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The feasible region of a Linear Programming Problem is the set of points that satisfy all the given constraints simultaneously.Constraint 1 requires the points to be in the region where the sum of $x$ and $y$ is less than or equal to 4.Constraint 2 requires the points to be in the region where the sum of $x$ and $y$ is greater than or equal to 6.Mathematically, a number cannot be both $\le 4$ and $\ge 6$ at the same time. Visually, looking at the graph, there is a clear gap between the shaded region $\Delta AOB$ and the shaded unbounded region above $PQ$. The two regions do not overlap.3. ConclusionSince there is no common region that satisfies all constraints, the feasible region is an empty set.Without a feasible region, there are no valid values for $x$ and $y$ to substitute into the objective function $Z$. Therefore, an optimal solution (maximum value) cannot be found.Answer:The correct option is (C) does not exist.

To find the maximum and minimum values of the objective function \(Z = 18x + 9y\), we must evaluate \(Z\) at each of the given corner points of the feasible region.The corner points \((x, y)\) are: \((2, 72)\), \((15, 20)\), and \((40, 15)\).

At \(\mathbf{(2, 72)}\):\[Z = 18(2) + 9(72)\]\[Z = 36 + 648 = \mathbf{684}\]At \(\mathbf{(15, 20)}\):\[Z = 18(15) + 9(20)\]\[Z = 270 + 180 = \mathbf{450}\]At \(\mathbf{(40, 15)}\):\[Z = 18(40) + 9(15)\]\[Z = 720 + 135 = \mathbf{855}\]Maximum Value: \(855\), which occurs at the point \(\mathbf{(40, 15)}\).Minimum Value: \(450\), which occurs at the point \(\mathbf{(15, 20)}\).

The correct conclusion is: \(Z\) is maximum at \((40, 15)\), minimum at \((15, 20)\).

The objective function maximizes the combined profit earned from selling products X and Y.