A car dealership fences in a rectangular
area behind their building to secure
unsold vehicles. One length will be
the back wall of the dealership. What is
the maximum parking area that can
be created if they have 2 km of fencing
to use?
2x+y = 2000
a = xy = x(2000-2x)
da/dx = 2000-4x
da/dx=0 when x=500
So, the maximum area is 500*1000 m^2
Note that this is achieved when the available fence is equally divided between lengths and widths.
To find the maximum parking area that can be created, we need to determine the dimensions of the rectangular area using the given 2 km of fencing.
Let's assume the back wall of the dealership is the longer side of the rectangular area. Let's call the length of the back wall L and the width of the rectangular area W.
Since the back wall is one length, the total length of the fencing required for the back and front walls is 2L.
The remaining fencing will be used for the two side walls, so the total length of the fencing required for the side walls is 2W.
According to the problem, we have a total of 2 km of fencing, which is equal to 2000 meters. Therefore, we can write this as an equation:
2L + 2W = 2000 (equation 1)
We want to maximize the parking area, which is given by the product of the length and width, A = L * W.
To find the maximum area, we can express one variable in terms of the other from equation 1 and substitute it into the area formula.
Rearranging equation 1:
2L = 2000 - 2W
L = (2000 - 2W)/2
L = 1000 - W
Substituting L = 1000 - W into the area equation:
A = (1000 - W) * W
Expanding the equation:
A = 1000W - W^2
To maximize the area, we need to find the maximum value of A. This can be done by finding the vertex of the parabolic equation.
The vertex of the parabola y = ax^2 + bx + c can be found using the formula:
x = -b / 2a
In our case, a = -1, b = 1000, and c = 0.
Using the formula, we get:
W = -1000 / (2 * (-1))
W = -1000 / -2
W = 500
Thus, the width of the rectangular area that will provide the maximum parking area is 500 meters.
Substituting this value back into equation 1 to find the length:
2L + 2W = 2000
2L + 2(500) = 2000
2L + 1000 = 2000
2L = 2000 - 1000
2L = 1000
L = 1000 / 2
L = 500
Therefore, the length of the rectangular area is also 500 meters.
The maximum parking area that can be created with 2 km of fencing is A = L * W = 500 * 500 = 250,000 square meters.
To find the maximum parking area that can be created, we need to determine the dimensions of the rectangular area that can be fenced in with the given amount of fencing.
Let's say the length of the rectangular area is L, and the width of the rectangular area is W.
We are told that one of the lengths will be the back wall of the dealership. This means that one side of the rectangle will be fixed, which is L.
Since the fencing material is used to enclose all sides of the rectangle, we can calculate the total length of the fencing required as follows:
Total fencing required = L (back wall) + 2W (two sides) = 2 km
From the given information, we know that the total fencing available is 2 km.
Now, we have a system of equations:
L + 2W = 2 km (Equation 1)
L = (2 km - 2W) (Rearranged Equation 1)
We want to maximize the parking area, which is given by the formula: Area = L * W.
Substituting the value of L from the rearranged Equation 1 into the area formula, we get:
Area = (2 km - 2W) * W = 2W - 2W^2
To find the maximum area, we need to find the value of W that maximizes the equation 2W - 2W^2.
The maximum value of a quadratic equation of the form ax^2 + bx + c is achieved at the vertex, which occurs when x = -b/2a.
In this case, a = -2, b = 2, and c = 0.
Applying the formula to find the value of W at the maximum area:
W = -b/2a = -2/(2*-2) = -2/(-4) = 1/2 km = 500 meters
Substituting the value of W back into Equation 1, we can find the value of L:
L = 2 km - 2W = 2 km - 2*(1/2) km = 1 km
Therefore, the dimensions of the rectangular area that will maximize the parking area are 1 km by 500 meters.
To find the maximum parking area, we substitute the values of L and W into the area formula:
Area = L * W = 1 km * 500 m = 500,000 square meters.
So, the maximum parking area that can be created with 2 km of fencing is 500,000 square meters.