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  • Maximise Z = 3x + 4y, subject to the constraints: x + y ≤ 1, x ≥ 0, y ≥ 0.

Maximise Z = 3x + 4y, subject to the constraints: x + y \leq 1, x \geq 0, y \geq 0. \\ \\

Answers (1)

Following is the answer

Z = 3x + 4y

It is subject to constraints

x + y $ \leq $ 1, x $ \geq $ 0, y $ \geq $ 0. \\ \\
Now let us convert the given inequalities into equation.
We obtain the following equation

\\x+y \leq 1 \\ \Rightarrow x+y=1 \\ x \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0

The part represented by x+y$ \leq $ 1:

One of the lines that is x+y=1 meets the axes (0,1) and (1,0) respectively. Then the lines are joined to obtain the line that is x+y=1. Therefore, it is clear that (0,0) the equation satisfies x+y$ \leq $ 1.

The region that is represented by x$ \geq $ 0 \and\ y$ \geq $ 0 is first quadrant, and further satisfies these inequations. The graphic plotting is given below:

The shaded region OBC shows the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.
Corner Points are $\mathrm{O}(0,0), \mathrm{B}(0,1)$ and $\mathrm{C}(1,0)$

When we substitute the values in Z, we get the following answer

\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(0,1) & Z=3(0)+4(1)=0+4=4 \rightarrow \max \\ C(1,0) & Z=3(1)+4(0)=3+0=3 \\ \hline \end{array}

 Hence, the maximum value of  Z  is 4  at the point (0,1).

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