All Possible Solutions: Diophantine EquationsDate: 12/06/2002 at 14:29:26 From: Chris Subject: Algebra problem I've tried everything that I can think of and I can't figure out how to do this. Could you please try it and let me know how to do it? I have the answer (16 Combos, 17 Hot dogs, 22 Hamburgers, 25 Sodas), but I do not know what method to use to solve it. Thanks! Lance and Mario were working in the snack bar at the Turbulent Tunas Concert. They sold hot dogs for $1.65, hamburgers for $2.35, sodas for $.85, and the combo plate with fries, salad, a hamburger; and soda for $3.89. They sold 80 items in an hour for $163.24. How many of each kind of food did they sell, and how much did they make on each kind of food? Date: 12/09/2002 at 00:23:51 From: Doctor Greenie Subject: Re: Algebra problem Hello, Chris - You apparently have one of many possible solutions to this problem. I have found several others and have not completed the task of finding all solutions. Following is a description of how you can go about finding all the solutions to this problem. You are selling the following items: $1.65 -- hot dog $2.35 -- hamburger $0.85 -- soda $3.89 -- combo 80 items were sold for a total of $163.24. Suppose we let D = # (hot) Dogs sold H = # Hamburgers sold S = # Sodas sold C = # Combos sold Then we have two equations; one says the total number of items sold is 80, and the other says the total cost of those items is $163.24. To avoid working with decimals, we write the second equation in terms of cents instead of dollars. The two equations are then (1) D + H + S + C = 80 (2) 165D + 235H + 85S + 389D = 16324 To limit the types of combinations we need to search for, notice that the prices of a hot dog, a hamburger, and a soda are all multiples of 5. That means the total cost of the hot dogs, hamburgers, and sodas is a multiple of 5, so that total cost of those items has final digit 0 or 5. Since the total cost of all items sold is $163.24, that means that the total cost of the combos has final digit either 4 or 9. This in turn means that the number of combos sold must be a number with final digit either 1 or 6. So we have the following possibilities at this point: # combos cost of combos # other items cost of other items ---------------------------------------------------------------- 1 3.89 79 159.35 6 23.34 74 139.90 11 42.79 69 120.45 16 62.24 64 101.00 21 81.69 59 81.55 26 101.14 54 62.10 31 120.59 49 42.65 36 140.04 44 23.20 41 159.49 39 3.75 The largest number of combos that could have been sold is 41, because with the next possible number, 46, the cost of the combos is by itself more than the total amount of the sales. Also, examination of the last two columns in the above table shows that there will be no solutions with either 36 or 41 combos, because the cheapest item is 85 cents, and you can't get the total in the last column by selling the number of items in the next-to-last column. And for the possible combination with 31 combos and 49 other items, the cost of 49 sodas is $41.65, so there won't be a solution with 31 combos either. However, it is possible and even likely that there are solution(s) for any of the other numbers of combos in the table. Following is how to find all the solutions with 6 combos sold; you can try to use the same process to find the solutions with other numbers of combos. We have 6 combos sold, for a total of $23.34; leaving 74 other items with a total cost of $139.90. So now, with C=6, we have the following equations: D+H+S = 74 165D+235H+85S = 13990 We can solve this pair of equations by eliminating the "S" variable - by factoring a constant of 5 out of the second equation and then multiplying the first equation by 17: 33D+47H+17S = 2798 17D+17H+17S = 1258 ------------------ 16D+30H = 1540 The approach at this point is to solve this equation for one of the variables in terms of the other and then to use the fact that the variables have integer values to determine the possible combinations of values for D and H. 16D = 1540-30H 16D = (1600-32H) + (2H-60) D = (100-2H) + (2H-60)/16 100 is an integer; and H is an integer, so 2H is an integer. For D to be an integer, then, (2H-60)/16 must be an integer, so (2H-60) must be a multiple of 16. We then get the following solutions with C=6: H D=(100-2H)+(2H-60)/16 S = 80-(C+H+D) ------------------------------------------------ 30 (100-60)+0 = 40 80-(6+30+40) = 4 38 (100-76)+1 = 25 80-(6+38+25) = 11 46 (100-92)+2 = 10 80-(6+46+10) = 18 The values chosen for H are those which make (2H-60) a multiple of 16. The next larger value of H would give a negative value for D, so this table shows all the solutions with C=6. We can check these three solutions to the problem: combos (C): 6 @ $3.89 = $23.34 hamburgers (H): 30 @ $2.35 = $70.50 hot dogs (D): 40 @ $1.65 = $66.00 sodas (S): 4 @ $0.85 = $ 3.40 ------- $163.24 combos (C): 6 @ $3.89 = $23.34 hamburgers (H): 38 @ $2.35 = $89.30 hot dogs (D): 25 @ $1.65 = $41.25 sodas (S): 11 @ $0.85 = $ 9.35 ------- $163.24 combos (C): 6 @ $3.89 = $23.34 hamburgers (H): 46 @ $2.35 =$108.10 hot dogs (D): 10 @ $1.65 = $16.50 sodas (S): 18 @ $0.85 = $15.30 ------- $163.24 You can follow the same procedure to find the other solutions to the problem. I did some quick calculations to verify that there is at least one solution with 1 combo sold and at least one solution with 26 combos sold; so I suspect there will also be at least one solution each with 11, 16, or 21 combos sold. Equations like those in this problem - where the number of unknowns is more than the number of equations, and where the values of the unknowns are integers - are called Diophantine equations. If you would like a better understanding of problems like this, try searching the Dr. Math archives using the keyword "Diophantine" and take a look at some of the pages that the search provides to you. I hope all this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
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