Question

A farmer is enclosing a rectangular area for a pigpen. He wants the length of the pen to be 20 ft longer than the width. The farmer can use no more than 100 ft of fencing. What is the pen’s greatest possible length?

Let w represent the width of the pen. What expression represents the length?

Enter your response in the box.

l=

Answer 1

2l+2w=100
2(w+20)+2w=100
2w+40+2w=100
4w+40=100
4w=60
w=15
l=w+20
l=15+20
Greatest length is 35ft.

Answer 2

a) The greatest possible length is 35.

b) l = w + 20

Explanation:

The farmer wants the length of the pen to be 20ft longer than the width. If we use w to represent the width. Then the length would be w + 20

The problem first ask us to find the greatest possible length considering the fencing will be no more than 100ft.

We know that the perimeter of a rectangular area is: 2l + 2w (where l is the length and w the width)

Therefore for the fencing we would need 2l + 2w ≤ 100.

We are going to substitute the values for l and w that we mentioned at the beginning and solve for w:

2l + 2w ≤ 100

2(w+20) +2w ≤ 100

2w + 40 +2w ≤ 100

4w ≤ 100 - 40

4w ≤60

w ≤ 15

Therefore the width has to be less or equal to 15.

Since the length is 20 ft longer than the width, the greatest possible length happens when the width has its greatest value.

So when w = 15, the length is 35 and this is the greatest possible length of it.

b) The expression that represents the length (as we said before) is w + 20