The Biggest Possible Area
Lesson 3 of 7
Objective: SWBAT determine maximum rectangular area for a given perimeter using graphical methods.
In class over the last couple of days, the students learned how length and area change as a function of width (when perimeter is constant). In this lesson we will focus on obtaining the maximum area and making connections with the graph of the length-area relation.
I begin by telling students that this time they have exactly 28 feet of fencing to build a rectangular garden. Their job is to maximize the area contained within the fence. I hand out the worksheet, Garden Sketches and Table. All students need to do here is finish the sketches of possible rectangular gardens and complete the table. I expect that students will have little difficulty with this task.
Once each pair of students is done completing the table on the worksheet, I ask if anyone recognizes a pattern in the table: Was there a pattern that made it easier to complete the table?
If necessary, I ask these followup questions:
- Do you see any pattern in the area values as the length changes by one?
- What is the sum of the length and width pairs? What does this sum represent?
- What is the product of the length and width pairs? What does this product represent?
- What information do we now have that can help you complete the table faster?
Then, I ask each pair of students to go to Desmos.com and open the graphing calculator by clicking "Launch Calculator." Students should then click the dropdown (plus sign) in top left hand corner and select "table". They should then enter all the length-area table values and watch the curve being plotted on main window. Students should see the parabola being formed as they enter the coordinates. I then ask these questions:
- What is the largest area that can be formed with 28 feet of fence? Where are these coordinates located on the graph. (students can place their pointer on this point and see the coordinates pop up. In this case the vertex point will read (7, 49)
- What kind of rectangle will the garden have? (square)
- Find the x intercepts. What do these coordinates mean again? Are they feasible dimensions?
Lastly, I ask students to use the Desmos calculator and do the same with a different perimeter, say 60 feet of fencing. This time the students should be able to work more quickly. I usually get learners eagerly wanting to express what they are discovering as they graph and determine the maximum area with a different perimeter.
Once they are investigating other possible areas, I plan to gather some of their comments and write these on the board. I expect to hear comments like:
- The graph is again a parabola
- The parabola again opens down
- The maximum area is at the vertex again
- The garden is a square again
The next obvious thing I do is ask students to explain why these comments are true, leading into the closure section of the lesson.
To end this lesson I want students to summarize and make generalizations on their work. I hand each student a set of Closing questions to answer. I allow them to discuss their answers with their partner.
If time permits, I ask student volunteers to share some of their answers. Either way, I collect each paper as students leave so I can read and obtain information on their grasp of the main points of the lesson.
Suppose you had to make a partition in the middle of the garden dividing it into two equal rectangular parts. This partition would be the same width as the garden (see figure below). However, you must only use your original 28 feet of fencing for the garden, including this middle partition.
Now, find the largest possible area of the entire garden using your available 28 feet of fencing.
As a start, I ask students to make a chart of their work, like they did in the first task in the Launch section. I also encourage them to use the graphing capabilities of Desmos to explore a solution.
Determining the largest area may be challenging here because the maximum does not have whole number coordinates. Therefore, students cannot visually determine the coordinates of the vertex. Many students will employ symmetry or use their calculators to find it, which is great. If they seem to be struggling, hint, asking students to consider the x-intercepts. This may be enough of a tip. In this case the maximum area should be approximately 32.66 sq. ft. with a partition of 4.66 feet and a length of 7 feet.
If students don't realize that the length is 7 feet, ask about this. What does this imply? I would probe here for a conjecture. Expect some students to want to go further and make two partitions with the 28 feet of fencing, instead of one.
With a length of 7 and two partitions, what is the greatest possible area?