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  • Maximise Z = x + y subject to x + 4y ≤ 8, 2x + 3y ≤ 12, 3x + y ≤ 9, x ≥ 0, y ≥ 0.

Q: Maximise Z = x + y  subject to     \\ x + 4y \leq 8, 2x + 3y \leq 12, 3x + y \leq 9, x \geq 0, y \geq 0.\\

Answers (1)

It is given that:

Z = x + y

And it is also subject to constraints that is given below:

\\ x + 4y \leq $ 8\\2x + 3y $ \leq $ 12\\3x + y $ \leq $ 9\\ x $ \geq $ 0\\ y $ \geq 0.\\

We have to maximize Z, we are subject to the constraints above.

We need to convert the inequalities into equation to get the following equation:

\\ \\ x + 4y \leq $ 8\\ \\ $ \Rightarrow $ x + 4y = 8\\ \\ 2x + 3y $ \leq $ 12\\ \\ $ \Rightarrow $ 2x + 3y = 12\\ \\ 3x + y $ \leq $ 9\\ \\ $ \Rightarrow $ 3x + y = 9\\ \\ x $ \geq $ 0\\ \\ $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0\\ \\ $ \Rightarrow y=0\\

The region that represents x + 4y \leq 8 is explained below:

The line x + 4y = 8 meets the coordinate axes (8,0) and (0,2) respectively. We will join these points to obtain the line x + 4y = 8. It is clear that (0,0) satisfies the inequation x + 4y \leq 8. So, the region containing the origin represents the solution set of the inequation x + 4y \leq 8

The region that represents 2x + 3y \leq 12:

The line that is 2x + 3y = 12 then meets the coordinate axes respectively to get the answer. When we join the points we obtain the line 2x + 3y = 12. It is clear that (0,0) satisfies the inequation 2x + 3y \leq 12. So, the region containing the origin represents the solution set of the inequation2x + 3y \leq 12.

The region that represents 3x + y \leq 9:

The line meets the coordinate axes that is 3x + y = 9 meets (3,0) and (0,9) respectively. After joining the lines, we get 3x + y = 9 and then it is clear that (0,0) satisfies the inequation. The region that contains the origin is represented by the solution set of 3x + y \leq 9.

The graph for the same is given below and also the final answer:

\\$Feasible region is ABCD\\ Value of $Z$ at corner points $A, B, C$ and $D-

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=x+y \\ \hline \text { A }(2,0) & z=2+0=2 \\ \hline B(2.54,1.36) & Z=2.54+1.36=3.90 \rightarrow \max \\ \hline \text { C }(3,0) & z=3+0=3 \\ \hline \text { D }(0,0) & Z=0+0=0 \\ \hline \end{array}

\text { So, value of } Z \text { is maximum at } B(2.54,1.36), \text { the maximum value is } 3.90 .

 

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infoexpert22

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