Question

A rectangular parking lot must have a perimeter of 440 feet and an area of at least 8000 square feet. Describe the possible lengths of the parking lot.

Answer 1

To determine the possible lengths, use the perimeter equation 2l + 2w = 440 to get l + w = 220. For the area, the inequality l × w ≥ 8000 must be satisfied. Lengths can vary so long as the product with corresponding width meets the area requirement.

Step-by-step explanation:

To determine the possible lengths of a parking lot with a perimeter of 440 feet and an area of at least 8000 square feet, we can set up two equations based on the given information.

The perimeter (P) of a rectangle is given by P = 2l + 2w, where l is the length and w is the width of the rectangle. In this case, P = 440, so we have 2l + 2w = 440. Simplifying this equation, we get l + w = 220.

The area (A) of a rectangle is given by A = l × w. We are told the area must be at least 8000 square feet, so we have l × w ≥ 8000.

Converting the perimeter equation in terms of width, we can express the width as w = 220 - l. Substituting this into the area inequality gives us l × (220 - l) ≥ 8000. This is a quadratic inequality in terms of l, which can be solved to find the range of possible lengths the parking lot can have while still meeting the conditions.

However, without completing the square or using the quadratic formula, we cannot provide specific lengths. For an area of at least 8000 square feet and a fixed perimeter of 440 feet, lengths and widths can vary. The values for l will be such that when multiplied by its corresponding width (220 - l), it results in an area that is equal to or greater than 8000 square feet.

Answer 2

The possible parking lengths are 45.96 feet and 174.031 feet

Step-by-step explanation:

Let x be the length of rectangular plot and y be the breadth of rectangular plot

A rectangular parking lot must have a perimeter of 440 feet

Perimeter of rectangular plot =2(l+b)=2(x+y)=440

2(x+y)=440

x+y=220

y=220-x

We are also given that an area of at least 8000 square feet.

So,
`xy \leq 8000`

So,
`x(220-x) \leq 8000`

`220x-x^2 \leq 8000`

So,
`220x-x^2 = 8000\\-x^2+220x-8000=0`

General quadratic equation :
`ax^2+bx+c=0`

Formula :
`x=(-b \pm √(b^2-4ac))/(2a)`

`x=(-220 \pm √(220^2-4(-1)(-8000)))/(2(-1))\\x=(-220 + √(220^2-4(-1)(-8000)))/(2(-1)) , (-220 - √(220^2-4(-1)(-8000)))/(2(-1))\\x=45.96,174.031`

So, The possible parking lengths are 45.96 feet and 174.031 feet