A college is planning to construct a retangular parking lot on land bordered on one side by a highway. The plan is to use 680 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?
To find the dimensions of the rectangular parking lot that will enclose the maximum area with 680 feet of fencing, we can apply optimization techniques.
Let the length of the parking lot be L and the width be W. Since there are three sides to be fenced (the two widths and one length), we have:
2W + L = 680 ...(1)
The area of the rectangular parking lot is given by A = LW.
To maximize the area, we need to express the area in terms of just one variable, either L or W. We can do this by solving equation (1) for L:
L = 680 - 2W
Substituting this value into the area equation, we have:
A = W(680 - 2W)
A = 680W - 2W²
To maximize A, we can take the derivative of A with respect to W, set it equal to zero, and solve for W:
dA/dW = 680 - 4W = 0
4W = 680
W = 170
Now we can substitute the value of W back into equation (1) to find L:
2(170) + L = 680
L = 680 - 340
L = 340
Therefore, the dimensions of the rectangular parking lot that will enclose the maximum area are 170 feet by 340 feet.
To find the dimensions of the rectangular parking lot that will maximize the enclosed area, we can use the concept of optimization.
Let's assume the length of the parking lot is L (in feet) and the width is W (in feet). Since the parking lot is bordered on one side by a highway, the highway side doesn't require any fencing. The other three sides will be fenced, which will use up the 680 feet of fencing available.
Given that the perimeter of a rectangle is calculated by adding up all its side lengths, we can determine the perimeter of the parking lot as follows:
Perimeter = 2L + W
Since the available fencing is 680 feet, we can write the equation:
2L + W = 680 .....(Equation 1)
Now, let's consider the area of the parking lot. The area of a rectangle is calculated by multiplying its length by its width:
Area = L * W
In order to maximize the enclosed area, we need to find the maximum value of Area. To do this, we can solve for one variable in Equation 1 and substitute it into the area equation.
Rearrange Equation 1 to solve for W:
W = 680 - 2L
Substitute this value of W into the area equation:
Area = L * (680 - 2L)
Area = 680L - 2L^2
Now we have the area in terms of L alone. To find the maximum area, we can take the derivative of the area equation with respect to L, set it equal to zero, and solve for L.
d(Area)/dL = 680 - 4L
Setting the derivative equal to zero:
680 - 4L = 0
4L = 680
L = 170
So, L = 170. We can substitute this value back into Equation 1 to find the width:
2L + W = 680
2 * 170 + W = 680
340 + W = 680
W = 340
Therefore, the dimensions that maximize the enclosed area would be a length of 170 feet and a width of 340 feet.
Let the side parallel to the highway be y ft
let each of the other two sides be x ft
So 2x + y = 680
y = 680-2x
area = xy = x(680-2x) = 680x - 2x^2
which is a parabola opening downwards, and you want its vertex.
Since you are doing this type of question, you must have learned how to find the vertex.
Go for it.