VIEWS: 39 PAGES: 7 POSTED ON: 3/22/2011 Public Domain
The Cookie Problem This semester my colleagues and I used a simple modeling problem as one of the questions on our Precalculus final exam. The question was, "The Fresh Bakery currently sells 1000 large chocolate chip cookies each week for $0.50 each. The bakery would like to increase its revenue from the cookie sales by increasing this price. A survey convinced the manager that for every $0.10 increase, the bakery would sell 70 fewer cookies each week. At what price should the cookies be sold to maximize the revenue to the bakery?” The Anticipated Solution The revenue, R, is the product of the price charged, p, and the number of cookies sold, C. So, R p C . The number of cookies sold is a linear function of price, since for each 10 cent increase in price, the demand is reduced by 70 cookies. The slope of the line is 70 cookies cookies m 7 . 10 cent increase cent increase Since the bakery presently sells 1000 cookies at 50 cents each, the point 50,1000 lies on the line. So C 7 p 50 1000 7 p 1350 The revenue, then, is given by R p p 7 p 350 . This is a quadratic function and the optimum price is represented by the p-coordinate of the vertex of this parabola. 4 7.5 10 Revenue in cents 4 5 10 R( p ) 4 2.5 10 0 0 40 80 120 160 200 p P rice in cent s We expected that students would find the vertex in one of three ways. Recall from Algebra the location of the vertex of the quadratic y ax 2 bx c is b 1350 x , or in this case, p 96.43 . 2a 14 Remember the geometry of a quadratic function with the vertex at the midpoint of the zeros, and recognize the given quadratic is factored, so the zeros are easily found at 1350 1350 0 p 0 and p . The vertex is at p 7 96.43 . 7 2 Graph the function R on their calculators and seek the numerical maximum, which is reported as p 9643. . Regardless of the technique, we expected to have the solution reported as the optimum price is p 96 cents. We expected several students to consider the optional prices of 95 cents or $1.00, since these are more practical values and the difference in revenue is negligible. We also thought we might find some solutions adjusting the price so that, with tax, the total would be exactly $1.00. My colleagues and I expected some variability in the process used to locate the vertex and in the numerical value presented as the solution, but we never expected the variaty of approaches used in setting up the problem. In all, we had seven different approaches to the modeling aspect of the problem. Student Solutions Solution 1 was the anticipated solution, and was a fairly common solution path. Solution 2 is a modification of the anticipated solution. The students first developed a table of price and demand. Price 50 60 70 80 90 100 110 120 Demand 1000 930 860 790 720 650 580 510 They then used their calculators to fit the regression line relating Price to Demand. This line is y 7x 1350. Next, they created the revenue function R x y x 7 x 1350 . Solutions 3 and 4 carried this data analytic approach one step farther. These students multiplied the two rows in the table together and created a table with revenue. Price 50 60 70 80 90 100 110 120 Demand 1000 930 860 790 720 650 580 510 Revenue 50000 55800 60200 63200 64800 65000 63800 61200 Some students simply looked on the chart and reported the optimum price as $1.00 and the expected revenue as $650.00. A more common approach was to use the calculator to fit a quadratic function to the data Revenue using price, quadratic regression. 4 7 10 4 Revenue in cents 6.375 10 R 4 i 5.75 10 4 5.125 10 4 4.5 10 40 56 72 88 104 120 P i P rice in cent s This quadratic is y 7 x 2 1350x . From here they found the vertex of the quadratic model. The first four solutions used price as the independent variable. Some students used units of cents while others used dollars, but the solutions generally followed those outlined above. If the independent variable is price, then the algebraic task for the students is to write the demand function in terms of price. However, the other solutions used different aspects of the problem as the independent variable. Solution 5 used the number of dime increases, d, as the independent variable. In this case, the product of price and demand is more complicated to write. Both price and demand must be written in terms of the number of dime increases. The price is written as p .50 01d , . and the demand is written as D 1000 70d , so the revenue function is . R d .50 01d 1000 70 d . Students who located the vertex as the midpoint of the zeros used this form of R, otherwise they re-wrote it as R d 7d 2 65d 500 . The vertex of this parabola is at 65 d 4.64 . 14 As you might suspect, there were some solutions suggesting the best price is 46 cents, but most students who got this far realized the solution represented 4.64 dime increases for a price of 96 cents. Solution 6 was the most difficult of all the solutions. In this approach, the student let the independent variable be the number of cookies sold. There were very few correct solutions with this method. In this case, we need to write the price in terms of the number of cookies sold rather than cookies in terms of price. One correct solution used the model R C C 05 . 1000 C 700 . It was difficult to determine the student's reasoning in developing this model. Here's how it goes: “Since we lose 70 cookies with each 10 cent increase, we lose 700 cookies for each dollar increase. The number of cookies lost, L, is 700 times the increase in price measured in dollars, I, so L L 700 I and I . 700 The price we charge is 0.50 I . If C is the number of cookies sold, then the number of cookies lost due to increasing the price is L 1000 C , and the price we charge is 1000 C price 05 . 700 . Putting this together generates the function R C C 05 . 1000 C 700 .” This is also a quadratic function. 700 525 R( b ) 350 175 Revenue in Dollars 0 0 200 400 600 800 1000 b Number of Cookies Sold The vertex is located at C 635 . This means the best price to charge is 1000 635 price 05 . 700 0.96 . The final solution was the most difficult to understand. Perhaps you will see immediately what the student did. We certainly didn’t. As an exercise, work out the student’s thought processes in developing this model. The model given for revenue was .0570. 01. x R x x 1000 What is x, and what line of reasoning was used to create this particular function? Is this solution better or worse than others given? My colleagues and I were surprised by the variety and creativeness used in solving what I thought would be a relatively simple, straight-forward problem. The issues of assessing student solutions is a difficult one. Do you think all of these solutions are equally good? Should some receive more credit than others? Which solutions do you think illustrate a greater understanding of principles of precalculus than others? I would appreciate any thoughts and comments on your assessment of these solutions. Perhaps we can create another article from your ideas on the assessment of these problems. Let us know what you think. Send your ideas and suggestion by e-mail to: compton@odie.ncssm.edu or teague@odie.ncssm.edu or write to: Helen Compton and Dan Teague North Carolina School of Science and Mathematics 1219 Broad Street Durham, North Carolina 27715