Problem: You walk into the new Herizon Wireless store, which just opened in the mall. They offer two different plans for voice (the data and text plans are separate): * The Unlimited Plan costs $100 per month for as many minutes as you care to use. * The Pay-Go Plan costs 30 cents per minute but has no monthly charge. Which types of customers should choose which plan? Introduction to the Solution: To solve this problem we need to calculate which plan is more cost efficient for a customer based on how many minutes they talk per month. To solve this problem mathematically, we first need to set up an equation for each plan. We can use the slope-intercept form, y = mx+b, as a model for each plan s equation. Since we are looking for the cost of each plan based off of how many minutes a customer talks per month, we will let the variable y equal the cost of each plan per month and let the variable x equal how many minutes a customer talks per month. For the Unlimited Plan, the cost is always $100 whether you talk 1,000,000 minutes or 0 minutes. For the Pay-Go Plan, you have no monthly charge but instead you pay 30 cents for each minute you talk. So in order to get the monthly cost on the Pay-Go plan we must multiply how many minutes we talk in a month by 30 cents (.30). So, Equation for Unlimited Plan: Equation for Pay-Go Plan: TI-Nspire Solution: Y Cost of plan & X Minutes talked Y = 100 Y =.30*X Now that we have the equations for both plans we must now find which plan is more cost efficient for how many minutes a customer may talk. One way we can calculate this is by creating a graph using the TI-Nspire CAS app.
First we create a graph for the Unlimited Plan, Y = 100. Next, we create the graph for the Pay-Go plan, Y =.30*X We can see in the image above that the graphs intersect. The point at which the graphs intersect is where the X and Y values of each equation are equal. Where these values are equal means that not only are the number of minutes talked equal but so is the price of each plan. We can then use
the tool menu, seen as the wrench in the blue bar at the top of each image, to find the point of intersection. First click the wrench, then Analyze Graph in the drop down menu and then the option Intersection. After selecting Intersection a vertical line will appear on the graph representing the lower bound. After placing the lower bound at its desired point we then can drag the upper bound over the point of intersection in order for the app to give us our intersection point. We can then see that the intersection point of the two equations lies at (333,100). This means at 333 minutes both plans will cost $100.
By analyzing the graphs we can also notice that for every x-value less than 333 the y-values for the Pay-Go Plan (f2) graph are less than they y-values for the Unlimited Plan (f1). This means for any amount of minutes less than 333 the Pay-Go plan is going to be cheaper. Also, for every x-value greater than 333 the y-values for the Pay-Go plan are greater than the y-values of the Unlimited Plan. This means for any amount of minutes greater than 333 the Pay-Go Plan is going to more expensive than the Unlimited Plan. Through this data we can then advise a customer which plan they should choose based on how many minutes they talk each month. This can also be represented by the following inequality: Pay-Go Plan 333 Minutes Unlimited Plan If a customer talks less than 333 minutes per month then they should choose the Pay-Go Plan and if they talk more than 333 minutes per month they should choose the Unlimited Plan. Now, if they talk exactly 333 minutes a month the plan will both cost $100, so either plan will be fine. However, Herizon advises they better not risk it and just go for the Unlimited Plan. Problem Continued: In response to popular demand, Herizon decided to offer a new plan that combines the best of their two existing plans: * The Select Plan only costs $40 per month and you pay only 13 cents per minute. Which types of customers should switch to the new plan? TI-Nspire Solution Continued: We now have another plan to take into account when it comes to advising customers in which plan to choose. As we did for the previous two plans we must also create an equation for the Select Plan. Again, we will use the slope-intercept form, y = mx+b, as a model. The y-value will again represent the cost of the plan and the x-value will represent the number of minutes talked.
However, the Select Plan has a regular monthly cost of $40 regardless of how many minutes you talk, plus an addition 13 cents (.13) for each minute you talk. Equation for Select Plan: Y =.13*X + 40 The.13*X represents the price for how many minutes you talk and the 40 represents the charge for each month of service. So, if a customer talks for 100 minutes one month their phone bill will be 13 + 40 = $53.00. We can now add the Select Plan s equation to the graph and repeat our steps we used in the first part of the problem to find which customers should choose the Select Plan. We create the graph for the Select Plan, Y =.13*X + 40. We can now visualize the Select Plan s graph (f3) intersecting both of Herizon s other plan s graphs. Next, we will repeat the same process we used before in order to find our two new intersection points. First, select the wrench, then select Analyze Graph, and lastly Intersection. We then set our lower and upper bounds so the intersection point lies between them in order to find the exact intersection point.
After finding our two new intersection points we see that the Select Plan intersects the Pay-Go Plan at the point (235, 70.6). We also see that it intersects the Unlimited Plan at the point (462, 100). By further analyzing the graph we can see that the y-values for the Pay-Go Plan (f1) are less than the y-values of the Select Plan (f3) and the Unlimited Plan (f1) for all x-values less than 235. This means that for all minute values less than 235 the Pay-Go Plan will be the least expensive. We also notice that for the x-values between 235 and 462 the y-values are the least for the Select Plan (f3). Lastly, for all x-values greater than 462 the y-values are the least for the Unlimited Plan (f1). We can express the new data by the following inequality: Pay-Go Plan 235 Minutes Select Plan 462 Minutes Unlimited Plan This inequality states that if a customer talks less than 235 minutes per month they should stick with the Pay-Go plan. If a customer talks between 235 minutes and 462 minutes per month they should choose the new Select Plan. If the customer is a Chatty Cathy and talks more than 462 minutes per month they should select the Unlimited Plan. Microsoft Excel Solution: Not only can we solve this problem by using graphs and calculators but we can also solve it using spreadsheets such as Microsoft Excel. Excel is great for analyzing large amounts of data. Using Excel we can calculate and see how much each plan will cost side by side for any given value of minutes a customer may talk. We start our spreadsheet by creating a column for the minutes and each of the three plans we were given. We then want to create a large list of minute values to analyze the cost of each plan at those values. We ll start at zero and increase by increments of 5. To do so, we start by simply typing zero into cell A2. Instead of typing 5 into cell A3 we can create a formula to make the listing process easier.
In Excel we can create formulas by using the names of cells that we may want to build off of. Here we want to get a list of minutes in increments of 5. To get 5 in cell A3 we type =A2+5 into the box f(x). The equals sign is very important and if left out Excel will not calculate the following cell. Now that we have the formula that gives us 5 we can then use another handy tool Excel offers instead of typing =A3+5 into the cell A4. By clicking and dragging the small black box in the lower left hand corner we can apply our created formula to the following cells. This tool is called fill down. To get a larger amount of minute values we will fill these columns all the way down to row 100. Next we need to create the formula for the Unlimited Plan. Our formula we create for the Unlimited Plan was Y = 100. So no matter how many minutes a customer may use the cost of the Unlimited Plan will always be $100. We simply type in 100 into cell B2. We then apply the fill down tool directly from this cell all the way to B100 to get 100 into each cell in column B.
Next we need to create our values for the Pay-Go Plan. The Pay-Go Plan costs 30 cents per minute and the formula to calculate its cost is Y =.30*X. Instead of using x we want to use the corresponding minute value in column A. To get our value of the Pay-Go Plan for the minute value of 0 we type in the f(x) box, =.3*A2. To make our job easier we then use the fill down tool again and drag the C column all the way to row 100.
After we fill down we can click on any given cell and check our f(x) box to make sure our formulas are working. As seen in the image above, clicking on cell C4 shows that it calculated the cost using the minute value in A4 as it should. We then repeat this same process for our Select Plan column using its equation y =.13*x+40. As we did for the Pay-Go Plan we use the cells in column A instead of x as seen in the f(x) box. Again, we fill down all the way to row 100. For the sake of space, here is an example of what the spreadsheet should now look like for minute values up to 80.
To solve the problem, we need to analyze the table and see what plan is cheapest for the range of minutes. If we look at the start of the table we see that the Pay-Go Plan is the cheapest for the lower minute values. As we scroll down we notice that between rows 49 and 50 the Select Plan becomes cheaper than the Pay-Go Plan. However, since we increased our minutes by increments of 5 we don t exactly where the Select Plan becomes cheaper. In order to do so we need to create another table. This time we will start our minute s column at 235 and increase them by a value of 0.1.
As we see in the image about the Select Plan becomes cheaper than the Pay-Go plan at a minute value of 253.3. As you may have noticed this value is more accurate than the one given in TI- Nspire. However, cell phone companies such as Herizon Wireless aren t that generous and always round up to the next whole minute value. So, if you have the Pay-Go plan and talk for 235.1 minutes you will be charged as if you talked for 236 minutes. We can use an inequality again to express which plan customers should pick: Pay-Go < 235 Minutes < Select This inequality represents the fact that if a customer talks less than 235 minutes they should choose the Pay-Go plan and if the talk more than that they should choose the Select Plan. However, we also notice that the Select plan is still increasing and the Unlimited Plan stays at 100 so we continue to scroll down to see if the Select Plan becomes more expansive than the Unlimited Plan. Again, we notice that somewhere between row 94 and 95 the Select Plan becomes more expensive than the Unlimited Plan. We now need to decrease the increments in which the minutes increase to see exactly where it becomes more expensive.
By changing the minutes to increase by.1 again instead of 5 we get a more accurate representation. By analyzing this graph we notice that the select plan becomes more expensive at the minute value of 461.6. But remember, Herizon always rounds up to the next whole minute value. By creating another inequality we get: Select Plan < 462 Minutes < Unlimited Plan So if a customer talks more than 462 minutes they should then go with the unlimited plan for the better deal. When we add this new inequality to the previous one we get: Pay-Go Plan < 235 Minutes < Select Plan < 462 Minutes < Unlimited Plan By comparing our answer from Excel to TI-Nspire we notice all the same except one difference. The signs in the inequality are less than as opposed to less than or equal to. As we noticed in our more accurate table the Pay-Go and Select Plans were not equal at 235 minutes, the Select Plan was 5 cents more. In final conclusion, since Herizon Wireless always rounds up to the next whole minute value, if a customer talks less than 235 minutes they should choose the Pay-Go Plan. If the talk between 235 and 462 minutes they should choose the Select Plan and if they talk more than 462 minutes they should go with the Unlimited Plan. Teacher Discussion: This activity addresses a number of standards in the Alabama Course of Study, particularly at the 7 th and 8 th grade levels. For the 7 th grade level the activity addresses standards 7-EE4 and 7- EE4a.
Standard 7-EE4 states: Use variables to represent quantities in a real-world or mathematical problem, and construct simple equations and inequalities to solve problems by reasoning about the quantities. Standard 7-EE4a: Solve word problems leading to equations of the form px + q = r and p(x + q) = r, where p, q, and r are specific rational numbers. Solve equations of these forms fluently. Compare an algebraic solution to an arithmetic solution, identifying the sequence of the operations used in each approach. In this cellphone example we used these standards in the introduction where the formulas for the cellphone plans were created. We let the variables x and y represent minutes and cost respectively and used formulas similar to the format px + q = r. The activity also met the standards 8-EE7(a) and 8-EE8(a,b,c). Standard 8-EE7: Solve linear equations in one variable. Standard 8-EE7a: Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms until an equivalent equation of the form x=a, a=a, or a=b results (where a and b are different numbers). This standard particularly applies to the equation of the Unlimited Plan in which y = 100. It did not depend on any value of x and always remained equal to 100. Standard 8-EE8a: Understand that solutions to a system of two linear equations in two variables correspond to points of intersections of their graphs because points of intersection satisfy both equations simultaneously. This applied directly to one of the main ideas used in the TI-Nspire solution. We found where each plan was more or less expensive by evaluating the graph before and after the intersection points. Standard 8-EE8b: Solve systems of two linear equations in two variables algebraically, and estimate solutions by graphing the equations. Solve simple cases by inspection. This standard was used as the main approach in the TI-Nspire solution. We created graphs of the three plans and analyzed them to come to our solution. Standard 8-EE8c: Solve real world mathematical problems leading to two linear equations in two variables. This standard was met in the problem itself. Although the problem could be expressed completely algebraically, it was stated in a real world approach that applied it to picking out a
cellphone plan that was cost efficient for a type of customer. These types of problems answer the question When am I ever going to use this? The activity also met Standard 8-F2: Compare properties of two functions, each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). In the activity we used both graphs and tables, in the form of spreadsheets, to compare the costs of each of the three plans based on how many minutes a customer uses. This activity could be used most effectively in an 8 th grade algebra class. It is great for learning what algebraic formulas can represent, in this case, the cost of a cellphone bill. The activity can be used as a direct answer for the question When will we ever use this? By using the technologies such as TI-Nspire and Excel the students are able to not only come to an answer to the problem but visualize why that answer is the right one. They are also able to visualize and understand why the wrong answer is wrong. This particularly satisfies the Technology Principle and the fact that technology enhances mathematics learning. The students are able to see answers for large amounts of values quickly, as we did in the spreadsheet, without having to do countless numbers of problems which take up lots of time just to see more data. Overall, the activity was extremely successful in solving linear equations and teaching the students how to represent linear equations and what their representations mean, especially when multiple equations are placed on a graph or table together.