Find the general flow pattern of the network.

20<---A
|\B__>x(4)
|/
80<---C
I believe the system of equations for A,B,&C is-A: x(1)+x(3)=20 B: x(2)=x(3)+x(4) C: 0=x(1)+x(2)
Assuming flows are all nonnegative, what is the largest possible value for x(3). Sorry, this would be a lot easier if I could post a picture. But I got to the answers x(1)=20-x(3) x(2)=x(3)-100 x(4)=-100 x(3) is free. I'm not sure what to do from here.

To find the largest possible value for x(3), we need to analyze the constraints and relationships between the variables in the system of equations.

From the given flow pattern, we have the following equations:
A: x(1) + x(3) = 20
B: x(2) = x(3) + x(4)
C: 0 = x(1) + x(2)

Let's rearrange these equations to express x(3) in terms of the other variables:

From equation A, we have: x(3) = 20 - x(1)
Substitute this into equation B: x(2) = (20 - x(1)) + x(4)
From equation C, we have: 0 = x(1) + (20 - x(1)) + x(4)
Simplifying equation C: 0 = 20 + x(4)

Now, let's substitute x(3) back into equation B and simplify further:
x(2) = (20 - x(1)) + x(4)
= 20 - x(1) + x(4)

Since we are looking for the largest possible value for x(3), we want to maximize its value by making x(1) and x(4) as small as possible.

From equation C, we know that x(1) + x(2) = 0, which means x(1) = -x(2)
Similarly, from equation C, we know that x(1) + x(2) + x(4) = 0, which means x(4) = -x(1) - x(2)

Substitute these values into the equation for x(2):
x(2) = 20 - x(1) + x(4)
= 20 - x(1) + (-x(1) - x(2))

Simplifying further:
x(2) = 20 - 2x(1) - x(2)

Rearrange the equation:
2x(2) = 20 - 2x(1)

Now, let's consider the non-negativity constraint: x(1), x(2), x(3), x(4) ≥ 0

Since we want to find the largest possible value for x(3), we want to minimize the values of x(1), x(2), and x(4). Therefore, we set x(1) = 0 and x(2) = 0 to obtain the minimum values for these variables.

So, when x(1) = 0 and x(2) = 0:
2(0) = 20 - 2(0)
0 = 20

This is a contradiction, which means there is no valid solution where x(1), x(2), and x(4) are all non-negative.

Therefore, there is no largest possible value for x(3) that satisfies the given constraints.