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  • Need solution for RD Sharma maths Class 12 Chapter 29 Linear Programming Exercise Multiple Choice Question Question 15 textbook solution.

Need solution for RD Sharma maths Class 12 Chapter 29 Linear Programming Exercise Multiple Choice Question Question 15 textbook solution.

Answers (1)

Answer : (d) (40,15)

Hint :

Convert the given inequalities into equation

Given:

Z_{max}=x+y  subject to the constraints x+2 y \leq 70,2 x+y \leq 95, x, y \geq 0

Solution:

Let us consider the mentioned constraints as equations for a while,

x+2y=70                                                                                            ....(i)

2x+y=95                                                                                            ....(ii)

Now, graph the equations by transforming the equations to intercept form of line.

Equation (i) dividing throughout by 70

\begin{aligned} &\frac{x}{70}+\frac{2 y}{70}=\frac{70}{70} \\ &\frac{x}{70}+\frac{y}{35}=1 \end{aligned}

The line x+2y=70 can be plot in the graph as a line passing through the points, (70,0) and (0,35) as 70 and 35 are the intercepts of the line on the x-axis and y-axis respectively.

Similarly, Equation (ii) can be divided by 95

\begin{aligned} &\frac{2 x}{95}+\frac{y}{95}=\frac{95}{95} \\ &\frac{x}{\frac{95}{2}}+\frac{y}{95}=1 \end{aligned}

The line 2x+y=95can be plot in the graph as a line passing through the points, \left ( \frac{95}{2},0 \right ) and (0,95) as \frac{95}{2} and 95 are the intercepts of the line on the x-axis and y-axis respectively.

By considering the constraints x\geq 0 and y\geq 0, thus clearly shows that the region can only be in the first quadrant. The graph of inequalities will look like,

The points OABC is the feasible region of LPP

Now, form the points O,A,B and C the vertices of polygon formed by the constraints one of the points will provide the maximum solution z=x+y

Now, checking the points, O, A, B, and C by substituting in z=x+y

\begin{aligned} &z_{\text {at }} O(0,0)=0+0=0 \\ \end{aligned}

z_{\text {at }} A(0,35)=0+35=35 \\

z_{\text {at }} B(40,15)=40+15=55 \\

z \text { at } C\left(\frac{95}{2}, 0\right)=\frac{95}{2}+0=\frac{95}{2}=47.5

From the values, it is cler that z maximized at B(40,15)

 

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