Linear ProgrammingClass 12 Mathematics NCERT Solutions

Given: Objective function: Maximise $Z = 3x + 4y$ Constraints:

Solution: First, we will plot the feasible region determined by the constraints. The constraints $x \geq 0$ and $y \geq 0$ mean that the feasible region is in the first quadrant. The inequality $x + y \leq 4$ can be represented by the line $x + y = 4$ and the region below it. The line passes through the points $(4,0)$ and $(0,4)$.

The feasible region is a bounded triangle, let's call it OAB, where O is the origin $(0,0)$, A is $(4,0)$ and B is $(0,4)$. These are the corner points of the feasible region.

Now, we evaluate the objective function Z at each corner point:

From the table, the maximum value of Z is 16.

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

Given: Objective function: Minimise $Z = -3x + 4y$ Constraints:

Solution: The feasible region is in the first quadrant. We plot the lines $x + 2y = 8$ (passes through (8,0) and (0,4)) and $3x + 2y = 12$ (passes through (4,0) and (0,6)). The feasible region is the area satisfying all inequalities.

The corner points of the bounded feasible region are:

The corner points are O(0,0), A(4,0), B(2,3), and C(0,4). Now, we evaluate Z at these points:

From the table, the minimum value of Z is -12.

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

Given: Objective function: Maximise $Z = 5x + 3y$ Constraints:

Solution: The feasible region is in the first quadrant. We plot the lines $3x + 5y = 15$ (passes through (5,0) and (0,3)) and $5x + 2y = 10$ (passes through (2,0) and (0,5)). The feasible region is the area satisfying all inequalities.

The corner points of the bounded feasible region are:

The corner points are O(0,0), A(2,0), B$(\frac{20}{19}, \frac{45}{19})$, and C(0,3). Now, we evaluate Z at these points:

Comparing the values, $\frac{235}{19} \approx 12.37$, which is the maximum value.

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

Given: Objective function: Minimise $Z = 3x + 5y$ Constraints:

Solution: The feasible region is in the first quadrant. We plot the lines $x + 3y = 3$ (passes through (3,0) and (0,1)) and $x + y = 2$ (passes through (2,0) and (0,2)). Since the inequalities are 'greater than or equal to', the feasible region is the area above both lines, which is an unbounded region.

The corner points of the feasible region are:

The corner points are A(3,0), B$(\frac{3}{2}, \frac{1}{2})$, and C(0,2). Now, we evaluate Z at these points:

The smallest value of Z is 7. Since the feasible region is unbounded, we must check if this is the minimum value. We graph the open half-plane $3x + 5y < 7$. This region is the area below the line $3x + 5y = 7$. We can see that this open half-plane has no points in common with the feasible region. Therefore, the smallest value is indeed the minimum value.

Final Answer: The minimum value of Z is 7, which occurs at the point $(\frac{3}{2}, \frac{1}{2})$.

Given: Objective function: Maximise $Z = 3x + 2y$ Constraints:

Solution: The feasible region is in the first quadrant. We plot the lines $x + 2y = 10$ (passes through (10,0) and (0,5)) and $3x + y = 15$ (passes through (5,0) and (0,15)). The feasible region is the area satisfying all inequalities.

The corner points of the bounded feasible region are:

The corner points are O(0,0), A(5,0), B(4,3), and C(0,5). Now, we evaluate Z at these points:

From the table, the maximum value of Z is 18.

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

Given: Objective function: Minimise $Z = x + 2y$ Constraints:

Solution: The feasible region is in the first quadrant. We plot the lines $2x + y = 3$ (passes through (1.5,0) and (0,3)) and $x + 2y = 6$ (passes through (6,0) and (0,3)). The feasible region is the unbounded area above both lines.

The corner points of the feasible region are:

The corner points are A(6,0) and B(0,3). Now, we evaluate Z at these points:

The minimum value of Z is 6. Since the region is unbounded, we check the open half-plane $x + 2y < 6$. This region has no points in common with the feasible region (which is defined by $x + 2y \geq 6$ and other constraints). Thus, 6 is the minimum value.

To show that the minimum occurs at more than two points, we observe that the objective function $Z = x + 2y$ is parallel to the constraint line $x + 2y = 6$. When this happens, the optimal value occurs at every point on the line segment of the feasible region's boundary that lies on this line. In this case, it is the line segment joining the corner points A(6,0) and B(0,3).

For example, let's take the midpoint of the segment AB, which is $(\frac{6+0}{2}, \frac{0+3}{2}) = (3, 1.5)$. For the point $(3, 1.5)$, $Z = 3 + 2(1.5) = 3 + 3 = 6$. This confirms that any point on the line segment connecting (6,0) and (0,3) gives the same minimum value.

Final Answer: The minimum value of Z is 6. This minimum value occurs at all points on the line segment joining the points $(6, 0)$ and $(0, 3)$. This shows that the minimum of Z occurs at more than two points.

Given: Objective function: Minimise and Maximise $Z = 5x + 10y$ Constraints:

Solution: We graph the feasible region based on the constraints. The region is a bounded polygon. Let's find the corner points by finding the intersections of the boundary lines:

The corner points of the feasible region are A(40, 20), B(60, 30), C(60, 0), and D(120, 0). Now, we evaluate Z at these points:

From the table: The minimum value of Z is 300, which occurs at the point (60, 0). The maximum value of Z is 600, which occurs at two corner points, B(60, 30) and D(120, 0). This means the maximum value is attained at every point on the line segment joining B and D.

Final Answer: The minimum value of Z is 300 at the point (60, 0). The maximum value of Z is 600 at all points on the line segment joining the points (60, 30) and (120, 0).

Given: Objective function: Minimise and Maximise $Z = x + 2y$ Constraints:

Solution: We graph the feasible region based on the constraints. The region is a bounded polygon. Let's find the corner points by finding the intersections of the boundary lines:

The corner points of the feasible region are A(20, 40), B(50, 100), C(0, 50), and D(0, 200). Now, we evaluate Z at these points:

From the table: The minimum value of Z is 100. This occurs at two corner points, A(20, 40) and C(0, 50). This means the minimum value is attained at every point on the line segment joining A and C. The maximum value of Z is 400, which occurs at the point D(0, 200).

Final Answer: The minimum value of Z is 100 at all points on the line segment joining the points (20, 40) and (0, 50). The maximum value of Z is 400 at the point (0, 200).

Given: Objective function: Maximise $Z = -x + 2y$ Constraints:

Solution: We graph the feasible region. The region is unbounded. The lower boundary of the feasible region is formed by segments of the lines $x=3$, $x+y=5$, and $x+2y=6$. Let's find the corner points:

The feasible region is an unbounded region with corner points A(3,2) and B(4,1). Now, we evaluate Z at these points:

The largest value of Z at a corner point is 1. However, since the feasible region is unbounded, we must check if Z has a maximum value. We graph the open half-plane $-x + 2y > 1$. This inequality can be written as $y > \frac{x+1}{2}$. Let's pick a point in the feasible region, for example, (5, 5). This point satisfies all constraints: $5 \geq 3$, $5+5 \geq 5$, $5+10 \geq 6$. At (5,5), the value of Z is $-5 + 2(5) = 5$, which is greater than 1. Since the open half-plane $-x + 2y > 1$ has points in common with the feasible region, the objective function Z does not have a maximum value.

Final Answer: The objective function Z has no maximum value.

Given: Objective function: Maximise $Z = x + y$ Constraints:

Solution: Let's analyze the constraints. Constraint 1 can be rewritten as $y \geq x + 1$. Constraint 2 can be rewritten as $y \leq x$.

We need to find points $(x, y)$ in the first quadrant ($x \geq 0, y \geq 0$) that simultaneously satisfy: $y \geq x + 1$ $y \leq x$

Let's combine these two inequalities: $x + 1 \leq y \leq x$. This implies that $x + 1 \leq x$, which simplifies to $1 \leq 0$. This is a contradiction. There are no points $(x, y)$ that can satisfy both $y \geq x + 1$ and $y \leq x$ at the same time. Therefore, there is no common region that satisfies all the given constraints.

Final Answer: There is no feasible region. Hence, the linear programming problem has no solution.