CBSE Class 12 Mathematics: Linear Programming NCERT Solutions

We are asked to maximise the objective function $Z\; =\; 3x\; +\; 4y$ subject to the constraints $x\; +\; y\; \backslash le\; 4$, $x\; \backslash ge\; 0$, and $y\; \backslash ge\; 0$.

First, we identify the feasible region defined by these constraints. The constraint $x\; \backslash ge\; 0$ and $y\; \backslash ge\; 0$ restrict our attention to the first quadrant of the coordinate plane.

The line $x\; +\; y\; =\; 4$ passes through the points (4, 0) and (0, 4). Since the inequality is $x\; +\; y\; \backslash le\; 4$, the feasible region lies below or on this line.

The feasible region is a triangle with vertices at the origin O (0, 0), point A on the x-axis, and point B on the y-axis.

To find the coordinates of A, we set $y\; =\; 0$ in $x\; +\; y\; =\; 4$, which gives $x\; =\; 4$. So, A is (4, 0).

To find the coordinates of B, we set $x\; =\; 0$ in $x\; +\; y\; =\; 4$, which gives $y\; =\; 4$. So, B is (0, 4).

The corner points of the feasible region are O(0, 0), A(4, 0), and B(0, 4).

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(4, 0): $Z\; =\; 3(4)\; +\; 4(0)\; =\; 12$
• At B(0, 4): $Z\; =\; 3(0)\; +\; 4(4)\; =\; 16$

Comparing the values of Z, the maximum value is 16, which occurs at the point B(0, 4).

Therefore, the maximum value of Z is 16 at the point (0, 4).

We need to minimise the objective function $Z\; =\; -3x\; +\; 4y$ subject to the constraints $x+2y\; \backslash le\; 8$, $3x+2y\; \backslash le\; 12$, $x\; \backslash ge\; 0$, and $y\; \backslash ge\; 0$.

The constraints $x\; \backslash ge\; 0$ and $y\; \backslash ge\; 0$ restrict the feasible region to the first quadrant.

Consider the boundary lines:

1. $x\; +\; 2y\; =\; 8$: This line passes through (8, 0) and (0, 4).
2. $3x\; +\; 2y\; =\; 12$: This line passes through (4, 0) and (0, 6).

The feasible region is the area that satisfies all inequalities. We find the intersection points of these lines.

The corner points are:

• The origin O(0, 0).
• The x-intercept of $3x\; +\; 2y\; =\; 12$ is when $y=0$, so $3x\; =\; 12\; \backslash Rightarrow\; x\; =\; 4$. This point is A(4, 0).
• The y-intercept of $x\; +\; 2y\; =\; 8$ is when $x=0$, so $2y\; =\; 8\; \backslash Rightarrow\; y\; =\; 4$. This point is C(0, 4). Note that (0,4) satisfies $3x+2y\; \backslash le\; 12$ as $3(0)+2(4)=8\; \backslash le\; 12$.
• The intersection of $x\; +\; 2y\; =\; 8$ and $3x\; +\; 2y\; =\; 12$. Subtracting the first equation from the second gives $(3x\; +\; 2y)\; -\; (x\; +\; 2y)\; =\; 12\; -\; 8$, which simplifies to $2x\; =\; 4$, so $x\; =\; 2$. Substituting $x\; =\; 2$ into $x\; +\; 2y\; =\; 8$ gives $2\; +\; 2y\; =\; 8$, so $2y\; =\; 6$, and $y\; =\; 3$. This intersection point is B(2, 3).

The corner points of the feasible region are O(0, 0), A(4, 0), B(2, 3), and C(0, 4).

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

Comparing these values, the minimum value of Z is -12, which occurs at the point A(4, 0).

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

We are asked to maximise the objective function $Z\; =\; 5x\; +\; 3y$ subject to the constraints $3x\; +\; 5y\; \backslash le\; 15$, $5x\; +\; 2y\; \backslash le\; 10$, $x\; \backslash ge\; 0$, and $y\; \backslash ge\; 0$.

The constraints $x\; \backslash ge\; 0$ and $y\; \backslash ge\; 0$ restrict the feasible region to the first quadrant.

We consider the boundary lines for the inequalities:

1. $3x\; +\; 5y\; =\; 15$: This line passes through (5, 0) and (0, 3).
2. $5x\; +\; 2y\; =\; 10$: This line passes through (2, 0) and (0, 5).

The feasible region is the area in the first quadrant that lies below or on both these lines.

The corner points of the feasible region are:

• The origin O(0, 0).
• The x-intercept of $5x\; +\; 2y\; =\; 10$ is when $y=0$, so $5x\; =\; 10\; \backslash Rightarrow\; x\; =\; 2$. This point is A(2, 0). Note that (2,0) satisfies $3x+5y\; \backslash le\; 15$ as $3(2)+5(0)=6\; \backslash le\; 15$.
• The y-intercept of $3x\; +\; 5y\; =\; 15$ is when $x=0$, so $5y\; =\; 15\; \backslash Rightarrow\; y\; =\; 3$. This point is C(0, 3). Note that (0,3) satisfies $5x+2y\; \backslash le\; 10$ as $5(0)+2(3)=6\; \backslash le\; 10$.
• The intersection of $3x\; +\; 5y\; =\; 15$ and $5x\; +\; 2y\; =\; 10$. To find this point, we can use elimination or substitution. Let's use elimination: Multiply the first equation by 2 and the second by 5: $2(3x\; +\; 5y)\; =\; 2(15)\; \backslash Rightarrow\; 6x\; +\; 10y\; =\; 30$ $5(5x\; +\; 2y)\; =\; 5(10)\; \backslash Rightarrow\; 25x\; +\; 10y\; =\; 50$ Subtracting the first new equation from the second: $(25x\; +\; 10y)\; -\; (6x\; +\; 10y)\; =\; 50\; -\; 30$ $19x\; =\; 20\; \backslash Rightarrow\; x\; =\; \backslash frac\{20\}\{19\}$ Substitute $x\; =\; \backslash frac\{20\}\{19\}$ into $5x\; +\; 2y\; =\; 10$: $5\backslash left(\backslash frac\{20\}\{19\}\backslash right)\; +\; 2y\; =\; 10$ $\backslash frac\{100\}\{19\}\; +\; 2y\; =\; 10$ $2y\; =\; 10\; -\; \backslash frac\{100\}\{19\}\; =\; \backslash frac\{190\; -\; 100\}\{19\}\; =\; \backslash frac\{90\}\{19\}$ $y\; =\; \backslash frac\{45\}\{19\}$ So, the intersection point is B $\backslash left(\backslash frac\{20\}\{19\},\; \backslash frac\{45\}\{19\}\backslash right)$.

The corner points of the feasible region are O(0, 0), A(2, 0), B

\left(\frac{20}{19}, \frac{45}{19}\right), and C(0, 3).

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

• At O(0, 0): $Z\; =\; 5(0)\; +\; 3(0)\; =\; 0$
• At A(2, 0): $Z\; =\; 5(2)\; +\; 3(0)\; =\; 10$
• At B $\backslash left(\backslash frac\{20\}\{19\},\; \backslash frac\{45\}\{19\}\backslash right)$: $Z\; =\; 5\backslash left(\backslash frac\{20\}\{19\}\backslash right)\; +\; 3\backslash left(\backslash frac\{45\}\{19\}\backslash right)\; =\; \backslash frac\{100\}\{19\}\; +\; \backslash frac\{135\}\{19\}\; =\; \backslash frac\{235\}\{19\}\; \backslash approx\; 12.37$
• At C(0, 3): $Z\; =\; 5(0)\; +\; 3(3)\; =\; 9$

Comparing these values, the maximum value of Z is \frac{235}{19}, which occurs at the point B

\left(\frac{20}{19}, \frac{45}{19}\right).

Therefore, the maximum value of Z is \frac{235}{19} at the point \left(\frac{20}{19}, \frac{45}{19}\right).