the perimeter of a paved area being built at a playground is going to be 46 meters. if the length and width are each a whole number of meters, what should the length and width of the area be to cover the greatest possible area? draw a picture and make an organized list to solve
First, a square gives the greatest area, but the length and width are whole numbers. 4 doesn't go equally into 46, but since it does in 48, the results should be a close to 12 X 12 as possible, so the size is 12 meters by 11 meters
2l + 2w =p
2(12) + 2(11) = p
24 + 22 = 26
To solve this problem, let's start by drawing a rectangle to represent the paved area. Let's say the length of the area is L meters, and the width is W meters. We can create an organized list to solve this problem step by step.
1. Start by representing the length and width of the paved area as variables. Let L represent the length and W represent the width.
2. We know that the perimeter of a rectangle is given by the formula: Perimeter = 2*(Length + Width). In our case, the perimeter is 46 meters. So we can write the equation as: 46 = 2*(L + W).
3. Simplify the equation by dividing both sides by 2: 23 = L + W.
4. Since L and W are whole numbers, we can now list the possible combinations of L and W that satisfy the equation:
- L = 1, W = 22
- L = 2, W = 21
- L = 3, W = 20
- L = 4, W = 19
- L = 5, W = 18
- L = 6, W = 17
- L = 7, W = 16
- L = 8, W = 15
- L = 9, W = 14
- L = 10, W = 13
- L = 11, W = 12
- L = 12, W = 11
- L = 13, W = 10
- L = 14, W = 9
- L = 15, W = 8
- L = 16, W = 7
- L = 17, W = 6
- L = 18, W = 5
- L = 19, W = 4
- L = 20, W = 3
- L = 21, W = 2
- L = 22, W = 1
5. We can calculate the area for each combination of L and W using the formula: Area = Length * Width.
6. After calculating the area for each combination, we find that the greatest possible area is when L = 11 meters and W = 12 meters. In this case, the area would be 11 * 12 = 132 square meters.
Therefore, to cover the greatest possible area, the length should be 11 meters and the width should be 12 meters.