Mang Lino is planning to construct a rectangular coop with the width 2 meter shorter that thrice the length. He wants to use at least 4 meters of chicken wire but at most 20 meters of the material. What are the possible integral dimensions of the coop? What is the inequality statement?
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

W = width
L = length
which are some positive whole numbers. The units for each are in meters.

The width is 2 meters shorter than thrice the length, meaning
W = 3L-2
since 3L is thrice, or three times, the length. Then we subtract off 2 from that result to get the width W.

The perimeter of the rectangle with length L and width W is
P = 2L+2W
we add up two copies of L and W each to get the total distance around the rectangle. This is the amount of fencing needed for the rectangular coop.

He wants to use at least 4 meters but at most 20 meters
This places boundaries on how low and how high P can get
P = 4 is the smallest possible perimeter
P = 20 is the largest possible perimeter
That leads us to 4+%3C=+P+%3C=+20

Let's replace P with 2L+2W to get
4+%3C=+P+%3C=+20

4+%3C=+2L%2B2W+%3C=+20

4+%3C=+2%28L%2BW%29+%3C=+20

2+%3C=+L%2BW+%3C=+10
In the last step, I divided everything by 2

Now let's replace W with 3L-2 and isolate L like so
2+%3C=+L%2BW+%3C=+10

2+%3C=+L%2B3L-2+%3C=+10

2+%3C=+4L-2+%3C=+10

2%2B2+%3C=+4L-2%2B2+%3C=+10%2B2 Adding 2 to all sides

4+%3C=+4L+%3C=+12

4%2F4+%3C=+4L%2F4+%3C=+12%2F4 Dividing all sides by 4

1+%3C=+L+%3C=+3

This tells us that L = 1 is the smallest length possible while L = 3 is the largest length possible. L = 2 is also possible as it's right in between. These are the only integer lengths possible that meet the required conditions.

If L = 1, then
W = 3L-2
W = 3(1)-2
W = 1
So L = 1 leads to W = 1
The perimeter would be P = 2L+2W = 2(1)+2(1) = 4
So this is the case where he uses the least amount of wire possible (4 meters of it).

If L = 2, then,
W = 3L-2
W = 3(2)-2
W = 4
So L = 2 leads to W = 4
The perimeter would be P = 2L+2W = 2(2)+2(4) = 12
He is now using 12 meters of wire. This fits the inequality 4+%3C=+P+%3C=+20 because 4+%3C=+12+%3C=+20 is a true statement.

Lastly, if L = 3, then
W = 3L-2
W = 3(3)-2
W = 7
So L = 3 leads to W = 7
The perimeter would be P = 2L+2W = 2(3)+2(7) = 20
Telling us that he's using the most wire he's alloting or budgeting out.

To recap everything:
  • The 1 m by 1 m coop needs 4 m of wire
  • The 2 m by 4 m coop needs 12 m of wire
  • The 3 m by 7 m coop needs 20 m of wire
These are the only three possible cases if we want integral (ie integer) dimensions to the rectangle.