600 feet of fencing to enclose a rectangular plot
If I don't fence one side what is the length and width of the plot that maximized the area?
it is useful to know that a square is the figure of maximum area for a given perimeter.
Similarly, a cube is the figure of maximum volume for a given surface area.
Formally, if the width is x, then the length is (600-2x)/2 = 300-x
So, the area is given by
a = x(300-x) = 300x-x^2
da/dx = 300-2x
da/dx=0 at x = 150
so, the rectangle is 150x150
To find the dimensions of the plot that maximize the area given the constraint of using 600 feet of fencing and not fencing one side, we can use a optimization technique called calculus.
Let's consider the length of the plot as "L" and the width as "W."
Now, let's break down the information we have:
- We have 4 sides of the rectangular plot, 3 of which are fenced with a total of 600 feet of fencing.
- One side is not fenced, which means that side does not contribute to the perimeter (fencing length).
Since the total fencing length is 600 feet, the three fenced sides must add up to 600 feet:
2L + W = 600 -------- (Equation 1)
The area of a rectangle is given by length times width:
Area = L * W
We want to maximize the area. To do this, we need to express the area in terms of a single variable, either L or W. In this case, let's express it in terms of L.
Using Equation 1, we can rearrange it to express W in terms of L:
W = 600 - 2L
Substituting this value of W in the area equation, we get:
Area = L * (600 - 2L)
Area = 600L - 2L^2
Now, to find the maximum area, we need to differentiate the area equation with respect to L and set it equal to 0. So, let's differentiate the area equation:
d(Area)/dL = 600 - 4L
Setting this derivative equal to 0 and solving for L:
600 - 4L = 0
4L = 600
L = 600/4
L = 150
Now that we have the value of L, we can substitute it back into Equation 1 to find W:
2L + W = 600
2(150) + W = 600
300 + W = 600
W = 600 - 300
W = 300
Therefore, the length that maximizes the area is 150 feet, and the width is 300 feet.