Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • The feasible region for a LPP is shown in Fig. 12.9. Find the minimum value of Z = 11x + 7y.

The feasible region for a LPP is shown in Fig. 12.9. Find the minimum value of Z = 11x + 7y.

 

Answers (1)

The following is given that:

Z = 11x + 7y.

It is subject to constraints

x+y \leq 5, x+3 y \geq 9, x \geq 0, y \geq 0

Now let us convert the given inequalities into equation
We obtain the following equation

\\ x+y \leq 5 \\ \Rightarrow x+y=5 \\ x+3 y \geq 9 \\ \Rightarrow x+3 y=9 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0

The region represented by x + y ≤ 5:

The line that shows x+y=5, then meets the axes that coordinate (5,0) and (0,5) respectively. We need to join the points in order to obtain the line x+y=5. Hence, it is clarified that (0,0) then satisfies the inequation x + y $ \leq $ 5. Therefore, the region that has the origin represents the solution set of x + y $ \leq $ 5.

The region that is represented by x +3y $ \geq $ 9:\\

The line that is x+3y=9 meets the coordinate axes (9,0) and (0,3) respectively to get the final answer. When we join the points we get the line x+3y=9. Therefore, it is then clear that the region doesn’t contain the origin and represents the solution set of the inequation.

The graph is further given below:

The shaded region BEC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
Corner Points are B(0,3), E(0,5) and C(3,2)
Now we will substitute these values in Z at each of these corner points, we get

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline B(0,3) & Z=11(0)+7(3)=0+21=21 \rightarrow \min \\ E(0,5) & Z=11(0)+7(5)=0+35=35 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \\ \hline \end{array}

Hence, the minimum value of Z  is 21 at the point (0,3)

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question