In the solution region (x\geq 0), (y\geq 0), (x+2y\leq 10), and (2x+y\leq 12), what is the maximum value of (x-y)?

Correct answer: (6)

The direct answer is B, 6. The feasible region lies in the first quadrant and satisfies two upper bounds. To maximize the linear expression \(x-y\), inspect the corner points of the region. On the x-axis, y=0, the constraints give \(x\leq10\) and \(2x\leq12\), so the farthest feasible x-axis point is \((6,0)\). At this point, \(x-y=6-0=6\). Other relevant corners give smaller values: \((0,0)\) gives 0, \((0,5)\) gives -5, and the intersection of the two boundary lines gives \((14/3,8/3)\), where \(x-y=2\). Thus 6 is the maximum. Option A, 5, is less than the attainable value. Option B, 6, is correct. Option C, 4, and option D, 7, are not the maximum; 7 is not even feasible because x cannot exceed 6 when y is nonnegative. In linear programming, checking all vertices is the reliable method.