Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Determine the maximum value of Z = 3x + 4y if the feasible region (shaded) for a LPP is shown in Fig.12.7.

Q: Determine the maximum value of   Z = 3x + 4y  if the feasible region (shaded) for a LPP is shown in Fig.12.7.

 

Answers (1)

From the question, it is given that 

Z= 3x+4y
The figure that is given above, from that we can come to a constraint that

\\ x + 2y \leq $ 76, 2x +y $ \leq $ 104, x $ \geq $ 0, y $ \geq 0\\

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

\\ x+2 y \leq 76 \\ \Rightarrow x+2 y=76 \\ 2 x+y \leq 104 \\ \Rightarrow 2 x+y=104 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0

The region represented by x + 2y \leq 76:\\

We can say that the line  x + 2y=76 meets the coordinate axes (76,0) and (0,38) respectively. When we join the points to further get the required line we get the line x + 2y=76. Then, we can say that it is clear that (0,0) satisfies the inequation x + 2y \leq 76  . And then origin is represented by the solution set of inequationx + 2y \leq 76

The region represented by 2x +y \leq 104:\\

The line that has 2x +y=104 then meets the other coordinate axes (52,0) and (0,104) simultaneously. Then we need to join the points to get the result of 2x +y=104. The origin then represents the solution set further of the inequation 2x +y \leq 104:\\

The first quadrant of the region represented is x\geq $ 0 and y$ \geq 0. \\ \\

The graph of the equation is given below:

The shaded region ODBA is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
Corner Points are O(0,0), D(0,38), B(44,16) and A(52,0)
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=3 x+4 y \\ \hline O(0,0) & Z=3(0)+4(0)=0+0=0 \\ B(0,38) & Z=3(0)+4(38)=0+152=152 \\ C(44,16) & Z=3(44)+4(16)=132+64=196 \rightarrow \max \\ D(52,0) & Z=3(52)+4(0)=156+0=156 \\ \hline \end{array}

Hence, the maximum value of Z  is  196  at the point (44,16) 

Posted by

infoexpert22

View full answer

Similar Questions

Latest Question