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  • Maximise and Minimise Z = 3x – 4y subject to x – 2y ≤ 0 – 3x + y ≤ 4 x – y ≤ 6 x, y ≥ 0

Maximise and Minimise Z = 3x - 4y   

subject to

\\ x - 2y \leq 0 \\- 3x + y \leq 4 \\x - y \leq 6

x,y \geq 0

Answers (1)

We need to maximize and minimize Z, subject to the given constraints.

Now let us convert the given inequalities into equations.

We obtain the following equation

We have constraints,

\\ x - 2y $ \leq $ 0\\ \\ - 3x + y $ \leq $ 4\\ \\ x - y $ \leq $ 6\\ \\ x, y $ \geq $ 0\\ \\ Z = 3x - 4y\\ \\

We need to maximize and minimize Z, subject to the given constraints.

Now let us convert the given inequalities into equations.

We obtain the following equation

\\x-2 y \leq 0 \\ \Rightarrow x-2 y=0 \\ -3 x+y \leq 4 \\ \Rightarrow-3 x+y=4 \\ x-y \leq 6

\\ $ \Rightarrow $ x - y = 6\\ \\ x $ \geq $ 0\\ \\ $ \Rightarrow $ x=0\\ \\ y $ \geq $ 0\\ \\ $ \Rightarrow $ y=0\\

The further explanation of the same is given below:

The region represented by x – 2y ≤ 0:

The line x - 2y = 0 meets the coordinate axes at origin and slope of the line is \frac{1}{2}. We will construct a line passing through origin and whose slope is \frac{1}{2}. As point (1,1) satisfies the inequality. So, the side of the line which contains (1,1) is feasible. Hence, the solution set of the inequation x – 2y ≤ 0 is the side which contains (1,1).

The region represented by – 3x + y ≤ 4:

The line – 3x + y = 4 meets the coordinate axes (-\frac{4}{3},0) and (0,4) respectively. We will join these points to obtain the line -3x + y = 4. It is clear that (0,0) satisfies the inequation – 3x + y ≤ 4. So, the region containing the origin represents the solution set of the inequation – 3x + y ≤ 4.

The region represented by x – y ≤ 6:

The line x – y = 6 then meets the coordinate axes (6,0) and (0,-6) respectively. We will join these points to obtain the line x – y = 6. It is clear that (0,0) satisfies the inequation x – y ≤ 6. So, the region containing the origin represents the solution set of the inequation x – y ≤ 6.

The graph for the same is given below:

 

The feasible region is region between line -3 x+y=4 \: \: and \: \: x-y=6,above BC and to the right of y axis as shown.
Feasible region is unbounded.
Corner points are A, B, C

Value of Z at corner points A, B, C and D –

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=3 x-4 y \\ \hline A(0,4) & Z=0-(4)(4)=-16 \rightarrow \text { min } \\ \hline B(0,0) & Z=0+0=0 \\ \hline C(12,6) & Z=3(12)-4(6)=12 \rightarrow \max \\ \hline \end{array}

So, maximum value of Z at corner points is 12 at C and minimum is -16 at A.

So, to check if the solution is correct, we plot 3 x-4 y>12 and 3 x-4 y<-16 for maximum and minimum respectively.
The region represented by 3 x-4 y>12

The line that is 3x – 4y = 12 meets the coordinate axes (4,0) and (0,-3) respectively. When we join these points, we get the line 3x – 4y > 12. It is clear that (0,0) does not satisfy the inequation 3x – 4y > 12. So, the region not containing the origin represents the solution set of the inequation 3x – 4y > 12.

The region represented by 3x – 4y <-16:

The line 3x – 4y = -16 meets the coordinate axes (-\frac{16}{3},0) and (0,4) respectively. We will join these points to obtain the line 3x – 4y <-16. It is clear that (0,0) does not satisfy the inequation 3x – 4y <-16. So, the region not containing the origin represents the solution set of the inequation 3x – 4y <-16.

We get the following answer:

Clearly, 3 x-4 y=12 has no point inside feasible region, but 3 x-4 y=-16 passes through the feasible region.
Therefore, Z has no minimum value it has only a maximum value which is 12.

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