Step 1: Formal Definition of Constraints

The objective function to be optimized is $Z = x + 2y$. The constraints are defined as follows:

• $C_1: x + 2y \geq 100$
• $C_2: 2x - y \leq 0 \implies y \geq 2x$
• $C_3: 2x + y \leq 200$
• $C_4: x \geq 0, y \geq 0$ (Non-negativity constraints)

Step 2: Determining Boundary Lines and Intersection Points

To identify the feasible region, we solve for the boundary lines by converting inequalities to equalities:

Intersection Calculations:

• Point A (Intersection of $L_1$ and $L_2$): $x + 2(2x) = 100 \implies 5x = 100 \implies x = 20, y = 40$. Point: $(20, 40)$.
• Point B (Intersection of $L_2$ and $L_3$): $2x + (2x) = 200 \implies 4x = 200 \implies x = 50, y = 100$. Point: $(50, 100)$.
• Point C (Intersection of $L_1$ and $L_3$): $x + 2y = 100$ and $2x + y = 200$. Multiply $L_1$ by 2: $2x + 4y = 200$. Subtract $L_3$: $3y = 0 \implies y = 0, x = 100$. Point: $(100, 0)$.

Step 3: Graphical Representation

Step 4: Evaluating the Objective Function at Corner Points

The feasible region is the polygon defined by vertices $(20, 40), (50, 100),$ and $(100, 0)$. We evaluate $Z = x + 2y$ at these points [Per the Corner Point Theorem]:

Step 5: Final Analysis

The value of $Z$ is 100 at both $(20, 40)$ and $(100, 0)$. Since the objective function is linear, $Z$ will be 100 at all points on the line segment connecting $(20, 40)$ and $(100, 0)$.

Final Solution: The minimum value of $Z$ is 100 at all points on the line segment joining $(20, 40)$ and $(100, 0)$; the maximum value of $Z$ is 250 at the point $(50, 100)$.

Step 1: Formal Representation of Constraints

The objective function to be maximized is $Z = -x + 2y$. The constraints are defined as follows:

• $L_1: x \geq 3$
• $L_2: x + y \geq 5$
• $L_3: x + 2y \geq 6$
• $L_4: y \geq 0$

Step 2: Identification of Boundary Lines and Intersections

To identify the feasible region, we convert inequalities into equations to find the boundary lines:

Determining Intersection Points:

• Intersection of $L_1 (x=3)$ and $L_2 (x+y=5)$: $3+y=5 \implies y=2$. Point: $(3, 2)$.
• Intersection of $L_1 (x=3)$ and $L_3 (x+2y=6)$: $3+2y=6 \implies y=1.5$. Point: $(3, 1.5)$.
• Intersection of $L_2 (x+y=5)$ and $L_3 (x+2y=6)$: Subtracting eqns: $(x+2y)-(x+y) = 6-5 \implies y=1$. Substituting $y=1$ into $x+y=5 \implies x=4$. Point: $(4, 1)$.

Step 3: Geometric Representation of the Feasible Region

Step 4: Evaluating the Corner Points

The feasible region is unbounded, extending toward infinity in the positive direction of $x$ and $y$. However, we test the vertices of the boundary:

Step 5: Analysis of Unbounded Region

Per the [Corner Point Theorem for unbounded regions], we check if any point in the feasible region yields a value of $Z$ greater than the maximum found at the vertices. For $Z = -x + 2y$, as $x \to \infty$ and $y \to \infty$ along the constraint $x+y=5$, $Z$ can increase indefinitely. However, check the constraint $Z > 1$: $-x + 2y > 1$. This region overlaps with the feasible region. Therefore, the function $Z = -x + 2y$ has no finite maximum value.

Final Solution: The Linear Programming Problem has no finite maximum value because the feasible region is unbounded and the objective function $Z$ can increase indefinitely.

Step 1: Formal Identification of Constraints and Objective Function

We aim to optimize the objective function $Z = 5x + 10y$ subject to the following system of linear inequalities:

• $L_1: x + 2y \leq 120$
• $L_2: x + y \geq 60$
• $L_3: x - 2y \geq 0$
• Non-negativity constraints: $x \geq 0, y \geq 0$

Step 2: Conversion to Equations for Boundary Plotting

To identify the feasible region, we define the boundary lines:

Step 3: Determination of the Feasible Region

The feasible region is the intersection of the half-planes defined by the constraints. [Per the Graphical Method for LPP], we test the origin $(0,0)$ or convenient points:

• For $x + 2y \leq 120$: Since $0 + 0 \leq 120$, the region includes the origin.
• For $x + y \geq 60$: Since $0 + 0 \ngeq 60$, the region is away from the origin.
• For $x - 2y \geq 0$: Testing $(60, 0)$, $60 - 0 \geq 0$ holds true. The region lies below the line $y = 0.5x$.

Step 4: Calculation of Corner Points

The vertices of the feasible region are found by solving the intersection of the boundary lines:

1. Point A (Intersection of $L_2$ and $L_3$): $x+y=60$ and $x=2y \Rightarrow 3y=60 \Rightarrow y=20, x=40$. Point: $(40, 20)$.
2. Point B (Intersection of $L_1$ and $L_3$): $x+2y=120$ and $x=2y \Rightarrow 4y=120 \Rightarrow y=30, x=60$. Point: $(60, 30)$.
3. Point C (Intersection of $L_1$ and $L_2$): $x+2y=120$ and $x+y=60 \Rightarrow y=60, x=0$. Point: $(0, 60)$.

Step 5: Evaluating the Objective Function $Z = 5x + 10y$

Final Solution: The minimum value of $Z$ is $400$ at point $(40, 20)$, and the maximum value of $Z$ is $600$ at all points on the segment joining $(60, 30)$ and $(0, 60)$.

Step 1: Establishing Constraints and Feasible Region

We are given the objective function $Z = -3x + 4y$ which is to be minimized, subject to the following system of linear inequalities:

• $x + 2y \leq 8$ [Constraint 1]
• $3x + 2y \leq 12$ [Constraint 2]
• $x \geq 0, y \geq 0$ [Non-negativity constraints]

Step 2: Identifying Boundary Lines

To determine the feasible region, we solve for the boundary lines by converting inequalities to equalities:

1. Line 1: $x + 2y = 8$. Intercepts: $(8, 0)$ and $(0, 4)$.
2. Line 2: $3x + 2y = 12$. Intercepts: $(4, 0)$ and $(0, 6)$.

Step 3: Determining the Intersection Point

Solving the system of linear equations to find the vertex where the two constraint lines meet:

$x + 2y = 8$ ... (Eq. i)

$3x + 2y = 12$ ... (Eq. ii)

Subtracting (Eq. i) from (Eq. ii): $(3x - x) + (2y - 2y) = 12 - 8 \implies 2x = 4 \implies x = 2$.

Substituting $x = 2$ into (Eq. i): $2 + 2y = 8 \implies 2y = 6 \implies y = 3$.

The point of intersection is $(2, 3)$.

Step 4: Evaluating the Objective Function at Vertices

The feasible region is bounded by the vertices $(0, 0)$, $(4, 0)$, $(2, 3)$, and $(0, 4)$. We evaluate $Z = -3x + 4y$ at each point:

Step 5: Determining the Minimum Value

Comparing the values obtained: $0, -12, 6, 16$. The minimum value is $-12$, which occurs at the point $(4, 0)$. [Per the Corner Point Theorem for Linear Programming].

Final Solution: The minimum value of $Z$ is $-12$ at the point $(4, 0)$.

Step 1: Formalization of Constraints

The objective function to be minimized is $Z = x + 2y$. The constraints are defined as follows:

• Constraint 1: $2x + y \geq 3$
• Constraint 2: $x + 2y \geq 6$
• Non-negativity constraints: $x \geq 0, y \geq 0$

Step 2: Determination of Boundary Lines

To plot the feasible region, we convert the inequalities into equations:

1. Line $L_1$: $2x + y = 3$.
   • If $x = 0$, $y = 3$. Point: $(0, 3)$
   • If $y = 0$, $x = 1.5$. Point: $(1.5, 0)$
2. Line $L_2$: $x + 2y = 6$.
   • If $x = 0$, $y = 3$. Point: $(0, 3)$
   • If $y = 0$, $x = 6$. Point: $(6, 0)$

Step 3: Identification of Intersection Points and Corner Points

Solving the system of equations $2x + y = 3$ and $x + 2y = 6$ simultaneously:

Multiply the first equation by 2: $4x + 2y = 6$. Subtracting the second equation ($x + 2y = 6$):

$(4x - x) = 6 - 6 \implies 3x = 0 \implies x = 0$. Substituting $x=0$ into $2x + y = 3$ yields $y = 3$.

The region is bounded by the axes and these lines. Testing $(1, 2)$ shows that the shaded area lies to the right/above the lines.

Step 4: Evaluating the Objective Function at Corner Points

Since the region is unbounded in the first quadrant, we check the corner points identified by the boundary lines:

Step 5: Analysis of Multi-point Optimality

At both $(0, 3)$ and $(6, 0)$, $Z = 6$. We check if any point on the line segment connecting $(0, 3)$ and $(6, 0)$ provides a value less than 6. The line segment lies on $x + 2y = 6$. For any point $(x, y)$ on this segment, $x + 2y = 6$. Thus, $Z = 6$ for all points on the segment $x + 2y = 6$ between $x=0$ and $x=6$.

Since the problem asks for the minimum, and any point on the segment gives $Z=6$ (which is consistent with the constraint $x + 2y \geq 6$), the minimum occurs at infinitely many points along this boundary.

Final Solution: The minimum value of $Z$ is $6$, which occurs at all points on the line segment joining $(0, 3)$ and $(6, 0)$.

Step 1: Establishing the Objective Function and Constraints

We are tasked with maximizing the objective function $Z = 5x + 3y$ subject to the following non-negative and linear constraints:

• Constraint 1: $3x + 5y \leq 15$
• Constraint 2: $5x + 2y \leq 10$
• Non-negativity: $x \geq 0, y \geq 0$

Step 2: Determining Boundary Lines and Intercepts

To identify the feasible region, we convert the inequalities into equations to find the line intercepts:

Step 3: Finding the Intersection Point of Constraints

Solve the system of linear equations to find the intersection of the two boundary lines [By Algebraic Elimination Method]:

Multiply the first equation by 2 and the second by 5:

$6x + 10y = 30$

$25x + 10y = 50$

Subtract the first from the second: $19x = 20 \implies x = \frac{20}{19} \approx 1.05$

Substitute $x$ into $5x + 2y = 10$: $5(\frac{20}{19}) + 2y = 10 \implies \frac{100}{19} + 2y = \frac{190}{19} \implies 2y = \frac{90}{19} \implies y = \frac{45}{19} \approx 2.37$

Step 4: Graphical Representation

Step 5: Corner Point Evaluation

The feasible region is bounded by the vertices $(0,0)$, $(2,0)$, $(\frac{20}{19}, \frac{45}{19})$, and $(0,3)$ [Per the Corner Point Theorem].

Final Solution: The maximum value of $Z$ is $\frac{235}{19}$ occurring at the point $(\frac{20}{19}, \frac{45}{19})$.

Step 1: Conversion of Inequalities to Linear Equations

To determine the boundary lines of the feasible region, we treat the given constraints as equations:

• Constraint 1: $x + 3y = 3$
• Constraint 2: $x + y = 2$
• Non-negativity constraints: $x = 0, y = 0$

Step 2: Determination of Intercepts

We calculate the coordinates for the boundary lines:

Step 3: Intersection Point of Boundary Lines

To find the vertex of the feasible region, solve the system of equations:

$x + 3y = 3$

$x + y = 2$

Subtracting the second equation from the first: $(x + 3y) - (x + y) = 3 - 2 \implies 2y = 1 \implies y = 0.5$.

Substituting $y = 0.5$ into $x + y = 2$: $x + 0.5 = 2 \implies x = 1.5$.

The intersection point is $(1.5, 0.5)$.

Step 4: Visualizing the Feasible Region

Step 5: Evaluating the Objective Function

The feasible region is unbounded and defined by the vertices formed by the constraints $x+3y \geq 3$ and $x+y \geq 2$ in the first quadrant. The corner points are:

• $A(0, 2)$
• $B(1.5, 0.5)$
• $C(3, 0)$

Step 6: Conclusion on Unbounded Region

For an unbounded region, we check if $3x + 5y < 7$ has any common points with the feasible region. Setting $3x + 5y < 7$, we find no intersection, as the minimum value $7$ at $(1.5, 0.5)$ satisfies the constraints and provides the absolute minimum.

Final Solution: The minimum value of $Z$ is $7$ at the point $(1.5, 0.5)$.

Step 1: Identifying Constraints and Objective Function

We are given the objective function $Z = 3x + 2y$ to be maximized, subject to the following system of linear inequalities (constraints):

• Constraint 1: $x + 2y \leq 10$
• Constraint 2: $3x + y \leq 15$
• Non-negativity: $x \geq 0, y \geq 0$

Step 2: Determining Boundary Line Intercepts

To plot the lines, we determine the intercepts for each equation:

Step 3: Calculating the Point of Intersection

To find the intersection of the two lines, we solve the system of linear equations:

1. $x + 2y = 10$
2. $3x + y = 15 \implies y = 15 - 3x$

Substituting $(2)$ into $(1)$:

Step 4: Visualizing the Feasible Region

Step 5: Evaluating the Objective Function at Corner Points

The feasible region is bounded by the vertices $(0, 0)$, $(5, 0)$, $(4, 3)$, and $(0, 5)$. We test these points in $Z = 3x + 2y$:

Comparing the values, the maximum value of $Z$ is $18$.

Final Solution: The maximum value of $Z$ is 18 at the point $(4, 3)$.

Step 1: Formalizing the Mathematical Model

We are given the objective function $Z = x + y$, which is to be maximized subject to the following system of linear inequalities (constraints):

• Constraint 1: $x - y \leq -1$
• Constraint 2: $-x + y \leq 0 \implies y \leq x$
• Non-negativity constraints: $x \geq 0, y \geq 0$

Step 2: Analyzing the Feasible Region

To determine the region, we analyze the boundary lines individually:

1. For $x - y = -1$: If $x=0$, then $y=1$. If $y=0$, then $x=-1$. The line passes through $(0, 1)$ and $(-1, 0)$. Testing the origin $(0,0)$: $0 - 0 \leq -1$ is $0 \leq -1$, which is false. Therefore, the region for this constraint lies on the side away from the origin.
2. For $-x + y = 0$ (or $y = x$): This is a line passing through the origin $(0,0)$ at a $45^\circ$ angle. Testing point $(1, 0)$: $-1 + 0 \leq 0 \implies -1 \leq 0$, which is true. The region lies below the line $y=x$.

Step 3: Evaluating the Feasible Region Existence

Observe the constraints simultaneously:

• $y \geq x + 1$ (from Constraint 1)
• $y \leq x$ (from Constraint 2)

These two inequalities imply $x + 1 \leq y \leq x$, which simplifies to $x + 1 \leq x$, or $1 \leq 0$. [By the law of trichotomy and basic algebraic inconsistency].

Step 4: Geometric Representation

Step 5: Logical Conclusion

Since the inequality $x - y \leq -1$ requires $y \geq x + 1$ and the inequality $-x + y \leq 0$ requires $y \leq x$, there exists no set of points $(x, y)$ that satisfies both conditions simultaneously. The region defined by the intersection of these half-planes is an empty set (null set). Consequently, there is no feasible region.

Final Solution: The Linear Programming Problem has no feasible solution, and therefore, no maximum value for $Z$ exists.

Step 1: Identification of Constraints and Objective Function

Given the objective function: $Z = 3x + 4y$ (to be maximized).

The constraints are defined as:

• Constraint 1: $x + y \leq 4$
• Non-negativity Constraints: $x \geq 0, y \geq 0$

Step 2: Determining Feasible Region Boundary Lines

We convert the inequality constraints into equations to determine the boundary lines of the feasible region:

1. For $x + y = 4$:
   • If $x = 0$, then $y = 4$. Point: $(0, 4)$
   • If $y = 0$, then $x = 4$. Point: $(4, 0)$

Step 3: Graphical Representation

Step 4: Identifying Corner Points of the Feasible Region

The feasible region is the triangle formed by the vertices $(0, 0)$, $(4, 0)$, and $(0, 4)$ [By the definition of the solution set for $x, y \geq 0$ and the half-plane $x + y \leq 4$].

Step 5: Evaluating the Objective Function $Z = 3x + 4y$

Step 6: Conclusion based on Fundamental Theorem of LPP

The Fundamental Theorem of Linear Programming states that the maximum value of the objective function occurs at a corner point of the feasible region. Comparing the values obtained at the vertices: $0, 12, 16$. The maximum value is $16$.

Final Solution: The maximum value of $Z$ is $16$ at the point $(0, 4)$.