Linear Programmig Problem

1. Every LPP admits an optimal solution.
   This statement is false. An LPP may not have an optimal solution if it is unbounded or infeasible.
2. Every LPP admits a unique solution.
   This statement is false. An LPP can have multiple optimal solutions, especially along an edge of the feasible region.
3. The optimal value does not occur at a corner point of the feasible region only.
   This statement is false. In LPP, if an optimal solution exists, it has to occur at a vertex or along an edge formed by vertices of the feasible region.
4. If a LPP admits an optimal solution at two points, then it has an optimal solution at an infinite number of points.
   This statement is true. If two distinct points in the feasible region are optimal, then the entire line segment connecting these points is also optimal, implying infinite optimal solutions along the segment.

To solve for the relationship between a and b given that the maximum value of Z = ax + by occurs at both (2, 4) and (4, 0), follow these steps:

1. Calculate the value of Z at point (2, 4):
   Z₁ = a(2) + b(4) = 2a + 4b.
2. Calculate the value of Z at point (4, 0):
   Z₂ = a(4) + b(0) = 4a.
3. Since both points provide the maximum value of Z, set Z₁ equal to Z₂:
   2a + 4b = 4a.
4. Simplify the equation:
   4b = 4a - 2a
5. Simplify further:
   4b = 2a
6. Divide both sides by 2:
   2b = a
7. Rearrange to find the relationship:
   a = 2b

This shows that the correct relationship between a and b is a = 2b.

• simplifies to

• For ,
• For ,

• For ,
• For ,

• Intersection of and :

• Substitute into :

• At ,
• At ,
• At ,
• At ,

1. Calculate at (0,10) :
   
2. Calculate at (0,20) :
   
3. Calculate at (5,5) :
   
4. Calculate at (15,15) :
   

1. A - III: The common region determined by all the linear constraints of a LPP is the feasible region, which is where all constraints overlap and potential solutions exist.
2. B - I: A point in the feasible region that is the intersection of two boundary lines is a corner point. These points are crucial as they may contain the optimal solution.
3. C - IV: The feasible region for an LPP is always a convex polygon when linear inequalities are involved, since this ensures that for any two points within it, the line segment connecting them is also within the region.
4. D - II: The constraints ensure that the variables are non-negative, as all solutions must lie in the first quadrant.

• Objective Function: Maximize
• Constraints:

1. Identify corner points of the feasible region by solving the system of linear inequalities.
2. Find intersections of boundary lines:

• Subtract the first equation from the second:
  
  
• Substitute into :
  
  

1. For :
2. For :
3. For : Invalid, as it doesn't satisfy .
4. For : Invalid, as it doesn't satisfy .

To find the maximum value of subject to the constraint and , we need to determine the feasible region and evaluate the objective function at the vertices of this region. The constraint can be rewritten as .

1. Graphical Representation:

Let's plot the line and consider the constraints:

The feasible region is a triangle with vertices at , , and .

2. Evaluate the Objective Function at Each Vertex:

3. Determine the Maximum Value:

The maximum value of is at the vertex .

Therefore, the maximum value of under the given constraints is .

• Step 1: Intersection Points on the Axes
  
  • For :
    • If , .
    • If , .
  • For :
    • If , .
    • If , .
• Step 2: Intersection of the Constraint Lines
  • Solve and simultaneously:
    • Multiply the second equation by 2: .
    • Subtract the first equation: .
    • Substitute into : .

    The intersection point is .

• Step 3: Feasible Region and Basic Feasible Solutions
  Considering non-negativity constraints, the feasible region is bounded by the lines intersecting at , , , as well as along the axes (subject to non-negativity). Therefore, basic feasible solutions include points forming the corners or intersections in this region: , , .

• At (0, 0):
• At (2, 0):
• At
• At (0, 3):

• Option 1:
  x≤5, y≤10
  This constraint is linear because both x and y are to the power of one.
• Option 2:
  2x+3y≤60
  This constraint is linear. The coefficients of the variables and the constant do not affect linearity.
• Option 3:
  x+2y≤40
  This is also a linear constraint for the same reason as above.
• Option 4:
  x²+y≤50
  This constraint is non-linear because x is squared. In a valid linear programming constraint, all variable terms should be of degree one.

• A LPP is an optimization problem.
  This statement is true because LPP focuses on optimizing (maximizing or minimizing) a linear objective function.
• In a LPP, the constraints and objective function are linear.
  True, as by definition, both the objective function and constraints are linear expressions.
• Max z=xy+2x+3y can be a valid objective function for a LPP.
  This statement is false. In LPP, the objective function must be a linear equation. Here, 'xy' introduces non-linearity, making it invalid for LPP.
• The optimal solution of a LPP is at one of the corner points of basic feasible solution.
  True. For LPPs, if an optimal solution exists, it is found at a vertex (corner point) of the feasible region.

• Wood:
• Labour:
• Non-negativity:

• x + y ≤ 300: Total number of passengers cannot exceed 300.
• x ≥ 40: At least 40 seats are reserved for executive class.
• y ≥ 5x: At least 5 times as many passengers must prefer economy class over executive class.

1. From y ≥ 5x, and x + y ≤ 300, substitute y in the inequality:
   x + 5x ≤ 300 → 6x ≤ 300 → x ≤ 50
2. Since x ≥ 40, x can be between 40 and 50.
3. Now check boundaries to find the maximum profit:
4. When x = 40, y = 300 - x = 300 - 40 = 260. Then calculate Profit: 1200(40) + 800(260) = ₹2,40,000.
5. When x = 50, y = 300 - 50 = 250. Then calculate Profit: 1200(50) + 800(250) = ₹2,60,000.

• Tailor A earns ₹150/day and can stitch 6 shirts and 4 pants per day.
• Tailor B earns ₹200/day and can stitch 10 shirts and 4 pants per day.

• At :
• At :

The solution to the given linear programming problem (LPP) involves determining the feasible region defined by the intersection of the constraints. We have the following constraints for the LPP:

To find the feasible region, we need to plot these constraints on a coordinate plane and look for the intersection that satisfies all the conditions:

1. For the constraint , rewrite as . This represents the region above the line .
2. The inequality simplifies to . This represents the region below the line .
3. Constraints and restrict the region to the first quadrant.

Combining all these, the feasible region is where these shaded areas overlap in the first quadrant. This region is bounded by the intersection of the lines , , the y-axis ( ), and the x-axis ( ).
The attached image accurately represents this feasible region, which shows the feasible area created by these constraints.