A college is planning to construct a rectangular parking lot on land bordered on one side by a highway. The plan is to use 400 feet of fencing to fence off the other three sides. What dimensions should the lot have if the enclosed area is to be a maximum?
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Part 1
The width is enter your response here feet. The length is enter your response here feet. (Let the longer side be the length and the shorter side the width.)
To find the dimensions that will maximize the enclosed area, we can use the formula for the area of a rectangle: A = length * width.
Let's assume the width is x feet. The length, which is the side bordering the highway, will then be 400 - 2x feet (since there are three sides to fence off and 400 feet of fencing total).
Now we can substitute these values into the formula for the area:
A = (400 - 2x) * x
A = 400x - 2x^2
To find the maximum area, we can take the derivative of this function and set it equal to zero:
dA/dx = 400 - 4x
400 - 4x = 0
4x = 400
x = 100
Therefore, the width of the parking lot should be 100 feet. The length, which is the side bordering the highway, will be 400 - 2(100) = 400 - 200 = 200 feet.
So, the dimensions for the lot that will maximize the enclosed area are: width = 100 feet and length = 200 feet.
To find the dimensions that will maximize the enclosed area, we need to write an equation for the perimeter of the parking lot using the given information.
Let's assume the width of the parking lot is 'w' feet and the length is 'l' feet.
The perimeter of the parking lot is the sum of the lengths of all sides, which consists of three sides (width, length, and width).
Given that the total fencing available is 400 feet, we can set up the equation:
2w + l = 400
We can rearrange the equation to solve for l:
l = 400 - 2w
Now, we need to find the area of the parking lot. The area is equal to the product of the width and length:
Area = w * l
Substituting the value of l from the previous equation:
Area = w * (400 - 2w)
To maximize the enclosed area, we need to find the maximum value of this equation. We can do this by finding the vertex of the parabola formed by this equation.
The vertex of a parabola with the equation ax² + bx + c is given by (-b/2a, f(-b/2a)), where f(x) is the equation of the parabola.
In our case, a = -2, b = 400, and c = 0.
The width at the vertex will give us the width that maximizes the area. So, we can find the width by using the formula:
width at vertex = -b/2a
width at vertex = -(400) / (2 * -2)
width at vertex = 100 feet
Now, substitute this value back into the equation for the length to find the length at the vertex:
l = 400 - 2w
l = 400 - 2(100)
l = 400 - 200
l = 200 feet
Therefore, the width of the parking lot is 100 feet, and the length is 200 feet.
To find the dimensions of the parking lot that would maximize its enclosed area, we can use the concept of optimization.
Let's denote the width of the parking lot as "w" and the length of the parking lot as "l". We are given that the total amount of fencing available is 400 feet, which will be used for three sides of the parking lot (excluding the side bordered by the highway).
Since the fencing will be used for three sides, we have the equation: 2w + l = 400. Rearranging this equation, we get: l = 400 - 2w.
The area of a rectangular shape is given by the formula: A = length × width. Therefore, we need to maximize the function A = lw.
To find the maximum value of A, we can substitute the expression for l into the equation for A: A = w(400 - 2w) = 400w - 2w^2.
Now, we need to find the critical point of this quadratic function to find the maximum. To do this, we take the derivative of the function: dA/dw = 400 - 4w.
Setting dA/dw equal to zero and solving for w, we get: 400 - 4w = 0, 4w = 400, w = 100.
Therefore, the width of the parking lot should be 100 feet.
Substituting this value back into the equation for l, we can find the length: l = 400 - 2w = 400 - 2(100) = 200.
Thus, the width of the parking lot should be 100 feet and the length should be 200 feet to maximize the enclosed area.