CBSE Class 12 Mathematics HOTs Linear Programming Set 02

Question. For a LPP, find min \( Z = 5x + 3y \) (where \( Z \) is the objective function) for the feasible region shaded in the given figure.
Answer: From the given graph, the corner points of the shaded feasible region are \( A(3,2) \), \( B(0,5) \), and \( C(0,3) \).
Evaluating the objective function \( Z = 5x + 3y \) at these corner points:

• At \( A(3,2) \): \( Z = 5(3) + 3(2) = 15 + 6 = 21 \)
• At \( B(0,5) \): \( Z = 5(0) + 3(5) = 15 \)
• At \( C(0,3) \): \( Z = 5(0) + 3(3) = 9 \) (Minimum)

Hence, the minimum value of \( Z \) is \( 9 \) at the point \( C(0,3) \).

Question. Maximise the following linear programming problem graphically:
\( Z = 2x+3y \)
Subject to the constraints:
\( x+2y \le 10, 2x+y \le 14 \) and \( x \ge 0, y \ge 0 \).
Answer: To solve the LPP graphically, we first convert the inequations into equations:

1. \( x + 2y = 10 \)
   This line passes through \( (10,0) \) and \( (0,5) \). On putting \( (0,0) \) in \( x + 2y \le 10 \), we get \( 0 \le 10 \) (which is true), so the half-plane contains the origin.
2. \( 2x + y = 14 \)
   This line passes through \( (7,0) \) and \( (0,14) \). On putting \( (0,0) \) in \( 2x + y \le 14 \), we get \( 0 \le 14 \) (which is true), so the half-plane contains the origin.

Since \( x \ge 0, y \ge 0 \), the feasible region lies entirely in the first quadrant.
Solving \( x + 2y = 10 \) and \( 2x + y = 14 \) simultaneously, we obtain the point of intersection \( B(6,2) \).
Thus, the corner points of the bounded feasible region are \( O(0,0) \), \( A(7,0) \), \( B(6,2) \), and \( D(0,5) \).

Evaluating the objective function \( Z = 2x + 3y \) at these corner points:

• At \( O(0,0) \): \( Z = 2(0) + 3(0) = 0 \)
• At \( A(7,0) \): \( Z = 2(7) + 3(0) = 14 \)
• At \( B(6,2) \): \( Z = 2(6) + 3(2) = 18 \) (Maximum)
• At \( D(0,5) \): \( Z = 2(0) + 3(5) = 15 \)

Hence, the maximum value of \( Z \) is \( 18 \) at the point \( B(6,2) \).

Question. Maximise the following linear programming problem graphically:
\( Z = 8000x + 12000y \)
Subject to the constraints:
\( 3x + 4y \le 60, x + 3y \le 30 \) and \( x \ge 0, y \ge 0 \).
Answer: Let us convert the inequalities into equations:

1. \( 3x + 4y = 60 \)
   This line passes through \( (20,0) \) and \( (0,15) \). On putting \( (0,0) \) in the inequality, we get \( 0 \le 60 \) (which is true), so the half-plane contains the origin.
2. \( x + 3y = 30 \)
   This line passes through \( (30,0) \) and \( (0,10) \). On putting \( (0,0) \) in the inequality, we get \( 0 \le 30 \) (which is true), so the half-plane contains the origin.

Since \( x \ge 0, y \ge 0 \), the feasible region lies in the first quadrant.
Solving the equations \( 3x + 4y = 60 \) and \( x + 3y = 30 \) simultaneously, we get the intersection point \( B(12,6) \).
Thus, the corner points of the bounded feasible region are \( O(0,0) \), \( A(20,0) \), \( B(12,6) \), and \( C(0,10) \).

Evaluating the objective function \( Z = 8000x + 12000y \) at these corner points:

• At \( O(0,0) \): \( Z = 0 \)
• At \( A(20,0) \): \( Z = 8000(20) + 12000(0) = 160000 \)
• At \( B(12,6) \): \( Z = 8000(12) + 12000(6) = 96000 + 72000 = 168000 \) (Maximum)
• At \( C(0,10) \): \( Z = 8000(0) + 12000(10) = 120000 \)

Hence, the maximum value of \( Z \) is \( 168000 \) when \( x = 12 \) and \( y = 6 \).

Question. Maximise the following linear programming problem graphically:
\( Z = x + y \)
Subject to the constraints:
\( x - y \le -1, -x + y \le 0 \) and \( x, y \ge 0 \).
Answer: Let us plot the constraints as lines:

1. \( x - y = -1 \implies y - x = 1 \)
   This line passes through \( (0,1) \) and \( (-1,0) \). On putting \( (0,0) \), we get \( 0 \le -1 \) (which is false), so the half-plane does not contain the origin. Thus, the region is \( y - x \ge 1 \).
2. \( -x + y = 0 \implies y = x \)
   This line passes through \( (0,0) \) and \( (1,1) \). On putting \( (1,0) \), we get \( -1 \le 0 \) (which is true), so the half-plane is the region below the line \( y = x \).

Since the regions defined by \( y \ge x + 1 \) and \( y \le x \) do not overlap anywhere in the first quadrant (\( x, y \ge 0 \)), there is no common region.
Hence, there is no feasible region, which means \( Z \) has no maximum value.

Question. Maximise the following linear programming problem graphically:
\( Z = -x + 2y \)
Subject to the constraints:
\( x \ge 3, x + y \ge 5, x + 2y \ge 6 \) and \( x, y \ge 0 \).
Answer: We plot the boundary lines of the given constraints:

1. \( x = 3 \): A vertical line passing through \( (3,0) \). The region \( x \ge 3 \) lies to the right of this line.
2. \( x + y = 5 \): This line passes through \( (5,0) \) and \( (0,5) \). Since \( x + y \ge 5 \), the region lies away from the origin.
3. \( x + 2y = 6 \): This line passes through \( (6,0) \) and \( (0,3) \). Since \( x + 2y \ge 6 \), the region lies away from the origin.

Since \( x, y \ge 0 \), the region lies in the first quadrant.
Evaluating the intersections of these boundary lines:

• The line \( x = 3 \) intersects \( x + y = 5 \) at \( C(3,2) \).
• The line \( x + y = 5 \) intersects \( x + 2y = 6 \) at \( B(4,1) \).
• The line \( x + 2y = 6 \) intersects the x-axis at \( A(6,0) \).

Thus, the corner points of the unbounded feasible region are \( A(6,0) \), \( B(4,1) \), and \( C(3,2) \).

Evaluating the objective function \( Z = -x + 2y \) at these corner points:

• At \( A(6,0) \): \( Z = -6 + 2(0) = -6 \)
• At \( B(4,1) \): \( Z = -4 + 2(1) = -2 \)
• At \( C(3,2) \): \( Z = -3 + 2(2) = 1 \) (Maximum)

Since the feasible region is unbounded, \( Z = 1 \) may or may not be the maximum value. We draw the graph of the inequality \( -x + 2y > 1 \).
We observe that the open half-plane \( -x + 2y > 1 \) has points in common with the feasible region (for example, the point \( (3,3) \) belongs to the feasible region and yields \( Z = 3 > 1 \)).
Hence, \( Z \) has no maximum value.

Question. Maximise the following linear programming problem graphically:
\( Z = 105x + 90y \)
Subject to the constraints:
\( x + y \le 50, 2x + y \le 80, x \ge 20 \) and \( x \ge 0, y \ge 0 \).
Answer: Let us convert the inequalities into equations:

1. \( x + y = 50 \)
   Passing through \( (50,0) \) and \( (0,50) \). For \( (0,0) \), \( 0 \le 50 \) is true, so the half-plane is towards the origin.
2. \( 2x + y = 80 \)
   Passing through \( (40,0) \) and \( (0,80) \). For \( (0,0) \), \( 0 \le 80 \) is true, so the half-plane is towards the origin.
3. \( x = 20 \)
   A vertical line passing through \( (20,0) \). Since \( x \ge 20 \), the region lies to the right of this line.

Since \( x, y \ge 0 \), the feasible region lies in the first quadrant.
The corner points of the bounded feasible region are:

• \( A(20,30) \) (the intersection of \( x = 20 \) and \( x + y = 50 \))
• \( B(30,20) \) (the intersection of \( x + y = 50 \) and \( 2x + y = 80 \))
• \( C(40,0) \) (the intersection of \( 2x + y = 80 \) with the x-axis)
• \( D(20,0) \) (the intersection of \( x = 20 \) with the x-axis)

Evaluating the objective function \( Z = 105x + 90y \) at these corner points:

• At \( A(20,30) \): \( Z = 105(20) + 90(30) = 2100 + 2700 = 4800 \)
• At \( B(30,20) \): \( Z = 105(30) + 90(20) = 3150 + 1800 = 4950 \) (Maximum)
• At \( C(40,0) \): \( Z = 105(40) + 90(0) = 4200 \)
• At \( D(20,0) \): \( Z = 105(20) + 90(0) = 2100 \)

Hence, the maximum value of \( Z \) is \( 4950 \) when \( x = 30 \) and \( y = 20 \).

Question. Minimise the following linear programming problem graphically:
\( Z = 2x + 3y \)
Subject to the constraints:
\( 2x + 5y \ge 100, 8x + 5y \le 200 \) and \( x, y \ge 0 \).
Answer: Let us convert the inequalities into equations:

1. \( 2x + 5y = 100 \)
   Passing through \( (50,0) \) and \( (0,20) \). For \( (0,0) \), \( 0 \ge 100 \) is false, so the half-plane lies away from the origin.
2. \( 8x + 5y = 200 \)
   Passing through \( (25,0) \) and \( (0,40) \). For \( (0,0) \), \( 0 \le 200 \) is true, so the half-plane is towards the origin.

Since \( x, y \ge 0 \), the region lies in the first quadrant.
Solving \( 2x + 5y = 100 \) and \( 8x + 5y = 200 \) simultaneously:
Subtracting the equations: \( 6x = 100 \implies x = \frac{50}{3} \).
Substituting \( x \): \( 2\left(\frac{50}{3}\right) + 5y = 100 \implies 5y = \frac{200}{3} \implies y = \frac{40}{3} \).
So, the intersection point is \( C\left(\frac{50}{3}, \frac{40}{3}\right) \).

The corner points of the bounded feasible region are \( A(0,20) \), \( B(0,40) \), and \( C\left(\frac{50}{3}, \frac{40}{3}\right) \).
Evaluating the objective function \( Z = 2x + 3y \) at these corner points:

• At \( A(0,20) \): \( Z = 2(0) + 3(20) = 60 \) (Minimum)
• At \( B(0,40) \): \( Z = 2(0) + 3(40) = 120 \)
• At \( C\left(\frac{50}{3}, \frac{40}{3}\right) \): \( Z = 2\left(\frac{50}{3}\right) + 3\left(\frac{40}{3}\right) = \frac{100}{3} + 40 \approx 73.33 \)

Hence, the minimum value of \( Z \) is \( 60 \) at the point \( A(0,20) \).

Question. Minimise the following linear programming problem graphically:
\( Z = x + y \)
Subject to the constraints:
\( 3x + 2y \ge 12, x + 3y \ge 11 \) and \( x \ge 0, y \ge 0 \).
Answer: Converting the inequalities into equations:

1. \( 3x + 2y = 12 \)
   Passing through \( (4,0) \) and \( (0,6) \). For \( (0,0) \), \( 0 \ge 12 \) is false, so the half-plane lies away from the origin.
2. \( x + 3y = 11 \)
   Passing through \( (11,0) \) and \( \left(0, \frac{11}{3}\right) \). For \( (0,0) \), \( 0 \ge 11 \) is false, so the half-plane lies away from the origin.

Since \( x, y \ge 0 \), the region lies in the first quadrant.
Solving \( 3x + 2y = 12 \) and \( x + 3y = 11 \) simultaneously:
From the second equation, \( x = 11 - 3y \). Substituting into the first equation: \( 3(11-3y) + 2y = 12 \implies 33 - 7y = 12 \implies 7y = 21 \implies y = 3 \).
Then \( x = 11 - 3(3) = 2 \). Thus, the point of intersection is \( B(2,3) \).

The corner points of the unbounded feasible region are \( A(0,6) \), \( B(2,3) \), and \( D(11,0) \).
Evaluating the objective function \( Z = x + y \) at these corner points:

• At \( A(0,6) \): \( Z = 0 + 6 = 6 \)
• At \( B(2,3) \): \( Z = 2 + 3 = 5 \) (Minimum)
• At \( D(11,0) \): \( Z = 11 + 0 = 11 \)

Since the feasible region is unbounded, we must check if \( 5 \) is the minimum value of \( Z \). We draw the dotted line representing the inequality \( x + y < 5 \).
Since the open half-plane \( x + y < 5 \) has no points in common with the feasible region, the minimum value of \( Z \) is indeed \( 5 \) at the point \( B(2,3) \).

Question. Solve the following LPP graphically.
Maximise \( Z = 3x + 2y \)
Subject to constraints are \( x + 2y \le 10, 3x + y \le 15 \) and \( x \ge 0, y \ge 0 \).
Also, determine the area of the feasible region.
Answer: Let us plot the lines of the given constraints:

1. \( x + 2y = 10 \)
   Passing through \( (10,0) \) and \( (0,5) \). On putting \( (0,0) \), we get \( 0 \le 10 \) (true), so the half-plane is towards the origin.
2. \( 3x + y = 15 \)
   Passing through \( (5,0) \) and \( (0,15) \). On putting \( (0,0) \), we get \( 0 \le 15 \) (true), so the half-plane is towards the origin.

Solving \( x + 2y = 10 \) and \( 3x + y = 15 \) simultaneously gives the intersection point \( B(4,3) \).
The corner points of the bounded feasible region are \( O(0,0) \), \( A(5,0) \), \( B(4,3) \), and \( C(0,5) \).

Evaluating the objective function \( Z = 3x + 2y \) at these corner points:

• At \( O(0,0) \): \( Z = 3(0) + 2(0) = 0 \)
• At \( A(5,0) \): \( Z = 3(5) + 2(0) = 15 \)
• At \( B(4,3) \): \( Z = 3(4) + 2(3) = 18 \) (Maximum)
• At \( C(0,5) \): \( Z = 3(0) + 2(5) = 10 \)

Thus, the maximum value of \( Z \) is \( 18 \) at the point \( B(4,3) \).

Area of the feasible region:
The feasible region is the quadrilateral \( OABC \). We can divide it into two triangles, \( \triangle OBC \) and \( \triangle OAB \): \[ \text{Area of feasible region} = \text{Area of } \triangle OBC + \text{Area of } \triangle OAB \] \[ = \left( \frac{1}{2} \times \text{base } OC \times \text{height } h_x \right) + \left( \frac{1}{2} \times \text{base } OA \times \text{height } h_y \right) \] \[ = \left( \frac{1}{2} \times 5 \times 4 \right) + \left( \frac{1}{2} \times 5 \times 3 \right) \] \[ = 10 + 7.5 = 17.5 \text{ sq units} \]

Question. A linear programming problem (LPP) along with the graph of its constraints is shown below.
The shaded portion represents the feasible region.
Maximise \( Z = 10x + 20y \)
Subject to constraints,
\( 2x + 3y \ge 6, 4x + y \ge 4 \) and \( x \ge 0, y \ge 0 \)
What can you conclude about the existence of optimal solution for the above LPP? Justify your answer.
Answer: From the given graph, the corner points of the unbounded feasible region are \( (0,4) \), \( \left(\frac{3}{5}, \frac{8}{5}\right) \), and \( (3,0) \).
Evaluating the objective function \( Z = 10x + 20y \) at these points:

• At \( (0,4) \): \( Z = 10(0) + 20(4) = 80 \)
• At \( (3,0) \): \( Z = 10(3) + 20(0) = 30 \)
• At \( \left(\frac{3}{5}, \frac{8}{5}\right) \): \( Z = 10\left(\frac{3}{5}\right) + 20\left(\frac{8}{5}\right) = 6 + 32 = 38 \)

The value of \( Z \) at \( (0,4) \) is \( 80 \). Since the feasible region is unbounded in the first quadrant, we must check whether this is the maximum value by drawing the graph of the open half-plane: \[ 10x + 20y > 80 \implies x + 2y > 4 \] Since the feasible region is unbounded and extends infinitely in the positive direction of both axes, the open half-plane \( x + 2y > 4 \) contains infinitely many points that lie within the feasible region.
As the open half-plane has common points with the feasible region, the values of \( Z \) can increase without bound.
Conclusion: No optimal (maximum) solution exists for this linear programming problem.

Question. Feasible region (shaded) for LPP is shown in the following figure, Maximise \( Z = 5x + 7y \). (With corner points \( O(0,0) \), \( A(7,0) \), \( B(3,4) \), and \( C(0,2) \))
Answer: Given the corner points of the feasible region as \( O(0,0) \), \( A(7,0) \), \( B(3,4) \), and \( C(0,2) \).
Evaluating the objective function \( Z = 5x + 7y \) at these corner points:

• At \( O(0,0) \): \( Z = 5(0) + 7(0) = 0 \)
• At \( A(7,0) \): \( Z = 5(7) + 7(0) = 35 \)
• At \( B(3,4) \): \( Z = 5(3) + 7(4) = 15 + 28 = 43 \) (Maximum)
• At \( C(0,2) \): \( Z = 5(0) + 7(2) = 14 \)

Hence, the maximum value of \( Z \) is \( 43 \) at the point \( B(3,4) \).

Question. Solve the following linear programming problem graphically.
Minimise \( Z = -3x + 4y \)
Subject to constraints,
\( x + 2y \le 8, 3x + 2y \le 12 \) and \( x, y \ge 0 \).
Answer: Let us convert the inequalities into equations:

1. \( x + 2y = 8 \)
   Passing through \( (8,0) \) and \( (0,4) \). For \( (0,0) \), \( 0 \le 8 \) is true, so the half-plane lies towards the origin.
2. \( 3x + 2y = 12 \)
   Passing through \( (4,0) \) and \( (0,6) \). For \( (0,0) \), \( 0 \le 12 \) is true, so the half-plane lies towards the origin.

Since \( x, y \ge 0 \), the region lies in the first quadrant.
Solving both equations simultaneously to find their point of intersection:
Subtracting \( x + 2y = 8 \) from \( 3x + 2y = 12 \), we get \( 2x = 4 \implies x = 2 \).
Substituting \( x = 2 \): \( 2 + 2y = 8 \implies 2y = 6 \implies y = 3 \).
So, the intersection point is \( B(2,3) \).

The corner points of the bounded feasible region are \( O(0,0) \), \( A(0,4) \), \( B(2,3) \), and \( C(4,0) \).
Evaluating the objective function \( Z = -3x + 4y \) at these corner points:

• At \( A(0,4) \): \( Z = -3(0) + 4(4) = 16 \)
• At \( B(2,3) \): \( Z = -3(2) + 4(3) = 6 \)
• At \( C(4,0) \): \( Z = -3(4) + 4(0) = -12 \) (Minimum)
• At \( O(0,0) \): \( Z = 0 \)

Hence, the minimum value of \( Z \) is \( -12 \) at the point \( C(4,0) \).

Question. Solve the following linear programming problem graphically.
Maximise \( Z = -3x - 5y \)
Subject to the constraints
\( -2x + y \le 4, x + y \ge 3 \),
\( x - 2y \le 2 \) and \( x \ge 0, y \ge 0 \).
Answer: Let us convert the inequalities into equations to plot the boundary lines:

1. \( -2x + y = 4 \)
   Passing through \( (-2,0) \) and \( (0,4) \). For \( (0,0) \), \( 0 \le 4 \) is true, so the half-plane lies towards the origin.
2. \( x + y = 3 \)
   Passing through \( (3,0) \) and \( (0,3) \). For \( (0,0) \), \( 0 \ge 3 \) is false, so the half-plane lies away from the origin.
3. \( x - 2y = 2 \)
   Passing through \( (2,0) \) and \( (0,-1) \). For \( (0,0) \), \( 0 \le 2 \) is true, so the half-plane lies towards the origin.

Since \( x, y \ge 0 \), the region lies in the first quadrant.
By finding the intersection points of these boundary lines in the first quadrant:

• The intersection of \( x + y = 3 \) and \( x - 2y = 2 \) is \( A\left(\frac{8}{3}, \frac{1}{3}\right) \).
• The other corner points on the y-axis are \( B(0,3) \) and \( C(0,4) \).

Thus, the corner points of the unbounded feasible region are \( A\left(\frac{8}{3}, \frac{1}{3}\right) \), \( B(0,3) \), and \( C(0,4) \).
Evaluating \( Z = -3x - 5y \) at these corner points:

• At \( A\left(\frac{8}{3}, \frac{1}{3}\right) \): \( Z = -3\left(\frac{8}{3}\right) - 5\left(\frac{1}{3}\right) = -8 - \frac{5}{3} = -\frac{29}{3} \) (Maximum)
• At \( B(0,3) \): \( Z = -3(0) - 5(3) = -15 \)
• At \( C(0,4) \): \( Z = -3(0) - 5(4) = -20 \)

Since the region is unbounded, we must check if \( -\frac{29}{3} \) is the maximum value of \( Z \). We draw the dotted line representing the inequality: \[ -3x - 5y > -\frac{29}{3} \] We observe that this open half-plane has no points in common with the feasible region.
Hence, the maximum value of \( Z \) is \( -\frac{29}{3} \) at the point \( A\left(\frac{8}{3}, \frac{1}{3}\right) \).

Question. Solve the following linear programming problem graphically.
Maximise \( Z = 6x + 3y \)
Subject to the constraints,
\( 4x + y \ge 80, 3x + 2y \le 150 \),
\( x + 5y \ge 115 \) and \( x \ge 0, y \ge 0 \).
Answer: Convert the inequalities into equations to determine the boundary lines:

1. \( 4x + y = 80 \)
   Passing through \( (20,0) \) and \( (0,80) \). The half-plane lies away from the origin.
2. \( 3x + 2y = 150 \)
   Passing through \( (50,0) \) and \( (0,75) \). The half-plane lies towards the origin.
3. \( x + 5y = 115 \)
   Passing through \( (115,0) \) and \( (0,23) \). The half-plane lies away from the origin.

By solving the boundary equations pairwise, we find the intersection points:

• The intersection of \( 4x + y = 80 \) and \( 3x + 2y = 150 \) is \( P(2,72) \).
• The intersection of \( 3x + 2y = 150 \) and \( x + 5y = 115 \) is \( R(40,15) \).
• The intersection of \( 4x + y = 80 \) and \( x + 5y = 115 \) is \( Q(15,20) \).

Thus, the feasible region is the triangle \( PQR \), with corner points \( P(2,72) \), \( Q(15,20) \), and \( R(40,15) \).

Evaluating \( Z = 6x + 3y \) at these corner points:

• At \( P(2,72) \): \( Z = 6(2) + 3(72) = 12 + 216 = 228 \)
• At \( Q(15,20) \): \( Z = 6(15) + 3(20) = 90 + 60 = 150 \)
• At \( R(40,15) \): \( Z = 6(40) + 3(15) = 240 + 45 = 285 \) (Maximum)

Hence, the maximum value of \( Z \) is \( 285 \) at the point \( R(40,15) \).

Question. Solve the following LPP graphically.
Maximise \( Z = 60x + 40y \)
Subject to the constraints,
\( x + 2y \le 12, 2x + y \le 12 \),
\( 4x + 5y \ge 20 \) and \( x, y \ge 0 \).
Answer: Converting the inequalities into equations:

1. \( x + 2y = 12 \)
   Passing through \( (12,0) \) and \( (0,6) \). The half-plane lies towards the origin.
2. \( 2x + y = 12 \)
   Passing through \( (6,0) \) and \( (0,12) \). The half-plane lies towards the origin.
3. \( 4x + 5y = 20 \)
   Passing through \( (5,0) \) and \( (0,4) \). The half-plane lies away from the origin.

Solving \( x + 2y = 12 \) and \( 2x + y = 12 \) simultaneously gives the intersection point \( D(4,4) \).
The corner points of the bounded feasible region are \( A(0,4) \), \( B(5,0) \), \( C(6,0) \), \( D(4,4) \), and \( E(0,6) \).

Evaluating \( Z = 60x + 40y \) at these corner points:

• At \( A(0,4) \): \( Z = 60(0) + 40(4) = 160 \)
• At \( B(5,0) \): \( Z = 60(5) + 40(0) = 300 \)
• At \( C(6,0) \): \( Z = 60(6) + 40(0) = 360 \)
• At \( D(4,4) \): \( Z = 60(4) + 40(4) = 240 + 160 = 400 \) (Maximum)
• At \( E(0,6) \): \( Z = 60(0) + 40(6) = 240 \)

Hence, the maximum value of \( Z \) is \( 400 \) at the point \( D(4,4) \).

Question. Solve the following linear programming problem graphically.
Maximise \( Z = 300x + 600y \)
Subject to the constraints,
\( x + 2y \le 12 \),
\( 2x + y \le 12 \),
\( x + \frac{5}{4}y \ge 5 \) and \( x \ge 0, y \ge 0 \).
Answer: The constraint \( x + \frac{5}{4}y \ge 5 \) is equivalent to \( 4x + 5y \ge 20 \). Therefore, the system of linear constraints and the feasible region are identical to the previous problem.
The corner points of the bounded feasible region are \( A(0,4) \), \( B(5,0) \), \( C(6,0) \), \( D(4,4) \), and \( E(0,6) \).

Evaluating \( Z = 300x + 600y \) at these corner points:

• At \( A(0,4) \): \( Z = 300(0) + 600(4) = 2400 \)
• At \( B(5,0) \): \( Z = 300(5) + 600(0) = 1500 \)
• At \( C(6,0) \): \( Z = 300(6) + 600(0) = 1800 \)
• At \( D(4,4) \): \( Z = 300(4) + 600(4) = 1200 + 2400 = 3600 \) (Maximum)
• At \( E(0,6) \): \( Z = 300(0) + 600(6) = 3600 \) (Maximum)

Since the maximum value of \( Z = 3600 \) occurs at two different corner points, \( D(4,4) \) and \( E(0,6) \), any point along the line segment joining \( D \) and \( E \) will also yield the maximum value of \( 3600 \).
Hence, the maximum value of \( Z \) is \( 3600 \) at the points \( D(4,4) \) and \( E(0,6) \).

Question. Solve the following linear programming problem graphically.
Maximise \( Z = 70x + 40y \)
Subject to constraints,
\( 3x + 2y \le 9 \),
\( 3x + y \le 9 \)
and \( x \ge 0, y \ge 0 \).
Answer: Converting the inequalities into equations:

1. \( 3x + 2y = 9 \)
   Passing through \( (3,0) \) and \( (0, 9/2) \). The half-plane lies towards the origin.
2. \( 3x + y = 9 \)
   Passing through \( (3,0) \) and \( (0,9) \). The half-plane lies towards the origin.

Since \( x, y \ge 0 \), the region lies in the first quadrant.
The boundary lines intersect at the point \( A(3,0) \).
The corner points of the bounded feasible region are \( O(0,0) \), \( A(3,0) \), and \( B(0, 9/2) \).

Evaluating \( Z = 70x + 40y \) at these corner points:

• At \( O(0,0) \): \( Z = 0 \)
• At \( A(3,0) \): \( Z = 70(3) + 40(0) = 210 \) (Maximum)
• At \( B(0, 9/2) \): \( Z = 70(0) + 40(9/2) = 180 \)

Hence, the maximum value of \( Z \) is \( 210 \) at the point \( A(3,0) \).

Question. The corner points of the feasible region determined by the system of linear constraints are \( A(0,8) \), \( B(4,10) \), \( C(6,8) \), \( D(6,5) \), \( E(4,0) \), and \( O(0,0) \).
Answer each of the following.
(i) Let \( Z = 3x - 4y \) be the objective function. Find the maximum and minimum value of \( Z \) and also the corresponding points at which the maximum and minimum value occurs.
(ii) Let \( Z = px + qy \), where \( p, q > 0 \) be the objective function. Find the condition on \( p \) and \( q \) so that the maximum value of \( Z \) occurs at \( B(4,10) \) and \( C(6,8) \). Also, mention the number of optimal solutions in this case.
Answer:
(i) Let us evaluate the objective function \( Z = 3x - 4y \) at each of the given corner points:

• At \( O(0,0) \): \( Z = 3(0) - 4(0) = 0 \)
• At \( A(0,8) \): \( Z = 3(0) - 4(8) = -32 \) (Minimum)
• At \( B(4,10) \): \( Z = 3(4) - 4(10) = 12 - 40 = -28 \)
• At \( C(6,8) \): \( Z = 3(6) - 4(8) = 18 - 32 = -14 \)
• At \( D(6,5) \): \( Z = 3(6) - 4(5) = 18 - 20 = -2 \)
• At \( E(4,0) \): \( Z = 3(4) - 4(0) = 12 \) (Maximum)

Thus, the maximum value of \( Z \) is \( 12 \), occurring at \( E(4,0) \), and the minimum value of \( Z \) is \( -32 \), occurring at \( A(0,8) \).

(ii) For the maximum value of \( Z = px + qy \) to occur at both \( B(4,10) \) and \( C(6,8) \), the values of the objective function at these two points must be equal: \[ Z(B) = Z(C) \] \[ 4p + 10q = 6p + 8q \] \[ 10q - 8q = 6p - 4p \] \[ 2q = 2p \implies p = q \] Thus, the required condition is \( p = q \).
Since the maximum value is attained at more than one corner point, all points on the line segment joining the points \( B(4,10) \) and \( C(6,8) \) will also yield the maximum value.
Therefore, the number of optimal solutions in this case is infinite.

Question. Solve the following linear programming problem graphically.
Minimise \( Z = 6x + 7y \)
Subject to constraints,
\( x + 2y \ge 240 \),
\( 3x + 4y \le 620 \),
\( 2x + y \ge 180 \)
and \( x, y \ge 0 \).
Answer: Converting the inequalities into equations to determine the boundary lines:

1. \( x + 2y = 240 \)
   Passing through \( (240,0) \) and \( (0,120) \). On putting \( (0,0) \), we get \( 0 \ge 240 \) (false), so the half-plane lies away from the origin.
2. \( 3x + 4y = 620 \)
   Passing through \( (0, 155) \) and \( \left(\frac{620}{3}, 0\right) \). On putting \( (0,0) \), we get \( 0 \le 620 \) (true), so the half-plane lies towards the origin.
3. \( 2x + y = 180 \)
   Passing through \( (90,0) \) and \( (0,180) \). On putting \( (0,0) \), we get \( 0 \ge 180 \) (false), so the half-plane lies away from the origin.

Since \( x, y \ge 0 \), the region lies in the first quadrant.
We find the intersection points by solving the equations pairwise:

• Solving \( 2x + y = 180 \) and \( 3x + 4y = 620 \) gives the point \( A(20,140) \).
• Solving \( x + 2y = 240 \) and \( 2x + y = 180 \) gives the point \( B(40,100) \).
• Solving \( x + 2y = 240 \) and \( 3x + 4y = 620 \) gives the point \( C(140,50) \).

Thus, the corner points of the bounded feasible region \( ABCA \) are \( A(20,140) \), \( B(40,100) \), and \( C(140,50) \).

Evaluating \( Z = 6x + 7y \) at these corner points:

• At \( A(20,140) \): \( Z = 6(20) + 7(140) = 120 + 980 = 1100 \)
• At \( B(40,100) \): \( Z = 6(40) + 7(100) = 240 + 700 = 940 \) (Minimum)
• At \( C(140,50) \): \( Z = 6(140) + 7(50) = 840 + 350 = 1190 \)

Hence, the minimum value of \( Z \) is \( 940 \) at the point \( B(40,100) \).