MaplePrimes - answers and comments on Question, Problem please help me !

Better title and specification of the problem.

Jarekkk — Mon, 20 Feb 2012 15:51:45 Z

It would be nice if you could titled differently your question and if your problem was better specified.

If I understand it correctly:

One boy eats 6 or 7 pieces of pizza, one girl eats 2 or 3 pieces of pizza. When we give them 4 pizzas (48 pieces) it is not enough for them, when we give them 5 pizzas (60 pieces), at least one piece of pizza remains. How many boys and girls are there?

Solution:

Let us denote x the number of boys and y the number of girls. Since 48 pieces are not enough, we can write the first inequality:

ineq1 := 6*x+2*y >= 49:

Since 60 pieces are more than enough we deduce the second inequality:

ineq2 := 7*x+3*y <= 59:

Now, there are several solvers for inequalities. In our case the isolve command for the integer solutions would suit it best. Unfortunately, it finds no solution. So we can use the general solve command instead.

solve({ineq1, ineq2});

There is also the possibility to use assumptions. Unfortunately again, it doesn't help. So now the one step by hand. From the obtained solution we see x has to be at least equal to 8 (since it is a whole number). On the other hand from the second inequality (ineq2) we see that x has to be less than 9 (since y is nonnegative), so x has to be equal to 8. And then it is easy to get the value of y with isolve.

x := 8: isolve({ineq1, ineq2});

range of possibilities

Christopher2222 — Mon, 20 Feb 2012 18:52:44 Z

There is a range of possibilities. What I understood was that 48 slices was not enough and 60 slices was too much.

7 boys could eat a minimum of 42 or max of 49
8 boys could eat 48 or 56 slices
9 boys could eat 54 or 63 slices

5 girls could eat 10 or 15,
4 - 8 or 12 etc...

So there could be 7, 8 or 9 boys.

If 7 boys (they would all have to eat 7 slices) then there could be 1 to 5 girls.
If 8 boys (at least one would have to eat 7 slices because says 48 slices not enough) again could be 1 to 5 girls. If all 8 ate 7 slices than there would be 1 girl and she would have to eat 2 slices.
If 9 boys (all eat 6 slices) there could be 1 or 2 girls.

The most likely answer is that there are 8 boys but still a range of 1 to 5 girls.

feasibility, LP

acer — Tue, 21 Feb 2012 10:39:05 Z

This can be considered as a feasibility question: given some linear constraints and the requirement of an integer-value solution then find a feasible point.

If one adds in a linear objective function then it can be seen as the familiar (I)LP. The Optimization package can handle this particular example, if we supply an (here inconsequential) objective.

Optimization:-Minimize(1,
                       {6*x+2*y>=49, 7*x+3*y=0, y>=0},
                       assume=integer)[2];

                         [x = 8, y = 1]

acer

Good point.

Jarekkk — Tue, 21 Feb 2012 14:10:03 Z

@acer That's a good point! Unfortunately, I have problems (not only) with your command. Maple tells me no solution found. This happens in every interface of Maple.

When I rewrite it into:

Optimization:-Minimize(7*x+3*y, {6*x+2*y >= 49}, assume = nonnegint);

which looks more understandable for me, it works well, but only in the Standard worksheet. I will post a question on this topic.

using isolve

Christopher2222 — Wed, 22 Feb 2012 03:32:07 Z

We can get the answer using isolve as well using seq as I found out here http://www.mapleprimes.com/questions/130964-Isolve-Wont-Give-All-Solutions-Here

ineq1:=6*x+2*y>=49:
ineq2:=7*x+3*y<=59:

seq(isolve({ineq1 , ineq2 , y<=k , y>=k }), k=0..5)

{x = 8, y = 1}

Thanks.

Jarekkk — Wed, 22 Feb 2012 23:13:09 Z

@Christopher2222 On the other hand it's "funny" that this

seq(isolve({ineq1, ineq2, y >= k, y <= k}), k = 0 .. 5);

works, but this

seq(isolve({ineq1, ineq2, y = k}), k = 0 .. 5);

which is actually the same, does not. :/

Using this command you also have to put k "large enough" to go through all possible solutions. In this case y=1 is the only solution, so k from 0..5 works, however it should be k=0..19 (since ineq2).