CBSE Class 12 Maths Exemplar Chapter 12: Linear Programming NCERT Solutions

We are asked to maximize the objective function $$Z\; =\; 11x\; +\; 7y$$ subject to the given constraints:

1. $2x\; +\; y\; \backslash le\; 6$
2. $x\; \backslash le\; 2$
3. $x\; \backslash ge\; 0$
4. $y\; \backslash ge\; 0$

First, we identify the feasible region by graphing these inequalities. The region is bounded by the lines $2x\; +\; y\; =\; 6$, $x\; =\; 2$, $x\; =\; 0$ (y-axis), and $y\; =\; 0$ (x-axis). The corner points of this feasible region (OABC) are found by the intersection of these lines.

The corner points are:

• O: Intersection of $x\; =\; 0$ and $y\; =\; 0$, which is (0, 0).
• A: Intersection of $y\; =\; 0$ and $x\; =\; 2$, which is (2, 0).
• B: Intersection of $x\; =\; 2$ and $2x\; +\; y\; =\; 6$. Substituting $x\; =\; 2$ into $2x\; +\; y\; =\; 6$ gives $2(2)\; +\; y\; =\; 6$, so $4\; +\; y\; =\; 6$, which means $y\; =\; 2$. Thus, point B is (2, 2).
• C: Intersection of $x\; =\; 0$ and $2x\; +\; y\; =\; 6$. Substituting $x\; =\; 0$ gives $2(0)\; +\; y\; =\; 6$, so $y\; =\; 6$. Thus, point C is (0, 6).

Now, we evaluate the objective function $Z\; =\; 11x\; +\; 7y$ at each corner point:

• At O(0, 0): $Z\; =\; 11(0)\; +\; 7(0)\; =\; 0$
• At A(2, 0): $Z\; =\; 11(2)\; +\; 7(0)\; =\; 22$
• At B(2, 2): $Z\; =\; 11(2)\; +\; 7(2)\; =\; 22\; +\; 14\; =\; 36$
• At C(0, 6): $Z\; =\; 11(0)\; +\; 7(6)\; =\; 42$

Comparing these values, the maximum value of Z is 42, which occurs at the point (0, 6).

Hence, the maximum value of Z is 42.

We need to maximize $$Z\; =\; 3x\; +\; 4y$$ subject to the constraints:

1. $x\; +\; y\; \backslash le\; 1$
2. $x\; \backslash ge\; 0$
3. $y\; \backslash ge\; 0$

The feasible region is determined by these inequalities. The boundary lines are $x\; +\; y\; =\; 1$, $x\; =\; 0$ (y-axis), and $y\; =\; 0$ (x-axis). This forms a bounded region OAB.

The corner points of the feasible region are:

• O: Intersection of $x\; =\; 0$ and $y\; =\; 0$, which is (0, 0).
• A: Intersection of $y\; =\; 0$ and $x\; +\; y\; =\; 1$. Substituting $y\; =\; 0$ gives $x\; +\; 0\; =\; 1$, so $x\; =\; 1$. Thus, point A is (1, 0).
• B: Intersection of $x\; =\; 0$ and $x\; +\; y\; =\; 1$. Substituting $x\; =\; 0$ gives $0\; +\; y\; =\; 1$, so $y\; =\; 1$. Thus, point B is (0, 1).

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

• At O(0, 0): $Z\; =\; 3(0)\; +\; 4(0)\; =\; 0$
• At A(1, 0): $Z\; =\; 3(1)\; +\; 4(0)\; =\; 3$
• At B(0, 1): $Z\; =\; 3(0)\; +\; 4(1)\; =\; 4$

Comparing these values, the maximum value of Z is 4, which occurs at the point (0, 1).

Hence, the maximum value of Z is 4.

We need to maximize $$Z\; =\; 11x\; +\; 7y$$ subject to the constraints:

1. $x\; \backslash le\; 3$
2. $y\; \backslash le\; 2$
3. $x\; \backslash ge\; 0$
4. $y\; \backslash ge\; 0$

The feasible region is defined by these inequalities. The boundary lines are $x\; =\; 3$, $y\; =\; 2$, $x\; =\; 0$ (y-axis), and $y\; =\; 0$ (x-axis). This forms a rectangle, which is a bounded region OABC.

The corner points of the feasible region are:

• O: Intersection of $x\; =\; 0$ and $y\; =\; 0$, which is (0, 0).
• A: Intersection of $y\; =\; 0$ and $x\; =\; 3$, which is (3, 0).
• B: Intersection of $x\; =\; 3$ and $y\; =\; 2$, which is (3, 2).
• C: Intersection of $x\; =\; 0$ and $y\; =\; 2$, which is (0, 2).

Now, we evaluate the objective function $Z\; =\; 11x\; +\; 7y$ at each corner point:

• At O(0, 0): $Z\; =\; 11(0)\; +\; 7(0)\; =\; 0$
• At A(3, 0): $Z\; =\; 11(3)\; +\; 7(0)\; =\; 33$
• At B(3, 2): $Z\; =\; 11(3)\; +\; 7(2)\; =\; 33\; +\; 14\; =\; 47$
• At C(0, 2): $Z\; =\; 11(0)\; +\; 7(2)\; =\; 14$

Comparing these values, the maximum value of Z is 47, which occurs at the point (3, 2).

Hence, the maximum value of Z is 47.