Since the decision variables are non-negative, the solution lies in the I-quadrant of the plane.Consider the equations
\({ x }_{ 1 }+3{ x }_{ 2 }=3\)

\({ x }_{ 1 }+{ x }_{ 2 }=2\)

The feasible region is ABC and its co-ordinates are A(3, 0) C(0,2) and B its the point of intersection of the lines
\({ x }_{ 1 }+3{ x }_{ 2 }=3\)  ...(1) and  \({ x }_{ 1 }+{ x }_{ 2 }=2\)   ...(2)
Verification of B:
\((1)\Rightarrow { x }_{ 1 }+3{ x }_{ 2 }=3\)
\( \quad \quad (-)\quad (-)\quad \quad (-)\)
\((2)\Rightarrow { x }_{ 1 }+{ x }_{ 2 }=2\)
\( \quad -----------\)
\(2{ x }_{ 2 }=1\Rightarrow { x }_{ 2 }=\frac { 1 }{ 2 } \)
\( From(2),\ { x }_{ 1 }+\frac { 1 }{ 2 } =2\ \Rightarrow { x }_{ 1 }=2-\frac { 1 }{ 2 } \Rightarrow \frac { 3 }{ 2 } \therefore B\quad is\quad \left( \frac { 3 }{ 2 } ,\frac { 1 }{ 2 } \right) \)

Minimum of Z occurs at B(3/2, 1/2)
Hence, the solution is x₁= 3/2, x₂ = 1/2 and Z_min = 7

Since the decision variables x₁ ,x₂ are non-negative, the solution lies in the I quadrant of the plane.
Consider the equations
\(-{ x }_{ 1 }+3{ x }_{ 2 }=10\)

\({ x }_{ 1 }+{ x }_{ 2 }=6\)

\({ x }_{ 1 }{ -x }_{ 2 }=2\)

The feasible region is OABCD and its co-ordinates are O(0, 0)A(2, 0) B(4, 2) C(2, 4) and D(0, 10/3)

Maximum of Z occurs at\(D\left( 0,\frac { 10 }{ 3 } \right) \). Hence, the solution is \({ x }_{ 1 }=0,{ x }_{ 2 }=\frac { 10 }{ 3 } \quad and\quad { Z }_{ max }=\frac { 20 }{ 3 } \)

Using the precedence relationships and following the rules of network construction, the required network diagram 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, 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

EFT and LFT are same in the activities.
1 - 2, 2 - 3, 3 - 4 and 4 - 5
Hence, the critical path is 1 - 2 - 3 - 4 - 5 and the duration of project completion is 48 days .

Let x₁cakes of the one kind and x₂ cakes of another kind are made. Let Z be the maximum number of cakes

Thus, the mathematical formulation of the LPP is Maximize Z = x₁+ x₂
Subject to the constraints
\(200{ x }_{ 1 }+100{ x }_{ 2 }\le 5000\)
\(25{ x }_{ 1 }+50{ x }_{ 2 } \le 1000\)
\({ x }_{ 1 },{ x }_{ 2 }\ge 0\)
Consider the equations
\(200{ x }_{ 1 }+100{ x }_{ 2 }= 5000\)

\(25{ x }_{ 1 }+50{ x }_{ 2 }=1000\)

The feasible region is OABC and its co-ordinates are O(0, 0) A(25, 0) C(O, 20) and B is the point of intersection of the lines
200x₁ + 100x₂ = 1000 .... (1)
and 25x₁ + 50x₂ = 1000 ... (2)
Verification of B:
\((1) \Rightarrow 200{ x }_{ 1 }+100{ x }_{ 2 }=5000\)
\( (-)\quad \quad \quad (-)\quad \quad (-)\)
\( (2)\times 5\Rightarrow 50{ x }_{ 1 }+100{ x }_{ 2 }=2000\)
\(------------------\)
\(150x_{ 1 }=3000 \Rightarrow { x }_{ 1 }=20\)
\(From(2), 25(20)+50{ x }_{ 2 }=1000\Rightarrow 500+50{ x }_{ 2 }=1000 \Rightarrow 50{ x }_{ 2 }=500\)
\(\Rightarrow { x }_{ 2 }=10\)
\(\therefore B\ is\ (20,10)\)

Maximum of Z occurs at B(20, 10)
Hence, the solution is x₁ = 20, x₂ = 10 and Z_max = 30.

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.

We find the feasible region using the given conditions.
Since both the decision variables x₁ and x₂ are non-negative, the solution lies in the first quadrant of the plane.
Write all the inequalities of the constraints in the form of equations.
Therefore we have the lines
x₁ + x₂ = 30; x₂ = 12; x₁ = 20
x₁ + x₂ = 30 is a line passing through the points (0,30) and (30,0)
x₂ = 12 is a line parallel to x₁–axis
x₁ = 20 is a line parallel to x₂–axis.
The feasible region satisfying all the conditions x₁ + x₂ ≤ 30; x₂ ≤ 12 ; x₁ ≤ 20 and x₁, x₂ ≥ 0 is shown in the following graph.

The feasible region satisfying all the conditions is OABCD.
The co-ordinates of the points are O(0,0); A(20,0); B(20,10); C(18,12) and D(0,12).

Maximum value of Z occurs at C. Therefore the solution is x₁ = 18 , x₂ = 12, Z_max = 72.

(i) Variables:
Let x₁, x₂ represents the first and second type of bonds respectively.
(ii) Objective function:
Let Z be the maximum return
\(\therefore \quad Z=\frac { 8 }{ 100 } { x }_{ 1 }+\frac { 10 }{ 100 } { x }_{ 2 } \Rightarrow Z=0.08{ x }_{ 1 }+0.1{ x }_{ 2 }\)
(iii) Constraints: 
\({ x }_{ 1 }+{ x }_{ 2 } \le 70,000\)
\({ x }_{ 1 } \ge 10,000\)
\( { x }_{ 2 } \le 30,000\)
(iv) Non-negative restrictions:
Since the number of first and second type of bonds cannot be negative,x₁, x₂ ≥ 0.
Hence, the mathematical formulation of the LPP is maximize \(Z=0.08{ x }_{ 1 }+0.1{ x }_{ 2 }\)
Subject to the constraints 
\({ x }_{ 1 }+{ x }_{ 2 } \le 70,000\)
\( { x }_{ 1 }\ge 10,000\)
\({ x }_{ 2 }\le 30,000\)
and x₁, x₂ ≥ 0.

(i) Variables:
Let x₁, x₂ represent bags of brand P and brand Q. 
(ii) Objective function:
Let Z be the amount of nitrogen. Since the amount of nitrogen is to be minimized, we have minimize Z = 3x + 3.5y
(iii) Constraints:
For Phosphoric acid,\({ x }_{ 1 }+2{ x }_{ 2 }\ge 240\)
For Potash, \({ 3x }_{ 1 }+{ 1.5x }_{ 2 }\ge 270\)
For Chlorine, \({ 1.5x }_{ 1 }+2{ x }_{ 2 }\ge 310\)
(iv) Non-negative restrictions:
Since the number of bags of brand P and Q, cannot be negative,\({ x }_{ 1 },{ x }_{ 2 }\ge 0\).
Here Mathematical form of the LPP is Minimize Z = 3x₁ + 3.5x₂ 
Subject to the constraints
\({ x }_{ 1 }+2{ x }_{ 2 }\ge 240\)
\({ 3x }_{ 1 }+{ 1.5x }_{ 2 }\ge 270\)
\( { 1.5x }_{ 1 }+2{ x }_{ 2 }\ge 310\)
and x₁, x₂ ≥ 0.

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