AP Calculus AB Exam Question: Optimization Problem

A farmer has 2000 meters of fencing to enclose a rectangular field. One side of the field is against a river, so the farmer only needs to put up three sides of fencing. What are the dimensions of the field that maximize the area, and what is the maximum area that can be enclosed?

Step-by-Step Solution:

Let's assume the length of the rectangular field is represented by "x" meters, and the width is represented by "y" meters.

To maximize the area, we need to set up an equation for the area in terms of "x" and "y". Since the farmer has 2000 meters of fencing available and only needs to fence three sides, the total length of fencing required is twice the width (since the width is against the river).

The equation for the perimeter of the field, using the length and width, is given by:

2x + y = 2000 ...............(1)

Now, let's solve equation (1) for "y" in terms of "x":

y = 2000 - 2x

The area of a rectangle is given by the product of its length and width. So, the equation for the area "A" of the field is:

A = xy

Substituting the value of "y" from equation (1) into the equation for the area, we have:

A = x(2000 - 2x) A = 2000x - 2x^2

Now, our goal is to find the maximum value of "A". To do this, we need to find the critical points of the function. Taking the derivative of "A" with respect to "x" and setting it equal to zero, we have:

A' = 2000 - 4x = 0 4x = 2000 x = 500

We have found a critical point when x = 500. To confirm if this is a maximum, we can take the second derivative:

A'' = -4

Since the second derivative is negative, it confirms that we have a maximum. Thus, the dimensions of the field that maximize the area are:

Length (x) = 500 meters Width (y) = 2000 - 2x = 2000 - 2(500) = 1000 meters

Substituting these values into the equation for the area, we can find the maximum area:

A = xy A = (500)(1000) A = 500000

Therefore, the maximum area that can be enclosed is 500,000 square meters.

Answer:

The dimensions of the rectangular field that maximize the area are: Length = 500 meters Width = 1000 meters

The maximum area that can be enclosed is 500,000 square meters.