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Provide solution for RD sharma maths class 12 chapter 29 Linear Programming exercise Fill in the blanks question 7

Answers (1)

Coordinates (10,15)

Hint:

First draw the graph and find point D.

Given:

Two points which provides the optimal solution of the linear programming problem

\begin{aligned} &z_{\max }=45 x+55 y \\ &6 x+4 y \leq 120 \\ &3 x+10 y \leq 180 \\ &x \geq 0, y \geq 0 \end{aligned}

has the coordinates _______

Solution:

3x+10y=180 and 6x+4y=120

(60,0) and (0,18) (20,0) and (0,30)

Now, we find point D

Subtracting the lines

\begin{aligned} &6 x+4 y \leq 120 \\ &(3 x+10 y \leq 180) \times 2 \\ &\overline{16 y=240} \\ &\therefore y=\frac{240}{16}=15 \end{aligned}

Putting y=15

Again,

\begin{aligned} &6 x+60=120 \\ &\therefore x=10 \\ &z=45 x+55 y \\ &z_{B}=0 \\ \end{aligned}

\begin{aligned} &z_{A}=0+55(18)=990 \\ &z_{C}=45(20)=900 \\ &z_{D}=450+825=1205 \end{aligned}

\begin{aligned} &z_{max}=1275 \\ \end{aligned} at \begin{aligned} (10,15) \end{aligned}

Hence, the coordinate is \begin{aligned} (10,15) \end{aligned}

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