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  • Refer to Exercise 7 above. Find the maximum value of Z.

Refer to Exercise 7 above. Find the maximum value of Z.

Answers (1)

It is given below that:

Z=11x + 7y

The figure that is given above, we can see the subject to constraints 

x + y $ \leq $ 5, x +3y $ \geq $ 9, x $ \geq $ 0, y $ \geq $ 0\\

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

\\ x+y \leq 5 \\ \Rightarrow x+y=5 \\ x+3 y \geq 9 \\ \Rightarrow x+3 y=9 \\ y \geq 0 \\ \Rightarrow x=0 \\ y \geq 0 \\ \Rightarrow y=0

The region which represents x + y $ \leq $ 5  is explained below:

The line that is x+y=5, when meets the coordinate axes we get (9,0) and (0,3) respectively. After joining these points, we get the further line that is x+3y=9. The region does not represent the solution set. 

The region that represents the first quadrant is x$ \geq $ 0 and y$ \geq $ 0. \\

The graph for the same is given below:

The shaded region BEC is the feasible region is bounded, so, minimum value will occur at a corner point of the feasible region.
Corner Points are B(0,3), E(0,5)  and C(3,2)
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=11 x+7 y \\ \hline B(0,3) & Z=11(0)+7(3)=0+21=21 \\ E(0,5) & Z=11(0)+7(5)=0+35=35 \\ C(3,2) & Z=11(3)+7(2)=33+14=47 \rightarrow \max \\ \hline \end{array}

\text { Hence, the maximum value of } Z \text { is } 47 \text { at the point }(3,2) \text { . }

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