A college is planning to construct a rectangular parking lot on land bordered on one side by a highway. The plan is to use 560 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?
Let's assume the length of the rectangular parking lot is x feet.
Then, the width of the rectangular parking lot will be (560 - 2x) / 2 feet, as there are two equal width sides and the other side is the highway.
The area of the rectangular parking lot is given by length x width, which is x * (560 - 2x) / 2 = (560x - 2x^2) / 2 = 280x - x^2.
To find the maximum area, we take the derivative of the area function and set it equal to zero.
The derivative of the area function is dA/dx = 280 - 2x.
Setting it equal to zero, we get 280 - 2x = 0, or x = 140.
Therefore, the length of the rectangular parking lot is 140 feet.
And the width of the rectangular parking lot is (560 - 2(140)) / 2 = 560 - 280 = 280 feet.
So, the dimensions of the lot that would result in a maximum area are 140 feet by 280 feet.
To find the dimensions of the rectangular parking lot that will maximize the enclosed area, we can apply the concept of optimization.
Let's assume the width of the parking lot is x feet. Since there are three sides that require fencing, the length of the parking lot will be (560 - 2x) feet.
The area (A) of the rectangular parking lot is given by multiplying its length and width: A = x(560 - 2x).
To maximize the area, we need to find the value of x that maximizes the equation A.
To find the maximum, we can take the derivative of A with respect to x and set it equal to zero. Then solve for x.
dA/dx = 560 - 4x
Setting dA/dx equal to zero:
560 - 4x = 0
4x = 560
x = 560/4
x = 140
Therefore, the width of the parking lot should be 140 feet.
Since the length is (560 - 2x), the length would be (560 - 2*140) = 280 feet.
So, the dimensions of the parking lot that will maximize the enclosed area are 140 feet for the width and 280 feet for the length.
To find the dimensions that would maximize the enclosed area of the parking lot, we can use the concept of optimization from calculus.
Let's assume that the length of the rectangular parking lot adjacent to the highway is 'x' feet. The width of the parking lot, which will have two sides each measured y feet, is unknown.
Given that the total fencing length is 560 feet, we know that the equation for the perimeter of the parking lot is:
Perimeter = 2x + y + y, which simplifies to:
560 = 2x + 2y
Now, let's solve the equation in terms of either x or y. We can rearrange it to solve for y:
y = 280 - x
The area of the rectangular parking lot can be calculated using the equation:
Area = Length × Width
Area = x × y
Area = xy
Since we have an equation for y in terms of x, we can substitute it into the area equation:
Area = x(280 - x)
Area = 280x - x^2
To maximize the area, we need to find the maximum value of the function Area = 280x - x^2.
To do that, we can take the derivative of the area equation with respect to x and set it equal to zero:
d(Area)/dx = 280 - 2x = 0
Solving this equation will give us the value of x that maximizes the area.
280 - 2x = 0
2x = 280
x = 140
Now that we have the value of x, we can substitute it back into the equation for y:
y = 280 - x
y = 280 - 140
y = 140
Therefore, the dimensions that maximize the enclosed area of the parking lot are 140 feet by 140 feet.