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  • Provide solution for RD Sharma maths Class 12 Chapter 29 Linear Programming Exercise Multiple Choice Question Question 9 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 29 Linear Programming Exercise Multiple Choice Question Question 9 textbook solution.

Answers (1)

Answer: (c) Infinite number of points

Hint: Firstly convert the inequality into equations

Given: z_{\max }=4 x+3 y  subject to the constraints 3 x+4 y \leq 24,8 x+6 y \leq 48, x \leq 5, y \leq 6, x, y \geq 0

Solution:

Let’s convert the given inequalities into equations, we obtain the following equations,

3 x+4 y=24,8 x+6 y=48, x=5, y=6, x=0, y=0

The line 3 x+4 y=24 meets the coordinate axis at A(8,0) and B(0,6) join these points to obtain the line 3 x+4 y=24

Clearly, (0,0)satisfies the inequality 3x+4y\leq 24. So, the region in xy plane that contains the origin represents the solution set of given equation.

The line 8x+6y=48 meets the coordinate axis at C(6,0) and D(0,8) join these points to obtain the line 8x+6y=48

Clearly, (0,0) satisfies the inequality 8x+6y\leq 48. So, the region in xy plane that contains the origin represents the solution set of given equation.

x=5 is the line passing through x=5 in y-axis region represented by x\geq 0 and y\geq 0. Since every point in the first quadrant satisfies these in inequations.

So, the first quadrant is the region represented by the inequations.

The corner points of the feasible region are O(0,0), G(5,0), F\left(5, \frac{4}{3}\right), E\left(\frac{24}{7}, \frac{24}{7}\right) and B(0,6)

The values of z at these corners points are as follows,

\text {corner point} \text{z}=\text {4x+3y}
O(0,0) 4(0)+3(0)=0
G(5,0) 4(5)+3(0)=20
F\left ( 5,\frac{4}{3} \right ) 4(5)+3\left ( \frac{4}{3} \right )=24
E\left ( \frac{24}{7},\frac{24}{7} \right ) 4(\frac{24}{7})+3(\frac{24}{7})=\frac{196}{7}=24
B(0,6) 4(0)+3(6)=18

We see that maximum value of the objective function z is 24 which is at F\left(5, \frac{4}{3}\right) and E\left(\frac{24}{7}, \frac{24}{7}\right) . Thus, the optimal value of z is 24.

Therefore, the given objective function can be subjected at an infinite number of points.

The correct option is (c)

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