Solving for Multiple UnknownsDate: 03/22/2003 at 02:57:55 From: Ralph Subject: Solving for multiple unknowns Three chickens and one duck cost as much as two geese. One chicken, two ducks and three geese cost $25. What is the cost of each bird? Date: 03/24/2003 at 12:09:41 From: Doctor Ian Subject: Re: Solving for multiple unknowns Hi Ralph, If c, d, and g are the prices of one chicken, one duck, and one goose, respectively, then we're told the following: [1] 3c + d = 2g [2] c + 2d + 3g = 2500 I'm representing the price in cents, so we can restrict ourselves to integers. We can double everything in [1] to get [1'] 6c + 2d = 4g [2] c + 2d + 3g = 2500 and subtract [2] from [1'] to get [3] 5c - 3g = 4g - 2500 which simplifies to 5c = 7g - 2500 c = 7g/5 - 500 which tells us a couple of things about g. First, g has to be a multiple of 5, or c can't be an integer. Second, unless they're paying people to take chickens (i.e., the price of a chicken is negative), it must be true that 7g/5 > 500 7g > 2500 g > 2500/7 g > 357 So 360 cents is the minimum possible cost of a goose. So the price of a goose is in the set {360, 365, 370, 375, ..., 2500} It would be nice to have an upper bound, to go along with our lower bound. Let's try something similar, except this time, let's eliminate chickens instead of ducks. If we triple everything in [2] we get [1] 3c + d = 2g [2'] 3c + 6d + 9g = 7500 Subtracting [1] from [2'], we get [4] 5d + 9g = 7500 - 2g which simplifies to 5d + 9g = 7500 - 2g 5d = 7500 - 11g d = 1500 - 11g/5 Again, the cost of a goose has to be a multiple of 5. But if the price has to be positive, then 1500 - 11g/5 > 0 1500 > 11g/5 7500 > 11g 7500/11 > g 681 > g So this gives us an upper bound on the cost of a goose, to go with our lower bound. The cost of a goose must be in the set {360, 365, 370, ..., 670, 675, 680} That's only 62 possibilities. Can we narrow that down some more? Well, one thing we might consider is that the price of each bird, expressed in dollars, might be an integer. That gives us only three possibilities for the price of a goose: 400 cents = $4 500 cents = $5 600 cents = $6 (Any other possibilities have been ruled out by our upper and lower bounds.) Now we can go back to our original equations, and use dollars instead of cents: [1] 3c + d = 2g [2] c + 2d + 3g = 25 And we can try this with g=4, g=5, and g=6. Each of those substitutions will give us two equations with two variables, and we can see if that generates integer values for c and d. But that sounds like a lot of work, doesn't it? Remember this equation? 5c = 7g - 2500 We can use it to do a quick test of our reduced set of values for g: 1) g = 400 5c = 7(400) - 2500 = 2800 - 2500 = 300 (which makes c equal to $0.60) 2) g = 500 5c = 7(500) - 2500 = 3500 - 2500 = 1000 (which makes c equal to $2) 3) g = 500 5c = 7(600) - 2500 = 4200 - 2500 = 1700 (which makes c equal to $3.40) So our _only_ hope for this to work out with even dollar costs is for g to be $5. I'll leave it for you to figure out whether $5 for a goose works. But if it doesn't, you'll have a strategy for finding a price that _does_ work: go back to using 2500 cents instead of 25 dollars in [2], and try all the possible values of g, i.e., {360, 365, 370, ..., 670, 675, 680} cents If the problem has a solution, this will find it. Or, to put that another way, if this doesn't find a solution, then the problem doesn't have one. (But the problem _does_ have a solution!) So this is a cute problem, but when you're done with it, what lessons can you take away from it? I can think of three: 1) A constraint doesn't have to be an equation, and doesn't even have to be stated explicitly. Simple rules like "you need at least as many equations as variables" are useful as rules of thumb, but they're not laws of nature. 2) Sometimes equations can take you right to where a solution is hiding. But sometimes all they can do is tell you where the solutions _aren't_ hiding, so you don't waste time looking there. 3) Try the easy cases first! I hope this helps. Write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 03/25/2003 at 04:43:40 From: Ralph Subject: Solving for multiple unknowns Dear Ian, First, I'd like to thank you for your excellent response concerning the "Birds" problem. After reviewing your answer and utilizing your suggestions I've come to some conclusions. You cannot solve this problem (3 unknowns), using conventional algebraic equations. This particular problem has numerous correct answer combinations (as long as you don't try to use logic when pricing your birds - i.e. a duck "should be" much more expensive than a chicken). Using methods you outlined, this is more of a "guess and test" problem. I think this exercise was meant to drive the person trying to solve it crazy! Your comments (and hopefully confirmation of my conclusions), would be greatly appreciated. Thanks again. Ralph Date: 03/25/2003 at 10:14:45 From: Doctor Ian Subject: Re: Solving for multiple unknowns Hi Ralph, I haven't actually worked out more than one solution, and I haven't thought about whether more than one solution is possible. The equations are [1] 3c + d = 2g [2] c + 2d + 3g = 2500 They can be rewritten as [1'] 3c + d - 2g = 0 [2'] c + 2d + 3g - 2500 = 0 If we change the variables, we get [1''] 3x + y - 2z = 0 [2''] x + 2y + 3z - 2500 = 0 When you write them this way, they look very familiar. Each equation defines a plane in three dimensions: Three dimensions: Planes - Analytic Geometry Formulas - Dr. Math FAQ http://mathforum.org/dr.math/faq/formulas/faq.ag3.html#threeplanes So, any solution to both equations at once will lie on the intersection of these planes, which is to say, on a line in three dimensions. So there is definitely an infinite number of 'solutions' if we allow the price of a bird to be any real number. But of course, money comes in finite denominations: cents, kopecks, agorot, and so on. So the only solutions that count are the ones where the prices are integers. Having found a solution, we know that the line passes through at least one point (x,y,z) where all three coordinates are integers. Could it pass through others? It's possible, but the only way to find out for sure is to try all the other possibilities. Have you done that? >Using methods you outlined,this is more of a "guess and test" problem. It's a very _constrained_ guess and test problem. Of 2498 possible prices for a goose ($.01 through $24.98), we managed to narrow it down to 62 possibilities (multiples of 5 from 360 to 680). If the combinations had been different, we might have been able to narrow it down further (for example, by showing that the price of a goose had to be a common multiple of two numbers). And if we made the simplifying assumption that the prices would all be whole-dollar prices, we had only three possibilities to check. So in fact, I'd say it's more of an 'identify and eliminate' problem, but that might be a distinction without a difference. >I think this exercise was meant to drive the person trying to solve >it crazy! It certainly wasn't intended to be easy. But I think the reason that problems like this come up so often is that many people will attack them by making up a pair of equations, and then concluding that there isn't enough information to solve the problem. Then it remains for someone to point out that there _is_ enough information, if you know where to look for it, and how to use it. And that, rather than the solution, or the solution method, is the take-home lesson. I would guess that the next time you're trying to solve a problem where you think that there isn't enough information, you'll remember this one, and decide to look a little harder before giving up. If so, then your time working on this problem was well spent, and whoever got you to work on the problem did you a favor. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/