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?
The maximum would be closest to a square.
2000m/3 = 667m
677^2 = ?
To find the maximum parking area that can be created with 2 km of fencing, we need to consider the shape of the parking area.
Let's assume the length of the back wall of the dealership is equal to L km. We can denote the width of the parking area as W km.
From the given information, we know that the total length of the fencing is 2 km. The back wall of the dealership will require L km of fencing, leaving us with 2 - L km of fencing for the remaining three sides.
Since the parking area is rectangular, it will have two equal sides (W) and one side (L) equal to the length of the back wall. Thus, the remaining three sides will be (W + W + L).
We can now set up an equation to represent the perimeter of the parking area:
2 - L + W + W + L = 2
Simplifying the equation:
2W = 2 - 2L
W = (2 - 2L) / 2
W = 1 - L
Now, we need to maximize the parking area, which is given by the product of length and width:
Area = L * (1 - L)
To find the maximum parking area, we need to find the maximum value of this quadratic equation.
To do that, we can find the vertex of the quadratic equation using the formula: x = -b / (2a)
In our case, a = -1, and b = 1:
L = -1 / (2 * -1) = 1/2
So, the length that maximizes the parking area is L = 1/2 km.
Substituting this value back into the equation for width:
W = 1 - L
W = 1 - 1/2
W = 1/2 km
Therefore, the maximum parking area that can be created with 2 km of fencing is given by:
Area = L * W
Area = (1/2 km) * (1/2 km)
Area = 1/4 km²
So, the maximum parking area that can be created is 1/4 square kilometer.