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? HELP PLS.
The maximum area is always a square.
Divide the 2km/3 since they only need it for 3 sides.
2/3km on each side.
To find the maximum parking area that can be created with 2 km of fencing, we need to determine the dimensions of the rectangular area.
Let's assume the length of the rectangular area is L km, and the width is W km.
According to the given information, one length will be the back wall of the dealership. This means that one side of the rectangular area is already known, which is L km.
To create the rectangular area and maximize the parking area, we can use the remaining fencing to construct the other three sides, which will be the two widths and one additional length.
The total length of the three remaining sides is W + W + L = 2W + L km.
Since the total length of the remaining sides is 2 km, we have the equation: 2W + L = 2.
Now, let's solve this equation to find the relationship between L and W.
1. Substitute the value of L (L = 2 - 2W) into the equation:
2W + (2 - 2W) = 2
2W + 2 - 2W = 2
2 = 2
This equation is always true, which means that we have an infinite number of solutions. This indicates that there is no constraint on the relationship between L and W.
2. Calculate the parking area:
Area = Length × Width
Area = L × W (since the length is fixed as the back wall of the dealership)
Since we want to maximize the parking area, we need to find the maximum value of L × W.
Since there is no constraint on the relationship between L and W, we can choose any values for L and W that sum up to 2. As a result, the maximum parking area can be achieved when L = 1 km and W = 1 km.
Therefore, the maximum parking area that can be created is:
Area = L × W = 1 km × 1 km = 1 km².
So, the maximum parking area that can be created with 2 km of fencing is 1 km².
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 2 km of fencing.
Let's assume the back wall of the dealership is the longer side of the rectangle, denoted by the length (L), and the remaining sides are the width (W).
According to the problem, the sum of all four sides of the rectangle, which includes the back wall of the dealership, is equal to 2 km. This can be represented as:
L + 2W + L = 2
Rearranging the equation, we can simplify it to:
2L + 2W = 2
Now, divide the equation by 2 to get:
L + W = 1
We have two variables and only one equation, so we can't find the exact values of L and W. However, we can use calculus to find the maximum area.
To find the maximum area, we need an equation for the area of the rectangle, which is given by:
A = L * W
Now, we can use the equation L + W = 1 to substitute for either L or W. Let's solve for L:
L = 1 - W
Substituting this into the area equation, we get:
A = (1 - W) * W
Expanding and simplifying further:
A = W - W^2
To maximize the area, take the derivative of A with respect to W and set it equal to zero:
dA/dW = 1 - 2W = 0
Solving for W:
1 - 2W = 0
2W = 1
W = 1/2
Substituting this back into the equation L + W = 1, we get:
L + 1/2 = 1
L = 1/2
So, the dimensions of the rectangular area that maximize the parking area are L = 1/2 km and W = 1/2 km.
Now, we can find the maximum parking area by substituting these values into the area equation:
A = (1/2 km) * (1/2 km)
A = 1/4 km^2
Therefore, the maximum parking area that can be created with 2 km of fencing is 1/4 km^2.