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MathGroup Archive 2006

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Re: solving a set of inequalities

  • To: mathgroup at smc.vnet.net
  • Subject: [mg65960] Re: solving a set of inequalities
  • From: Bill Rowe <readnewsciv at earthlink.net>
  • Date: Mon, 24 Apr 2006 06:02:33 -0400 (EDT)
  • Sender: owner-wri-mathgroup at wolfram.com

On 4/23/06 at 6:17 AM, reir at no.email.com (reiro) wrote:

>I'll be grateful if you tell me how to solve this set of
>inequalities in mathematica. I've been looking in the help file but
>it returns an error and I don't know how to sort it oout.

>I need to solve this set:

>3*a + 5*b + 9*c > 6*a + 6*b + 8*c
>3*a + 5*b + 9*c > a + 5*b + 12*c
>x + y + z = 1

The last of these three lines is not an equation in Mathematica syntax and clearly is completely independent of the first two lines. So, ignoring the last line, Reduce can be used as follows:

In[18]:=
Reduce[{3*a + 5*b + 9*c > 6*a + 6*b + 8*c, 
   3*a + 5*b + 9*c > a + 5*b + 12*c}, {a, b, c}]

Out[18]=
a â?? Reals && b < -((7*a)/3) && 3*a + b < c < (2*a)/3
  
If I assume the third line was intended to express a relationship between a,b, and c and change it to be valid Mathematica syntax, your problem becomes:

In[19]:=
Reduce[{3*a + 5*b + 9*c > 6*a + 6*b + 8*c, 
   3*a + 5*b + 9*c > a + 5*b + 12*c, a + b + c == 1}, 
  {a, b, c}]

Out[19]=
a < -(3/2) && (1/3)*(3 - 5*a) < b < (1/2)*(1 - 4*a) && 
  c == -a - b + 1

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