In order to solve the problem posed on the previous page, students must first identify the information given in the problem and determine what they need to find. After analyzing the problem, students can infer the following:
- The 120 meters of fencing represents the distance of the perimeter of the corral.
- The largest corral will be the corral with the greatest area.
- The dimensions of the corral will be given in the format Length (m) x Width (m).
Next, students should assess what they already know:
- The perimeter of a rectangle equals two times the length plus two times the width (P=2l+2w).
- The area of a rectangle equals the length times the width (A=lw), and the area involved in this problem will be measured in square meters.
Having analyzed the problem and deduced what information they need to find, students can then make a solving strategy in which they choose which technological tool(s) will most effectively aid in solving. In this analysis, we will examine a solving strategy in which a student uses a spreadsheet to calculate and organize possible solutions and then uses Desmos to verify which parameters will result in a corral with the greatest area.
Part I: A Corral Constructed with 120 Meters of Fencing
The first part of the problem asks us to consider a rectangular corral with a perimeter equal to 120 meters. In order to calculate possible lengths, widths, and areas of a rectangle that has a perimeter of 120 meters, we will utilize spreadsheet software according to the following steps:
- Label respective columns A1, B1, C1: Length (m), Width (m), and Area (m^2)
- In the Length column, enter the value 30 in the first cell (in this example, A2), since 120 meters of fencing divided equally among four sides of the rectangle allots 30 meters to each side.
- In cell A3, enter this formula: =A2-1.
- Copy and paste this formula down the Length column in order to generate a list of possible lengths, from 30 meters to 1 meter (we will only enter whole meters for simplicity).
- In the first cell (B2) of the Width column, enter this formula*: =(120-(2*A2))/2.
- Copy and paste this formula down the Width column in order to calculate widths of the rectangle which correspond to the possible lengths.
- In the first cell (C2) in the Area column, enter this formula: =A2*B2.
- Copy and paste this formula down the Area column in order to calculate the area corresponding to the various length and width combinations of the rectangle.
*To derive this formula, we reason that two times the length (A2) of the rectangle must be subtracted from the perimeter (120) in order to get the distance of fence remaining, which must be divided equally between the two wide sides of the rectangle.
It is important to recognize that “Length” and “Width” are interchangeable in terms of calculating area, since A=lw=wl. Thus, enumerating lengths of 31-59 in our spreadsheet is unnecessary because these cases are addressed in the Width column.
Examples of the above steps can be viewed below:
After generating a table of values, we can analyze the data in order to determine which dimensions result in the greatest area. As seen above in the highlighted cells, the size of the corral which results in the greatest area is a 30m x 30m corral, which has an area of 900 meters squared.
In order to verify this solution, we will next graph coordinates taken from our spreadsheet. The x-values will correspond to the length of the rectangle, and the y-value will correspond to the area, so our coordinates will be in the form (length, area). After plotting these coordinates, we notice that the points lie along a parabolic curve. Try to find the equation of the parabola that passes through the coordinates with the first page of this Desmos activity. Join the activity with the auto-filled code, then proceed with or without an account.
Applying the guess and check approach with the sliders in the above activity may or may not have resulted in determining the correct equation. However, it is possible to analytically find the equation of the parabola. In the activity, the two red points represent the x-intercepts of the parabola [(0,0), (60,0)]. The points are red because they do not actually exist in our data set (there is no area of zero), but the x-intercepts are important for determining the equation of the parabola. Our knowledge of parabolic equations tells us that for the parabola to open downwards, the leading coefficient must be negative. We also know that when the equation is factored, the value of the equation equals zero when at least one of its factors equal zero. So, based on our x-intercepts [(0,0), (60,0)] and the fact that the graph opens down, we can figure out the following equation:
As shown below, this equation equals zero when x=0 or x=60, and it indeed passes through the given coordinates.
The spreadsheet software can also be used to verify this equation. Using the graphing feature of the software, we can create a graph on which the coordinates are plotted. We will then have the software draw a line of best fit through the points and show the equation for that curve, as seen below.
Now we have the opportunity to make connections across the Desmos representation and the graph above. The spreadsheet software shows this equation for the parabola:
y= -x^2 + 60x + 1.079E-12
At first there appears to be a discrepancy between this and the equation on our Desmos graph, but upon closer inspection, we recognize that the value 1.079E-12 is extremely close to zero and therefore negligible. Further, we can distribute the equation found with Desmos so that both equations are in the form:
y= -x^2 + 60x
Having confirmed that our data lies on a parabolic curve, we see that the maximum area of the corral is represented by the peak of the parabola at the coordinate (30,900), which corresponds to the solution we determined by analyzing the data in our spreadsheet. This is significant because it proves that no intermediate lengths between 29 and 30 or 30 and 31 produce an area greater than 900 square meters.
Continue to the next page for Part II: A Corral Adjacent to a Barn.