If a man is fencing in a rectangular lot of grass next to the road and doesnt want to fence the side touching the road and has 248 feet of fence, whats the maximum area he can fence?
If we use h and w for length and width, with length parallel to the road,
perimeter p = 2w+h = 248
Area = w*h = w(248-2w) = 248w - 2w^2
We want to maximize the area, so we take the derivative and set it to zero.
248-4w = 0
w = 62
So, h = 124
Maximum area = 62*124 = 7688 sq ft
For a given perimeter, a square has maximum area. Here, we use the road as one side, so we get to enclose two squares instead of one.
To find the maximum area he can fence, we need to consider that one side of the rectangle is already fenced, leaving three sides to enclose the maximum area. Let's denote the length of the side parallel to the road as 'x' and the width as 'y'.
Given that the total fence length available is 248 feet, we can set up the equation:
2x + y = 248
This equation represents the sum of the three sides of the rectangle: two sides of length 'x' and one side of length 'y'.
Now, we need to express 'y' in terms of 'x' to get the area. Solving the equation for 'y', we have:
y = 248 - 2x
The area of the rectangle, A, is given by:
A = x * y
Substituting the value of 'y' obtained above, we have:
A = x * (248 - 2x)
To maximize the area, we can take the derivative of the area equation with respect to 'x' and set it equal to zero:
dA/dx = 0
Let's find the derivative:
dA/dx = 248 - 4x
Setting it equal to zero and solving for 'x':
248 - 4x = 0
4x = 248
x = 248/4
x = 62
Now that we have the value of 'x', we can substitute it back into the equation for 'y' to find the corresponding value:
y = 248 - 2x
y = 248 - 2*62
y = 248 - 124
y = 124
Therefore, the dimensions of the rectangle that maximize the area are 62 feet by 124 feet.
To find the maximum area, we can substitute these values back into the area formula:
A = x * y
A = 62 * 124
A = 7688 square feet
Hence, the maximum area he can fence is 7688 square feet.