RBSE Solutions Class 12 Maths Chapter 15 Linear Programming Exercise 15.1

Rajasthan Board RBSE Class 12 Maths Chapter 15 Linear Programming Ex 15.1

Solve the following linear programming problem by graphical method:

Question 1. Minimize \( Z = -3x + 4y \) subject to the constraints \( x + 2y \le 8 \), \( 3x + 2y \le 12 \) and \( x \ge 0, y \ge 0 \).
Answer: First, we convert the given inequalities into equations to find the boundary lines:
\( x + 2y = 8 \)
\( 3x + 2y = 12 \)

For the first equation, \( x + 2y = 8 \):

This means the line \( x + 2y = 8 \) passes through points A(8, 0) and B(0, 4). Joining these points gives us the line. For the inequality \( x + 2y \le 8 \), we check the origin (0, 0): \( 0 + 2(0) \le 8 \), which is \( 0 \le 8 \). This is true, so the region containing the origin represents the solution set for this inequality.

For the second equation, \( 3x + 2y = 12 \):

This means the line \( 3x + 2y = 12 \) passes through points C(4, 0) and D(0, 6). Joining these points gives us the line. For the inequality \( 3x + 2y \le 12 \), we check the origin (0, 0): \( 3(0) + 2(0) \le 12 \), which is \( 0 \le 12 \). This is true, so the region containing the origin represents the solution set for this inequality.

The constraints \( x \ge 0 \) and \( y \ge 0 \) mean that the solution must be in the first quadrant.

The common region satisfying all these inequalities is the feasible region OCEB. The corner points of this feasible region are O(0, 0), C(4, 0), E(2, 3), and B(0, 4). Point E(2, 3) is the intersection of \( x + 2y = 8 \) and \( 3x + 2y = 12 \).

Now, we evaluate the objective function \( Z = -3x + 4y \) at each corner point:

From the table, the minimum value of the objective function \( Z \) is -12, which occurs at point C(4, 0). The shaded region OCEB represents the common area that satisfies all the rules, which is called the feasible region. This means the problem has a clear solution.
In simple words: We find the area where all the given rules fit together on a graph. Then we check the corners of this area to find the smallest value for Z. The smallest value is -12 at point C(4, 0).

🎯 Exam Tip: Always draw clear lines and shade the feasible region accurately. Label all corner points and calculate the objective function value at each one to avoid errors.

Question 2. Maximize \( Z = 3x + 4y \) subject to the \( x + y \le 4 \) constraints \( x \ge 0, y \ge 0 \).
Answer: To solve this, we first change the inequality to an equation to find the boundary line:
\( x + y = 4 \)

For the equation \( x + y = 4 \):

The line \( x + y = 4 \) meets the coordinate axes at A(4, 0) and B(0, 4). We connect these points to draw the line. For the inequality \( x + y \le 4 \), we check with the origin (0, 0): \( 0 + 0 \le 4 \), which is \( 0 \le 4 \). This is true, so the region that includes the origin is the solution set for this inequality.

The constraints \( x \ge 0 \) and \( y \ge 0 \) mean that our solution must be in the first quadrant.

The common region satisfying all these inequalities is the feasible region OAB. The corner points of this feasible region are O(0, 0), A(4, 0), and B(0, 4). The graph shows this clearly.

Now, we calculate the value of the objective function \( Z = 3x + 4y \) at each corner point:

From the table, the maximum value of the objective function \( Z \) is 16, which occurs at point B(0, 4). The solution for this problem is \( x = 0, y = 4 \) with a maximum value of \( Z = 16 \). Linear programming helps find the best outcome in a mathematical model.
In simple words: We find the area on the graph where all conditions are met. Then we check the corners of this area. The biggest value for Z (which is 16) is found at point B(0, 4).

🎯 Exam Tip: Remember to always check the origin (0,0) as a test point for inequalities to correctly determine the solution region. This helps ensure you shade the correct side of the line.

Question 3. Minimize \( Z = -50x + 20y \) Subject to the constraints \( 2x - y \ge -5 \), \( 3x + y \ge 3 \), \( 2x - 3y \le 12 \) and \( x \ge 0, y \ge 0 \).
Answer: First, we convert the given inequalities into equations to define the boundary lines:
(i) \( 2x - y = -5 \)
(ii) \( 3x + y = 3 \)
(iii) \( 2x - 3y = 12 \)

For the first equation, \( 2x - y = -5 \):

The line \( 2x - y = -5 \) passes through points A(\( -5/2 \), 0) and B(0, 5). We draw this line. For the inequality \( 2x - y \ge -5 \), if we test with the origin (0, 0): \( 2(0) - 0 \ge -5 \), which is \( 0 \ge -5 \). This is true, so the region containing the origin is the solution set for this inequality.

For the second equation, \( 3x + y = 3 \):

The line \( 3x + y = 3 \) passes through points C(1, 0) and D(0, 3). We draw this line. For the inequality \( 3x + y \ge 3 \), if we test with the origin (0, 0): \( 3(0) + 0 \ge 3 \), which is \( 0 \ge 3 \). This is false, so the region *opposite* to the origin is the solution set for this inequality.

For the third equation, \( 2x - 3y = 12 \):

The line \( 2x - 3y = 12 \) passes through points E(6, 0) and F(0, -4). We draw this line. For the inequality \( 2x - 3y \le 12 \), if we test with the origin (0, 0): \( 2(0) - 3(0) \le 12 \), which is \( 0 \le 12 \). This is true, so the region containing the origin is the solution set for this inequality.

The constraints \( x \ge 0 \) and \( y \ge 0 \) mean that the solution must be in the first quadrant. However, the feasible region can extend beyond the first quadrant due to the negative y-intercept. The point of intersection for \( 2x - y = -5 \) and \( 3x + y = 3 \) is G(\( -2/5, 29/5 \)). The point of intersection for \( 2x - y = -5 \) and \( 2x - 3y = 12 \) is H(\( 21/11, -30/11 \)). The point of intersection for \( 3x + y = 3 \) and \( 2x - 3y = 12 \) is I(\( 21/11, -17/11 \)).

The shaded area is an open and common region that satisfies all the given rules. This area represents the feasible region for the problem. For an open feasible region, the minimum or maximum may or may not exist.

Let's consider the corner points where the objective function is evaluated. These points are typically vertices of the feasible region. From the graph, these points are B(0, 5), D(0, 3), C(1, 0), E(6, 0), and any points of intersection if they are part of the feasible region. Note that the feasible region is unbounded. Let's find the values of Z at the key points:

In this problem, the feasible region is unbounded. This means the objective function might not have a minimum value because the region extends infinitely in a direction where Z decreases. Based on the given values, -300 is the smallest value calculated at E(6,0). However, since it's an unbounded region, we cannot definitively say if this is the absolute minimum without further analysis of the open region. The given solution suggests these are the relevant points to check. In many unbounded LPPs, the minimum may not exist if the region extends indefinitely in a favorable direction for minimization. If the region is such that we can move further into it and get smaller Z values, then no minimum exists. Assuming these points are the vertices of the 'active' boundary for minimization, the minimum value is -300.
In simple words: We draw the lines for all the conditions and find the area where they all overlap. This area is open and goes on forever. We check Z values at the corners we found. The lowest Z value we see is -300 at E(6,0). Since the region is unbounded, we must be careful if a true minimum exists.

🎯 Exam Tip: For unbounded feasible regions, always check if the objective function can go to infinity (for maximization) or negative infinity (for minimization) within the region. Sometimes, no maximum or minimum exists in such cases.

Question 4. Minimize \( Z = 3x + 5y \) subject to the constraints \( x + 3y \ge 3 \), \( x + y \ge 2 \) and \( x \ge 0, y \ge 0 \).
Answer: First, we convert the given inequalities into equations:
(i) \( x + 3y = 3 \)
(ii) \( x + y = 2 \)

For the first equation, \( x + 3y = 3 \):

The line \( x + 3y = 3 \) passes through points A(3, 0) and B(0, 1). We draw this line. For the inequality \( x + 3y \ge 3 \), if we test with the origin (0, 0): \( 0 + 3(0) \ge 3 \), which is \( 0 \ge 3 \). This is false, so the region *opposite* to the origin is the solution set for this inequality.

For the second equation, \( x + y = 2 \):

The line \( x + y = 2 \) passes through points C(2, 0) and D(0, 2). We draw this line. For the inequality \( x + y \ge 2 \), if we test with the origin (0, 0): \( 0 + 0 \ge 2 \), which is \( 0 \ge 2 \). This is false, so the region *opposite* to the origin is the solution set for this inequality.

The constraints \( x \ge 0 \) and \( y \ge 0 \) mean that the solution must be in the first quadrant.

The common region satisfying all these inequalities is the feasible region AED. This region is open and unbounded. The corner points of this feasible region are A(3, 0), E(\( 3/2, 1/2 \)), and D(0, 2). Point E(\( 3/2, 1/2 \)) is the intersection of \( x + 3y = 3 \) and \( x + y = 2 \).

Now, we evaluate the objective function \( Z = 3x + 5y \) at each corner point:

From the table, the minimum value of \( Z \) is 7, which occurs at \( x = \frac{3}{2} \) and \( y = \frac{1}{2} \). Even though the feasible region is unbounded, we check if the minimum value line intersects the feasible region. It does not go below 7, so this is the minimum. The minimum value of Z is 7.
In simple words: We graph the lines and find the region where all conditions are met. This region is open. We test the corners of this region. The smallest Z value we find is 7, which happens at point E(\( 3/2, 1/2 \)).

🎯 Exam Tip: When dealing with "greater than or equal to" inequalities, the feasible region is often unbounded and lies away from the origin. Always identify all corner points, including intersections, to correctly evaluate the objective function.

Question 5. Evaluate maximum and minimum value, where \( Z = 3x + 9y \) Subject to the constraints \( x + 3y \le 60 \), \( x + y \ge 10 \) and \( x \ge 0, y \ge 0 \).
Answer: First, we convert the given inequalities into equations to find the boundary lines:
(i) \( x + 3y = 60 \)
(ii) \( x + y = 10 \)
(iii) \( x - y = 0 \) (which is \( x = y \))

For the first equation, \( x + 3y = 60 \):

The line \( x + 3y = 60 \) passes through A(60, 0) and B(0, 20). We draw this line. For the inequality \( x + 3y \le 60 \), if we test with the origin (0, 0): \( 0 + 3(0) \le 60 \), which is \( 0 \le 60 \). This is true, so the region containing the origin is the solution set.

For the second equation, \( x + y = 10 \):

The line \( x + y = 10 \) passes through C(10, 0) and D(0, 10). We draw this line. For the inequality \( x + y \ge 10 \), if we test with the origin (0, 0): \( 0 + 0 \ge 10 \), which is \( 0 \ge 10 \). This is false, so the region *opposite* to the origin is the solution set.

For the third equation, \( x = y \) (or \( x - y = 0 \)):

The line \( x = y \) passes through O(0, 0) and E(5, 5). We draw this line. For the inequality \( x - y \ge 0 \), if we test with a point not on the line, say (1, 0): \( 1 - 0 \ge 0 \), which is \( 1 \ge 0 \). This is true, so the region containing (1, 0) is the solution set.

The constraints \( x \ge 0 \) and \( y \ge 0 \) confine the solution to the first quadrant.

The shaded region BDEF represents the feasible region. The corner points of this feasible region are D(0, 10), E(5, 5), F(15, 15), and B(0, 20).
Point E(5,5) is the intersection of \( x+y=10 \) and \( x=y \).
Point F(15,15) is the intersection of \( x+3y=60 \) and \( x=y \).

Now, we evaluate the objective function \( Z = 3x + 9y \) at each corner point:

From the table, the minimum value of the objective function \( Z \) is 60, which occurs at point E(5, 5). The maximum value of \( Z \) is 180, which occurs at two points: F(15, 15) and B(0, 20). This means any point on the line segment FB would also yield the maximum value of 180.
In simple words: We find the area where all conditions are met on the graph. We then check the corners of this area. The smallest value for Z is 60 at point E(5, 5). The largest value for Z is 180, found at two points, F(15, 15) and B(0, 20), and along the line connecting them.

🎯 Exam Tip: When the maximum or minimum value occurs at more than one corner point, it means that value exists along the entire line segment connecting those points. Always mention all such points in your answer.

Question 6. Maximize \( Z = x + 2y \) Subject to the constraints \( 2x + y = 3 \), \( x + 2y = 6 \) and \( x \ge 0, y \ge 0 \).
Answer: First, we convert the given inequalities into equations:
(i) \( 2x + y = 3 \)
(ii) \( x + 2y = 6 \)

For the first equation, \( 2x + y = 3 \):

The line \( 2x + y = 3 \) passes through points A(\( 3/2 \), 0) and B(0, 3). We draw this line. This is a boundary line.

For the second equation, \( x + 2y = 6 \):

The line \( x + 2y = 6 \) passes through points C(6, 0) and D(0, 3). We draw this line. This is also a boundary line.

The constraints \( x \ge 0 \) and \( y \ge 0 \) mean that the solution must be in the first quadrant.

From the figure, the line joining point A(\( 3/2 \), 0) and B(0, 3) is parallel to the line joining point C(6, 0) and D(0, 3). Also, it seems point D(0,3) is the same as B(0,3). Let's re-evaluate the equations. The intersection of \( 2x+y=3 \) and \( x+2y=6 \) is found by multiplying the first equation by 2: \( 4x+2y=6 \). Subtracting the second equation \( x+2y=6 \) from this gives \( 3x=0 \implies x=0 \). Substituting \( x=0 \) into \( x+2y=6 \) gives \( 2y=6 \implies y=3 \). So, the intersection point is (0,3). This means point B and D are the same point, (0,3). The question is asking to maximize Z, subject to two *equality* constraints and the non-negativity constraints. If there are equality constraints, the feasible region is just the point(s) that satisfy all conditions. Here, the only point that satisfies both \( 2x+y=3 \) and \( x+2y=6 \) and \( x \ge 0, y \ge 0 \) is (0,3). So, the feasible region is a single point (0,3).

Now, we evaluate the objective function \( Z = x + 2y \) at this corner point:

Since the feasible region is a single point (0,3), both the maximum and minimum values of \( Z \) exist at this point. Thus, the maximum value of \( Z \) is 6, which occurs at \( x=0, y=3 \). The minimum value is also 6 at the same point.
In simple words: We draw the lines for both equations. We find that the only point that satisfies both equations and the x, y being positive is (0,3). So this single point is our solution area. At this point, the value of Z is 6. This means 6 is both the smallest and largest value Z can be.

🎯 Exam Tip: If the constraints are given as equalities (not inequalities), the feasible region is usually just a single point or a line segment. The maximum and minimum values of the objective function will then occur at those specific points.

Question 7. Find maximum and minimum value where \( Z = 5x + 10y \) Subject to the constraints \( x + 2y \le 120 \), \( x + y \ge 60 \), \( x - 2y \ge 0 \) and \( x \ge 0, y \ge 0 \).
Answer: First, we convert the given inequalities into equations to find the boundary lines:
(i) \( x + 2y = 120 \)
(ii) \( x + y = 60 \)
(iii) \( x - 2y = 0 \) (which is \( x = 2y \))

For the first equation, \( x + 2y = 120 \):

The line \( x + 2y = 120 \) passes through A(120, 0) and B(0, 60). We draw this line. For the inequality \( x + 2y \le 120 \), if we test with the origin (0, 0): \( 0 + 2(0) \le 120 \), which is \( 0 \le 120 \). This is true, so the region containing the origin is the solution set.

For the second equation, \( x + y = 60 \):

The line \( x + y = 60 \) passes through C(60, 0) and B(0, 60). We draw this line. For the inequality \( x + y \ge 60 \), if we test with the origin (0, 0): \( 0 + 0 \ge 60 \), which is \( 0 \ge 60 \). This is false, so the region *opposite* to the origin is the solution set.

For the third equation, \( x = 2y \):

The line \( x = 2y \) passes through the origin (0, 0) and F(60, 30). We draw this line. For the inequality \( x - 2y \ge 0 \), we test with a point not on the line, say (1, 0): \( 1 - 2(0) \ge 0 \), which is \( 1 \ge 0 \). This is true, so the region containing (1, 0) is the solution set.

The constraints \( x \ge 0 \) and \( y \ge 0 \) keep the solution in the first quadrant.

The feasible region is ACEF, as shown in the graph. The vertices (corner points) of this region are A(120, 0), C(60, 0), E(40, 20), and F(60, 30).
Point E(40, 20) is the intersection of \( x + 2y = 120 \) and \( x + y = 60 \).
Point F(60, 30) is the intersection of \( x + 2y = 120 \) and \( x = 2y \).

Now, we evaluate the objective function \( Z = 5x + 10y \) at each corner point:

From the table, the minimum value of the objective function \( Z \) is 300, which occurs at point C(60, 0). The maximum value of \( Z \) is 600, which occurs at two points: A(120, 0) and F(60, 30). This indicates that all points along the line segment AF will also yield the maximum value of 600. The boundaries of the feasible region form a polygon, and we evaluate the objective function at its corners to find the optimal values.
In simple words: We draw the lines for all conditions and find the overlapping region, which is ACEF. We check the Z values at its corners. The smallest Z value is 300 at C(60,0). The largest Z value is 600, found at A(120,0) and F(60,30), and anywhere on the line between them.

🎯 Exam Tip: Always label your lines and the feasible region clearly on the graph. Correctly identifying the corner points of the feasible region is essential for accurate calculation of minimum and maximum values.