Question

2. A farmer is planning to create a temporary ‘pen’ for livestock, by cutting a large section of plastic netting into strips, joining the strips together and then fixing the combined length to posts. He decides the fence must be 2 m high, and must form a shape that is nearly square: either a precise square or a rectangle whose length and width do not differ by more than 5 m.

If the length and width are both whole numbers of metres, what is the greatest area than can be created from 156 m2 of netting?

Answer 1

• The greatest area is 380m², using two strips 19 meters long and two strips 20 meters long, which is nearly a square.

Step-by-step explanation:

The area of the plastic netting is 156 m² and its height is told to be 2 m, so solving the equation of the area for the length, you can calcualte how many linear meters of plastic netting the farmer will use:

• Area = length × height

• 156 m² = length × 2 m ⇒ length = 156 m² / 2m = 78 m

So, you have to create the greatest possible area with 78 linear meters of plastic netting.

The greatest possible area enclosed by a fixed amount perimeter is a square.

If the figure were a precise square each side would be 1/4 of 78m, i.e. 78 m / 4 = 19.5 m.

But you need to use whole numbers that do not differ by more than 5 meters.

Suppose two sides measure 19 m and the other two sides measure 20 m. You can check that makes the perimeter equal to 78 m:

• 19 m + 19m + 20 m + 20m = 78 m.

That figure is a rectangle whose area is 19 m × 20 m = 380 m²

Since, that is the figure most similar to a square using wholde numbers you know that is the greatest area.

You can prove that using other numbers:

• 18m + 18m + 21m + 21m = 78m

• 18m × 21m = 378m²

As you see, the are for a rectanle 18 × 21 is smaller than the area for a rectangle 19 × 20. And the area will decrease as you move further away from the square shape.

Thus, the greatest area is 380m², using two strips 19 meters long and two strips 20 meters long, which is nearly a square.