Using the precedence relationships and following the rules of network construction, the required network is shown in following figure.

(i) Variables: Let x₁ be the number of days required to run plant C₁ and x₂ be the number of days required to run plant C₂
Objective function: Minimize Z = 600 x₁ + 400 x₂
(ii) Constraints: 3000 x₁ + 1000 x₂ ≥ 24000 (since there is a demand of 24000 bottles of drink A, production should not be less than 24000)
1000 x₁ + 1000 x₂ ≥ 16000
2000 x₁ + 6000 x₂ ≥ 48000
(iii) Non-negative restrictions: Since be the number of days required of a firm are non-negative, we have x₁, x₂ ≥ 0
Thus we have the following LP model.
Minimize Z = 600 x₁ + 400 x₂
subject to 3000 x₁ + 1000 x₂ ≥ 24000
1000 x₁ + 1000 x₂ ≥ 16000
2000 x₁ + 6000 x₂ ≥ 48000 and x₁, x₂ ≥ 0

(i) Variables:
Let x₁ and x₂ denote the number of tables and chairs respectively.
(ii) Objective function:
Profit on x₁ tables = 50 x₁
Profit on x₂ chairs = 15 x₂
Total profit = 50 x₁ + 15 x₂
Let Z = 50 x₁ + 15 x₂, which is the objective function.
Since the total profit is to be maximized, we have to maximize Z = 50 x₁ + 15 x₂
(iii) Constraints:
The dealer has a space to store atmost 60 pieces
x₁ + x₂ ≤ 60
The cost of x₁ tables = Rs. 500 x₁
The cost of x₂ tables = Rs. 200 x₂
Total cost = 500 x₁ + 200 x₂, which cannot be more than 10000
500 x₁ + 200 x₂ ≤ 10000
5 x₁+ 2 x₂ ≤ 100
(iv) Non-negative restrictions:
Since the number of tables and chairs cannot be negative, we have x₁ ≥ 0, x₂ ≥ 0
Thus, the mathematical formulation of the LPP is
Maximize Z = 50 x₁ + 15 x₂
Subject to the constrains
x₁ + x₂ ≤ 60
5x₁ + 2x₂ ≤ 100
x₁, x₂ ≥ 0

Since both the decision variables x₁, x₂ are non-negative, the solution lies in the first quadrant of the plane.
Consider the equations x₁ – x₂ = –1 and – x₁ + x₂ = 0
x₁ – x₂ = –1 is a line passing through the points (0,1) and (–1,0)
–x₁ + x₂ = 0 is a line passing through the point (0,0)
Now we draw the graph satisfying the conditions x₁ – x₂ ≤ –1; –x₁ + x₂ ≤ 0 and x₁, x₂ ≥ 0

There is no common region(feasible region) satisfying all the given conditions. Hence the given LPP has no solution.

Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant of the plane.
Consider the equations 4x₁ + x₂ = 40 and 2x₁ + 3x₂ = 90
4x₁ + x₂ = 40 is a line passing through the points (0,40) and (10,0). Any point lying on or above the line 4x₁ + x₂ = 40 satisfies the constraint 4x₁ + x₂ ≥ 40.
2x₁ + 3x₂ = 90 is a line passing through the points (0,30) and (45,0). Any point lying on or above the line 2x₁ + 3x₂ = 90 satisfies the constraint 2x₁ + 3x₂ ≥ 90.
Draw the graph using the given constraints.

The feasible region is ABC (since the problem is of minimization type we are moving towards the origin.

The minimum value of Z occurs at B(3,28).
Hence the optimal solution is x₁ = 3, x₂ = 28 and Z_min = 127.

The required network for the above information.

The required network for the above information.

Since there is no common area between the lines x₁ + x₂ ≤ 1 and 5x₁ + 5x₂ ≥ 0

Since x₁ = 2.5, x₂ = 35 satisfies all the Constraints