A farmer wants to divide a rectangular plot into five equal strips using 648 feet of fence. (A) draw a labeled diagram of the plot. (B) find the dimensions of the plot that give the maximum area and the maximum area. (C) state why the area is a maximum
We cannot draw a diagram on these posts. However, the maximum are would be closest to a square.
(A) To draw a labeled diagram of the plot, we can start by creating a rectangle and dividing it into five equal strips. Here's a representation:
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(B) Now, let's find the dimensions of the plot that give the maximum area and calculate the maximum area.
Let's assume the length of the rectangle is L and the width of each strip is W. Since we need to divide the plot into five equal strips, each strip will have a width of W, and the remaining length after the strips will be L - 4W.
Using the given information that we have 648 feet of fence, we can set up an equation to represent the total amount of fence used:
2L + 6W = 648
Simplifying the equation:
2L + 6W = 648
L + 3W = 324
Now, let's express the area of the plot in terms of L and W:
Area = L * W
We can rewrite this equation in terms of a single variable since we already have an equation relating L and W:
Area = (324 - 3W) * W
To find the maximum area, we can differentiate the area equation with respect to W and set it to 0 to find critical points:
d(Area)/dW = (324 - 3W) - 3W
d(Area)/dW = 324 - 6W
Setting d(Area)/dW = 0:
324 - 6W = 0
6W = 324
W = 54
Now, substitute the value of W back into the equation L + 3W = 324:
L + 3(54) = 324
L + 162 = 324
L = 324 - 162
L = 162
Therefore, the dimensions of the plot that give the maximum area are a length of 162 feet and a width of 54 feet, and the maximum area is given by:
Area = L * W = 162 * 54 = 8,748 square feet.
(C) The area is a maximum at these dimensions because we found the maximum by optimizing the area equation. By differentiating the area equation with respect to the width and setting it to 0, we found the critical point where the rate of change of the area with respect to the width is zero. At this point, the area reaches its maximum value.