a developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 336 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed and what are the dimensions that give this maximum area?
To find the largest area that can be enclosed, we can use the concept of optimization. Let's begin by assigning variables to the dimensions of the rectangular lot.
Let's assume that the length of the lot is L and the width is W.
The perimeter of the lot, excluding the side along the street, is given as 336 feet.
Considering that the side along the street is not fenced, we have two widths and one length contributing to the perimeter:
2W + L = 336
Now, let's solve this equation for L:
L = 336 - 2W
To find the area of the lot, we multiply the length by the width:
A = L * W
Substituting the value of L from above:
A = (336 - 2W) * W
Expanding this equation:
A = 336W - 2W^2
To find the maximum area, we need to find the value of W that maximizes the above equation. We can achieve this by using calculus.
Taking the derivative of A with respect to W and setting it equal to 0 to find the critical point:
dA/dW = 336 - 4W = 0
Solving for W:
336 - 4W = 0
4W = 336
W = 84
Now that we have the value of W, we can substitute it back into our equation for L:
L = 336 - 2W
L = 336 - 2(84)
L = 336 - 168
L = 168
So, the dimensions that will give the maximum area are a width of 84 feet and a length of 168 feet.
To find the maximum area, substitute these values back into the area equation:
A = (336 - 2W) * W
A = (336 - 2(84)) * 84
A = 168 * 84
A = 14,112 square feet
Therefore, the largest area that can be enclosed is 14,112 square feet, with dimensions of 84 feet by 168 feet.
To determine the largest possible area that can be enclosed, we need to find the dimensions of the rectangular lot. Let's denote the length of the lot as L and the width as W.
We are told that the overall perimeter of the lot, excluding the side along the street, is 336 feet. Since the two widths of the rectangle are adjacent to the street and are not fenced, we can deduce that the perimeter equation is as follows:
2L + W = 336.
To find the largest area, we need to find the dimensions that maximize the area formula A = L * W.
First, let's solve the perimeter equation for L:
2L = 336 - W.
Dividing both sides by 2, we get:
L = (336 - W) / 2.
Next, substitute this expression for L in the area formula:
A = L * W.
A = [(336 - W) / 2] * W.
Simplifying further:
A = (336W - W^2) / 2.
Now, we have the area formula in terms of W. To find the maximum possible area, we can take the derivative of A with respect to W, set it to zero, and solve for W.
dA / dW = 336/2 - 2W = 0.
168 - 2W = 0.
-2W = -168.
W = 84.
Now that we have the value of W, we can substitute it back into the equation for L:
L = (336 - W) / 2.
L = (336 - 84) / 2.
L = 252 / 2.
L = 126.
Therefore, the dimensions that yield the maximum area are a width of 84 feet and a length of 126 feet.
To find the maximum area, substitute these values into the area formula:
A = L * W.
A = 126 * 84.
A = 10,584 square feet.
So, the largest area that can be enclosed is 10,584 square feet with dimensions of 126 feet by 84 feet.