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  • Determine the maximum value of Z = 11x + 7y subject to the constraints: 2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0.

Determine the maximum value of Z = 11x + 7y  subject to the constraints: 2x + y \leq 6, x \leq 2, x \geq 0, y \geq 0.

Answers (1)

Given that:

Z = 11x+7y \\

It is subject to constraints

2x + y $ \leq $ 6, x $ \leq $ 2, x $ \geq $ 0, y $ \geq $ 0.

Now let us convert given inequalities into equation.

We obtain following equation

\\ 2 x+y \leq 6 \\ \Rightarrow 2 x+y=6 \\ x \leq 2 \\ \Rightarrow x=2 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0

The lines that represent 2x+y=6, then further meet the other axes respectively in order to get an answer. The points are to be joined to obtain the line 2x+y=6. It is then further clarified that the equation is satisfied. Then the region that contains the origin is then represented by the set of solutions of the inequation 2x+y$ \leq $ 6.\\

The region represented by x$ \leq $ 2:\\

The line which is parallel to the Y-axis then meets the X-axis which comes at X=2. Hence, it is clarified that (0,0) satisfies the inequation. 

After plotting the equation graphically, we get an answer:

\\$The shaded region OBDE is the feasible region is bounded, so, maximum value will occur at a corner point of the feasible region.\\ Corner Points are $\mathrm{O}(0,0), \mathrm{B}(0,6), \mathrm{D}(2,2)$ and $\mathrm{E}(2,0)$

Coming to the conclusion, when we substitute the values in Z at the corner points, we get the following answer:

\begin{array}{|l|l|} \hline \text { Corner Point } & \text { Value of } Z=11 x+7 y \\ \hline O(0,0) & Z=11(0)+7(0)=0+0=0 \\ B(0,6) & Z=11(0)+7(6)=0+42=42 \rightarrow \max \\ D(2,2) & Z=11(2)+7(2)=22+14=36 \\ E(2,0) . & Z=11(2)+7(0)=22+0=22 \\ \hline \end{array}

Therefore, the final answer is the maximum value of Z is 42 at the point (0,6).

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