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Say there is a cat that in three days eat 12 fishes.
Say that each day the cat eats more than the day before.
Say that the last day the cat has eaten less than the addition of the two previous days.

x+y+z = 12
z>y>x
z<x+y

I assume x -> first day, y -> second day, z -> third day.

How am I supposed to solve this? I have been taught to solve inequality systems with two variables (doing graphically), but I don't know how to do this one.

Thanks.

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First, your inequalities should be strict: $x<y<z$ and $z<x+y$. Second, if $x=4$ then $x+y+z\geq 3x+3 = 15$ so you know that $x<4$. If $x=3$ there is a unique choice of $y=4,z=5$. If $x=1,2$ then clearly either first or second inequality is violated. So, $(3,4,5)$ is the only solution. Your problem reminds a triangle a bit, so maybe there is another solution, more beautiful. Since the cat could not eat non-integer number of fish (cat it?), I don't think that graphical solution will help. –  Ilya Aug 19 '11 at 10:58

3 Answers 3

up vote 4 down vote accepted

To do a graphical solution, you can use the equality $z=12-x-y\ \ $ in the inequalities to get $12 \lt 2x+2y ,\ y \gt x ,\ 12 \gt 2y+x.\ \ $ You can plot these three as you are used to, shading the proper side of the lines. Then as you are looking for integer solutions you have to look for lattice points in the allowable region.

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+1. I think for the pure graphical solution there can be some doubts if some points are in the region or not (like $2,4,6$). So I will be also helpful for OP to make a check of each point. –  Ilya Aug 19 '11 at 14:04
    
@Gortaur: I just picked up on the graphical solution from the original post, trying to use the tools at hand. With numbers this small, points that appear on the boundary are really on the boundary. I agree that a specific check is valuable. –  Ross Millikan Aug 19 '11 at 14:08
    
Thanks for the answer. It helped. –  Nobita Aug 19 '11 at 16:08

Sometimes, unfortunately, the method you are "supposed" to use is gross overkill. But I will write out a linear inequalities approach, under the assumption that the number of fish eaten on any day is a non-negative integer.

Also, when your text said less than and your formula said $\le\;$, I assume it should be $\lt$. Similarly, if your text says more and your formula said $\ge\;$, I assume it should be $\gt$. So when there is conflict, I give priority to the words.

We have three variables, but one can be eliminated, using say $z=12-(x+y)$.

Since $x<y$, we have

(1) $x+1 \le y$.

Since $y \lt z$, we have $y+1 \le z$, and therefore $y+1 \le 12-(x+y)$. This can be rewritten as

(2) $x+2y \le 11$.

We have $z \lt x+y$, so $z+1 \le x+y$, and therefore $12-(x+y)+1 \le x+y$, giving

(3) $13 \le 2x+2y$.

Finally, we need to remember that $x$, $y$, and $12-(x+y)$ are all $\ge 0$. That gives the inequalities

(4) $x \ge 0$, $y \ge 0$, $12 \ge x+y$.
Now we can draw lines as usual.

Much too much work, since it is clear that $x=3$, $y=4$, $z=5$ is the only solution! This is because if $x\lt y \lt z$ and $x+y+z=12$, we must have $z\gt 12/3$, so $z \ge 5$. If $z\ge 6$, then $x+y \le 6$, but $x+y \le z$ is not allowed. So $z=5$. Now we have $7$ fish, to be distributed among the two days. But the second day must be $\le 4$. To get a total of $7$ we need $y=4$, $x=3$.

Note: If everywhere we use your formula inequalities, not your text inequalities, all the $+1$'s need to be removed. That changes the inequalities, and we get a few more solutions. Which ones are OK depends on interpretation of the wording. The solutions could include some or all of $(0,6,6)$, $(1,5,6)$, $(2,4,6)$, $(2,5,5)$, $(3,3,6)$, $(3,4,5)$, and $(4,4,4)$.

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Thanks André. I fail to see how you get this assumption: This is because if x<y<y and x+y+z=12, we must have z>12/4, so z≥5. Could you explain it? –  Nobita Aug 19 '11 at 16:11
    
@Nobita: Suppose that we had instead $z \le 4$. Then $y$ would be $\le 3$, and $x$ would be $\le 2$, giving total possible sum $x+y+z \le 9$. The way I put it says roughly the same thing, but in a fancier way. If you have $3$ different numbers whose average is $12/3=4$, on of the numbers at least must be $\gt 4$ –  André Nicolas Aug 19 '11 at 16:26
    
@Nobita: I see, there was a typo, it said $x\lt y \lt y$ when I meant $x \lt y \lt z$. Fixed. –  André Nicolas Aug 19 '11 at 16:31
    
Then it should be z>12/3, so z≥5, instead of z>12/4, so z≥5. Right? –  Nobita Aug 19 '11 at 16:34
    
@Nobita: Thanks! Right. –  André Nicolas Aug 19 '11 at 16:40

A general solution of a system of inequalities is very difficult (see e.g. mathoverfow.net). Your special case can be solved graphically as answered by @Ross. You can also argue as follows.

All variables must obviously be non-negative and less or equal 12. Combining the first equality and the last inequality we get $z \le 6$. $z$ must also satisfy $z \ge 4 $ because $z \lt 4$ would imply $y \lt 4$, $x \lt 4$, $x+y+z \lt 12$. Similarly, $x$ must satisfy $x \le 4 $ because $x > 4$ would imply $y \gt 4$, $z \gt 4$, $x+y+z \gt 12$.

So, all solutions are obtained by choosing $$x \in [0, 4]$$ $$z \in [4, 6]$$ $$y = 12 - (x+z)$$ If only positive solutions are allowed, then $x \in (0, 4]$.

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