Question

A parent has 32 feet of plastic dividers to create a rectangular fenced-in play area for a toddler. If the parent wants the play area to be as large as possible, what is the width, in feet, of

the fenced-in area the parent should set up?
feet

Answer 1

To determine the width of the fenced-in play area, we can set up an equation using the perimeter of the rectangle. By maximizing the width, we can find the largest possible play area. The width should be set to 5 feet.

Step-by-step explanation:

To determine the width of the fenced-in play area, we need to understand that it is a rectangle and that the perimeter is equal to the sum of all four sides. In this case, the sum of the two widths and the two lengths must be equal to 32 feet, which is the total amount of plastic dividers available.

Let's assume that the width of the play area is represented by 'w'.

Since a rectangle has two equal-length sides, we can set up the equation as follows: 2w + 2l = 32

Given that we want the play area to be as large as possible, we can maximize the width by considering that the length will be twice the width. So we can substitute '2w' for 'l' in the equation: 2w + 2(2w) = 32

Simplifying the equation: 6w = 32

Dividing both sides by 6, we get: w = 5.33 feet

Since it is not possible to have a fraction of a foot for the width, we can round down to the nearest whole number: w = 5 feet

Answer 2

To create the largest possible rectangular play area with 32 feet of plastic dividers, the area should be a square, making the width 8 feet.

Step-by-step explanation:

The question pertains to maximizing the area of a rectangle where the total perimeter is fixed at 32 feet. In order to maximize the area of a rectangle with a given perimeter, the rectangle must be a square.

Since the perimeter P of a rectangle is given by P = 2l + 2w, where l is the length and w is the width, we can set up the equation 32 = 2l + 2w for this problem.

To find the width when the rectangle is actually a square, we can use the fact that in a square, all sides are equal (l = w). Thus, we can rewrite the equation as 32 = 4w. Solving for w, we find that w = 8 feet.

Therefore, the width of the fenced-in area should be 8 feet.