CBSE Class 12 Mathematics Chapter 12 Linear Programming NCERT Solutions

We need to maximize 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 by graphing these inequalities.

The constraint $x\; +\; y\; \backslash le\; 4$ represents the region on or below the line $x\; +\; y\; =\; 4$. The constraints $x\; \backslash ge\; 0$ and $y\; \backslash ge\; 0$ restrict the region to the first quadrant. The feasible region is a triangle with vertices (corner points) at the origin and the intercepts of the line $x\; +\; y\; =\; 4$ with the axes.

The corner points of the feasible region are:

• O(0,0)
• A(4,0) (x-intercept, when y=0, x=4)
• B(0,4) (y-intercept, when x=0, y=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): $Z\; =\; 3(0)\; +\; 4(4)\; =\; 16$

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

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

We need to minimize 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 $y\; \backslash ge\; 0$. We will use the graphical method.

First, we graph the lines corresponding to the inequalities:

• Line 1: $x\; +\; 2y\; =\; 8$. Intercepts are (8,0) and (0,4).
• Line 2: $3x\; +\; 2y\; =\; 12$. Intercepts are (4,0) and (0,6).

The constraints $x\; \backslash ge\; 0$ and $y\; \backslash ge\; 0$ restrict us to the first quadrant. The feasible region is the area that satisfies all inequalities. We find the intersection point of Line 1 and Line 2:

Subtracting $x\; +\; 2y\; =\; 8$ from $3x\; +\; 2y\; =\; 12$, we get $2x\; =\; 4$, so $x\; =\; 2$. Substituting $x\; =\; 2$ into $x\; +\; 2y\; =\; 8$, we get $2\; +\; 2y\; =\; 8$, which gives $2y\; =\; 6$, so $y\; =\; 3$. The intersection point is (2,3).

The corner points of the feasible region are:

• O(0,0)
• A(4,0) (x-intercept of $3x\; +\; 2y\; =\; 12$, since it's more restrictive than (8,0) from $x+2y=8$ in the first quadrant)
• B(2,3) (intersection of the two lines)
• C(0,4) (y-intercept of $x\; +\; 2y\; =\; 8$, since it's more restrictive than (0,6) from $3x+2y=12$ in the first quadrant)

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): $Z\; =\; -3(0)\; +\; 4(4)\; =\; 16$

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

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

We need to maximize 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 $y\; \backslash ge\; 0$. We will use the graphical method.

First, we graph the lines corresponding to the inequalities:

• Line 1: $3x\; +\; 5y\; =\; 15$. Intercepts are (5,0) and (0,3).
• Line 2: $5x\; +\; 2y\; =\; 10$. Intercepts are (2,0) and (0,5).

The constraints $x\; \backslash ge\; 0$ and $y\; \backslash ge\; 0$ restrict the solution to the first quadrant. The feasible region is the area that satisfies all inequalities.

We find the intersection point of Line 1 and Line 2:

From $3x\; +\; 5y\; =\; 15$, multiply by 2: $6x\; +\; 10y\; =\; 30$.

From $5x\; +\; 2y\; =\; 10$, multiply by 5: $25x\; +\; 10y\; =\; 50$.

Subtracting the first modified equation from the second: $(25x\; +\; 10y)\; -\; (6x\; +\; 10y)\; =\; 50\; -\; 30$, which simplifies to $19x\; =\; 20$, so $x\; =\; \backslash frac\{20\}\{19\}$.

Substitute $x\; =\; \backslash frac\{20\}\{19\}$ into $5x\; +\; 2y\; =\; 10$: $5(\backslash frac\{20\}\{19\})\; +\; 2y\; =\; 10$ => $\backslash frac\{100\}\{19\}\; +\; 2y\; =\; 10$ => $2y\; =\; 10\; -\; \backslash frac\{100\}\{19\}\; =\; \backslash frac\{190\; -\; 100\}\{19\}\; =\; \backslash frac\{90\}\{19\}$ => $y\; =\; \backslash frac\{45\}\{19\}$.

The intersection point is ($\backslash frac\{20\}\{19\}$, $\backslash frac\{45\}\{19\}$).

The corner points of the feasible region are:

• O(0,0)
• A(2,0) (x-intercept of $5x\; +\; 2y\; =\; 10$, as it's more restrictive than (5,0))
• B($\backslash frac\{20\}\{19\}$, $\backslash frac\{45\}\{19\}$) (intersection of the two lines)
• C(0,3) (y-intercept of $3x\; +\; 5y\; =\; 15$, as it's more restrictive than (0,5))

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

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

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