A developer wants to enclose a rectangular grassy lot that borders a city street for parking. If the developer has 212 feet of fencing and does not fence the side along the street, what is the largest area that can be enclosed?
so you have one length and two widths
width -- x
length --- y
but y + 2x = 212
y = 212-2x
area = xy = x(212-2x) = -2x^2 + 212x
complete the square
area = -2(x^2 - 106x + 2809-2809)
= -2(x-53)^2 + 5618
the largest area is 5618 ft^2 when x = 53 and y= 106
check: 53x106 = 5618
one x lower: 52 x 108 =5616 , smaller
one x higher : 54 x 104 = 5616 smaller
my answer is correct.
Well, let's see here. The developer has 212 feet of fencing and doesn't need to fence the side along the street. So, that means the developer needs to fence the remaining three sides of the rectangular lot.
Now, since we want to maximize the area enclosed, we should aim for the most symmetrical shape possible. That would be a square! A square has equal length sides, so by dividing the remaining 212 feet by 3, we can find the length of each side.
212 feet divided by 3 equals approximately 70.67 feet. So, each side of the square would be roughly 70.67 feet.
Now, to find the area of the square, we simply square the length of one side. 70.67 feet squared is approximately 5,000 square feet.
Therefore, the largest area that can be enclosed is approximately 5,000 square feet. There you have it – the "square" deal!
To find the largest possible area that can be enclosed, we need to determine the dimensions of the rectangular lot.
Let's assume that the length of the lot is L and the width is W.
Given that the developer has 212 feet of fencing, we can form the equation:
2L + W = 212
Since the developer does not need to fence the side along the street, the perimeter of the rectangular lot will be:
2L + W + W = 212
2L + 2W = 212
This can be simplified to:
L + W = 106
Solving this equation for L, we have:
L = 106 - W
Now, we can find the area of the rectangular lot using the formula:
Area = length × width
Area = L × W
Area = (106 - W ) × W
To maximize the area, we need to find the value of W that will give us the maximum result.
We can achieve this by taking the derivative of the area equation with respect to W and setting it equal to zero, to find the critical point:
d(Area) / d(W) = (106 - 2W) = 0
Solving this equation, we get:
106 - 2W = 0
2W = 106
W = 53
Now we can substitute this value of W back into the area equation:
Area = (106 - W) × W
Area = (106 - 53) × 53
Area = 53 × 53
Area = 2809 square feet
Therefore, the largest area that can be enclosed is 2809 square feet.
To find the largest area that can be enclosed, we need to determine the dimensions of the rectangular lot. Let's call the width of the lot 'w' and the length of the lot 'L'.
We know that the developer has a total of 212 feet of fencing to enclose the remaining three sides of the lot.
Since the side along the street is not fenced, we have the following equation for the perimeter of the lot:
2w + L = 212 <-- equation (1)
To find the largest area, we need to determine the dimensions of the rectangle. We can do this by solving equation (1) for L:
L = 212 - 2w
The area of a rectangle is given by the equation:
A = width (w) * length (L)
Substituting the value of L from equation (1) into this equation:
A = w * (212 - 2w)
We can simplify this equation by expanding it:
A = 212w - 2w^2
To find the largest area, we need to find the vertex of this quadratic equation, which represents the maximum point on the graph. The x-coordinate of the vertex will give us the value of 'w' that maximizes the area.
The x-coordinate of the vertex of a quadratic equation in the form ax^2 + bx + c = 0 is given by:
x = -b / (2a)
In our case, a = -2, b = 212, and c = 0.
Substituting these values into the equation, we get:
w = -(212) / (2 * (-2))
w = -212 / -4
w = 53 feet
Therefore, the width of the lot should be 53 feet to maximize the area.
Now, we can substitute this value back into equation (1) to find the length (L) of the lot:
2(53) + L = 212
106 + L = 212
L = 212 - 106
L = 106 feet
So, the dimensions of the rectangular lot that will give the largest area are 53 feet (width) and 106 feet (length).
Finally, to find the largest area, we substitute the width and length into the area equation:
A = width * length
A = 53 * 106
A = 5,618 square feet
Therefore, the largest area that can be enclosed is 5,618 square feet.