You are not logged in.
Post a reply
Topic review (newest first)
- bobbym
- 2012-12-04 13:09:52
I did it by computer. It was too complicated to be done by hand. If you have a fixed formula, please post it.
- BarandaMan
- 2012-12-04 12:59:53
YAY!!!
Ok, I got the same answer, but basically by using the 'fixed formula' for the type of question we had which we were told in class.
I just want to know HOW i would solve this using the inequalities so that I know I can really do this without memorizing a formula...if that makes sense. Please can you show me how to solve this? 
- bobbym
- 2012-12-04 12:34:24
Hi;
Are those the last of the constraints?
With those I can get a maximum of
when
- BarandaMan
- 2012-12-04 12:30:00
I forgot to add that, 0 is less than or equal to each P which is less than or equal to 1. and P1+P2+P3+P4=1. Does this help? :d
- bobbym
- 2012-12-04 12:15:36
Hi;
Without even talking about a possible method of solution I am having a problem right there. I am getting no maximum with those constraints. Have you copied the problem exactly?
- BarandaMan
- 2012-12-04 12:11:23
Yes:P this is correct
- bobbym
- 2012-12-04 11:57:18
Hi;
Is this the problem?
Please check the constraints carefully!
- BarandaMan
- 2012-12-04 11:46:23
Unfortunately not, but it seems from my experience we are not meant to have...it is a very small part of the question (At the end)but obviously it is a big new thing to learn.
How can I solve this? Thanks bobbym
- bobbym
- 2012-12-04 11:28:33
Hi BarandaMan;
Looks like a linear programming problem. Have you studied them yet?
- BarandaMan
- 2012-12-04 10:18:15
Need to solve:
Maximise (8/3)P1 + 8P2 + 8P3 + 10P4 subject to........
1. (4/3)P1+6P2>(or equal to, for all) 2P1 + 5P2 .... SO P1 < (3/2) P2.
2. 2P3 + 5P4 > (4/3)P3 + 6P4.....so P4 < (2/3)P3
3. (4/3)P1 + 6P3 > 2P1 + 5P3....SO P1<(3/2)P3
3. 2P2 + 5P4 > (4/3) P2 + 6P4.....SO P4 < (2/3) P2
PLEASE help me solve this. How do I do it? What do you get? Help help help! 