hey guys, so here's the problem. i can't get my head

>>8504385
Yes, p=0 or n=1 lol

you can write x_3 - x_4 in terms of x_1 and x_2
plug that in the max
the solution becomes easier
ir just follow the simplex algorithm?

>>8504401
may i ask you to solve this for me and attach a picture?

>>8504385
First let's add lines 1 and 2 to get:
3x2+x3+x4=4
now we can replace the 3x2+x3 in the top with 4-x4 and we are left with x1+4-2x4. Since the last line implies that each variable must be greater than or equal to 0, it's obvious that in order to maximize the function x4=0. But if we do this, then the first line changes:
x1+2x2=3
Again, we want to maximize x1, but the smallest 2x2 can be is 0, so x1 at most can be 3. That means the function becomes 3+4-0=7.

First introduce a new variable t to get the problem:
max t, when
x_1 + 3 x_2 + x_3 - x_4 >= t
x_1 + 2 x_2 + x_4 = 3
- x_1 + x_2 + x_3 = 1
x_1 >= 0
x_2 >= 0
x_3 >= 0
x_4 >= 0


Now use the equations to eliminate x_3 and x_4 to get the problem:

max t, when
3 x_1 + 4 x_2 >= 2 + t
- x_1 - 2 x_2 >= -3
x_1 + x_2 >= -1
x_1 >= 0
x_2 >= 0

Now eliminate x_2:

max t, when
x_1 >= -4 + t
x_1 >= -5
-x_1 >= -3
x_1 >= 0

Now eliminate x_1:

max t, when
0 >= -7 + t
0 >= -8
0 >= -3

So you can see that t=7. With backwards substituition you get
x_1 = 3
x_2 = 0
x_3 = 4
x_4 = 0

wow guys. thanks you very much