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## 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!