Yukang
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Posts posted by Yukang
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1 hour ago, swansont said:
A1 + A2 = 25
B1 + B2 + B3 = 45
C1 + C2 + C3 +C4 +C5 = 100
(A1 + B1 +C1) = 2(A2 + B2 + C2) = 4(A3 + B3 + C3)
B1 + C1 <= 26
A2 + C2 >= 14
You could rewrite the third equation as C1 + C2 + C3 <= 100, in case that helps, since you don't need the C4 and C5 variables
Thanks, that looks about right. Looks like 8 equations with 9 unknowns after you throw away the C4 & C5 variables? Is 8 equations with 9 unknowns solvable? I thought you need 9 equations? This is assuming the 3 way equal equation can be arranged into 3 equations:
I think the 3 way equal equations may be reversed in their coefficient though
4(A1 + B1 +C1) = 2(A2 + B2 + C2)
2(A2 + B2 + C2) = (A3 + B3 + C3)
4(A1 + B1 +C1) = (A3 + B3 + C3)
How can you solve 8 equations with 9 unknowns? (A1, A2, A3, B1, B2, B3, C1, C2, C3)
I know from guess and check A1 = 7, A2= 18, A3 = 0, B1 = 26, etc. (and the rest can be filled out), but isn't it strange there are only 8 equations?
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Hi, I came up with the answer $44 through guess and check. Because the earnings are $26 for C & B on day 1, that leaves $18 for A on day 1 and 25-18=7 $7 dollars on day 2 for A. You want the contributions from Company C to be as low as possible so, $14 from C & A 2 days from now means the contributions from Company C is at least 14-7=7 $7. When setting the contributions for Company C in every other day to 0 until day 4 or 5, you get a total of $44 on day 1.
However, I have no idea how to solve the problem using step by step reasoning or even graphical analysis.
The problem is actually mine, and I tried to make it as hard as possible while not involving big numbers or difficult formulas.
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Hi, I came up with a word problem that I know the answer to but I have no idea why the solution is what it is or how to reason through it. I posted this on another forum before, but I did not get the answer (explanation) I was looking for.
Joe has investments in Company A, Company B, and Company C.
Joe is fated to earn $25.00 from Company A within 2 days from now.
Joe is fated to earn $45.00 from Company B within 3 days from now.
Joe is fated to earn $100.00 from Company C within 5 days from now.
Joe is fated to earn no more than $26.00 from Company C and Company B on day 1 (1 day from now).
Joe is fated to earn at least $14.00 from Company A and Company C on day 2 (2 days from now).
Joe has to earn twice the amount of money on the first day than the second day from Companies A, B, and C and twice the amount of money on the second day than the third day from Companies A, B, and C. This can be expressed algebraically as Joe earning x money on day 3 (3 days from now), 2x money on day 2 (2 days from now), and 4x money on day 1 (1 day from now).
Joe can earn whatever amount of money (that satisfies the other conditions) from Companies A, B, and C on day 4 and day 5 (4 and 5 days from now).
What is the lowest amount of money Joe can earn on day 1 (1 day from now) from Companies A, B, and C? Explain your reasoning.
P.S. How come there doesn't seem to be good formulas to use for this question?
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Problem Solving (Fated to earn money) Question that has no explanation yet
in Applied Mathematics
Posted · Edited by Yukang
Ok, my bad about day 1 income being twice as big as day 2, you're right. The question asks for the minimum on day 1, so I guess I look for the minimum of the range of values? I never done this type of math before. Are you saying all 9 variables will have a range of values for solutions and the answer would be the minimum of A1 + B1 + C1?
You helped a lot, thanks, but I still don't know what to do to reason through it 100%. There is only one answer right, because it asks for the minimum on day 1?
I also have no idea how to find a range of solutions for 8 equations with 9 unknowns, although I know how to find solutions for 9 equations with 9 unknowns.