As expected, the the second part of this problem can be solved using the strategy applied in Part I, although there are some minor differences, as demonstrated below.

Part II: A Corral Adjacent to a Barn

First, we must assess the implications of using a barn as one side of the corral:

- 120 meters of fencing will be shared among three sides of the rectangle.
- Logically, the barn should be used on one of the longer sides of the corral, in order to maximize the length of fencing remaining for the other sides.
- In order to calculate the width of the rectangle for a given length, we will use the equation w=(120-l)/2.
- The longest possible length for one side of the fence (in whole meters) is 118 meters, so that each remaining side has a width of 1 meter. It reasons that the list of possible lengths can be calculated by subtracting 2 from the previous length.

With these considerations, we can follow the steps listed in Part I to generate the following data table beside the table we created in Part I. (Notice the difference in equations used in the Length and Width columns.)

As seen with the highlighted cells above, analyzation of the data reveals that the largest dimensions for the corral adjacent to a barn is 60m x 30m, which yields 1800 square meters of area.

Next, we can plot the coordinates (length, area) taken from this data table using Desmos. As before, we notice that the points lie on a parabolic curve. Use the second page of the Desmos activity to find the equation of the curve that passes through the given points.

Analytically, we can use our knowledge that the graph opens down and the hypothetical x-intercepts of (0,0) and (120,0) to find the equation of the parabola. Using the method described in Part I, the first equation we might try is y=-x(x-120). However, when we graph this equation, we see that the vertex of this parabola is located at (60, 3600), which is twice the height of the vertex we want, (60, 1800). Intuitively, we adjust our equation as follows:

y=(-1/2)x^2(x-120)

As shown below, this equation passes neatly through the given coordinates.

Like before, we use the spreadsheet software to create the following graph of the coordinates, which verifies our equation.

As in Part I, our graphs verify our conclusion that the largest possible corral constructed next to a barn is 60m x 30 m with an area of 1800 square meters.

Comparing the graphs from Part I and Part II, we can see that the y-value of the vertex (60, 1800) of the second parabola is twice the y-value of the vertex (30, 900) of the first parabola. Correspondingly, we conclude that building the corral adjacent to a barn results in a corral with an area twice as big as the first corral.

Continue to the next page to see how this problem can be incorporated into the classroom.