i) variables:
Let x₁ and x₂ denote the product A and B respectively.
ii) Objective function:
Profit on x₁ of the product A = 3x₁
Profit on x₂ of the product B = 4x₂
Total profit = 3x₁ + 4x₂
Let Z = 3x₁ + 4x₂, which is the objective function. Since the total profit is to be maximized, we have to maximize Z = 3x₁ + 4x₂.
(iii) Constraints:
We make the following table from the given data

 x₁ + x₂ \(\le \) 450 [\(\because\) Machine M₁ is available for not more than 7 hrs 30 minutes]
 2x₁ + x₂ \(\le \) 600 [\(\because\) Machine M₂ is available for 10 hrs during any working day]
(iv) Non-negative restrictions:
Since the product A and product B are non-negative, we have x₁, x₂\(\ge \) 0.
Thus, we have the following linear programming model.
Maximize \(Z=3{ x }_{ 1 }+{ 4x }_{ 2 }\)
subject to the constraints 
\(x_{ 1 }+{ x }_{ 2 }\le 450\)
\(2x_{ 1 }+{ x }_{ 2 }\le 600\)
\(x_1,x_2\ge0\)

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

(i) variables: Let x₁ and x₂ represents the types of pen A and B respectively.
(ii) Objective function:
Profit on x₁ pens = 5x₁
Profit on x₂ pens = 3x₂
Total profit = 5x₁ + 3x₂
Let Z = 5x₁ + 3x₂ which is the objective function.
Since the total is to be maximized, we have to maximize Z = 5x₁ + 3x₂
(iii) Constraints:
The supply of raw material is sufficient only for 1000 pens per day.
2x₁ + x₂ ≤ 1000 [\(\because\) pen A is twice as that of pen B]
clips are available per day, for pen A: x₁ ≤ 400
clips are available per day, for pen B: x₂ ≤ 700
(iv) Non-negative restrictions: Since the number of pen A and pen B cannot be negative, we have x₁ ≥ 0, x₂ ≥ 0.
Thus, the mathematical formulation of the LPP is
Max Z = 5x₁ + 3x₂
Subject to the constraints
2x₁ + x₂ ≤ 1000
x₁ ≤ 400
x₂ ≤ 700
x₁, x₂ ≥ 0 (non-negative constraints)

(i) Variables: Let the mixture contains x₁ kg of food F₁ and x₂ kg of food F₂
(ii) Objective function: Cost of x₁ kg of food F₁ = 50 x₁
Cost of x₂ kg of food F₂ = 70x₂
The cost is to be minimized
Therefore minimize Z = 50 x₁ + 70x₂
(iii) Constraints:
We make the following table from the given data

4x₁ + 5x₂ ≥ 6 (since the mixture contains ‘atleast 6’ units of vitamin A, we have the inequality of the type ≥ )
6x₁ + 3x₂ ≥ 9 (since the mixture contains ‘atleast 9’ units of vitamin B, we have the inequality of the type ≥ )
(iv) Non-negative restrictions:
Since the number of kgs of vitamin A and vitamin B are non-negative, we have x₁, x₂ ≥ 0
Thus, we have the following linear programming model
Minimize Z = 50x₁ + 70x₂ subject to 4x₁+ 5x₂ ≥ 6
6x₁ + 3x₂ ≥ 9 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