If the solution region is (3x+y\leq 21), (x+3y\leq 21), (x\geq 0), and (y\geq 0), what is the maximum value of (x-y)?

Correct answer: (7)

The direct answer is option C, 7. A linear expression over a bounded polygon reaches its maximum at a vertex, so first identify the important corner points. On the x-axis, y=0; the inequality 3x+y≤21 gives x≤7, so (7,0) is feasible. At this point x-y=7-0=7. The other relevant corners are (0,0), (0,7), and the intersection of 3x+y=21 with x+3y=21, which is (5.25,5.25); their values of x-y are 0, -7, and 0. Thus 7 is the maximum. Option A, 0, occurs at several balanced points but is not largest. Option B, 5, is feasible in value but can be improved to 7. Option C, 7, is achieved at (7,0). Option D, 21, incorrectly treats the right side of an inequality as the value of x-y. Exam cue: to maximize x-y, prefer large x and small y, while still checking all constraints.