i) variables:
Let x₁, x₂ be the number of pills of size A and B respectively.
ii) Objective function:
The cost is to be minimized
∴ minimize Z = x₁ + x₂
iii) Constraints:
The data can be summarised as follows

\(2 x_1+x_2 \geq 12 \ [\because\) Chemical is found by users that it requires at 12 mgs of aspirin]
\(5 x_1+8 x_2 \geq 74 \ [\because\) Chemical is found by users that it requires at 74 mgs of Bicarbonate]
\(x_1+6 x_2 \geq 24 \ [\because\) Chemical is found by users that it requires at 24 mgs of codeine]
(iii) Non-negative restrictions:
Since the number of pills of size A and B are non negative, we have x₁ + x₂  ​\(\le \)​  0
Thus, we have the following linear programming model.
Minimize  \(z=x_1+x_2\)   
Subject to constraints 
\(\\ 2x_{ 1 }+{ x }_{ 2 }\ge 12\)
\(5x_{ 1 }+8{ x }_{ 2 }\ge 74\)
\(x_{ 1 }+6{ x }_{ 2 }\ge 24\)
\({ x }_{ 1 },{ x }_{ 2\ }\le \ 0\)

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

(i) variables:
Let x₁ and x₂ denote the types of product A and B respectively.
(ii) Objective function:
Profit on x₁ of the product A = 30x₁ 
profit on x₂ of product B = 40x₂
Total profit = 30x₁+ 40x₂
Since the total profit is to be maximized, we have to maximize z = 30x₁ + 40x₂
(iii) Constraints:
60x₁ + 120x₂ ≤ 12,000
8x₁ + 5x₂ ≤ 600
3x₁ + 4x₂ ≤ 500
(iv) Non-negative restrictions: Since the number of products A, B cannot be negative, we have x₁, x₂ ≥ 0
Max Z = 30x₁ + 40x₂
subject to the constraints
60x₁ + 120x₂ ≤ 12,000
8x₁ + 5x₂ ≤ 600
3x₁ + 4x₂ ≤ 500
x₁, x₂ ≥ 0

(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, x₂ and x₃ be the number of units of products P₁, P₂ and P₃ to be produced.
(ii) Objective function: Profit on x₁ units of the product P₁ = 20 x₁
Profit on x₂ units of the product P₂ = 25 x₂
Profit on x₃ units of the product P₃ = 15 x₃
Total profit = 20 x₁ + 25 x₂ + 15 x₃
Since the total profit is to be maximized, we have to maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Constraints: 6x₁ + 3x₂ + 12x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
Non-negative restrictions: Since the number of units of the products A, B and C cannot be negative, we have x₁, x₂, x₃ ≥ 0
Thus, we have the following linear programming model.
Maximize Z = 20 x₁ + 25 x₂ + 15 x₃
Subject to 6 x₁ + 3 x₂ + 12 x₃ ≤ 200
2x₁ + 5x₂ + 4x₃ ≤ 350
x₁ + 2x₂ + x₃ ≤ 100
x₁, x₂, x₃ ≥ 0