If the solution region is formed by (x\geq 2), (y\geq 1), and (x+y\leq 8), what is its area?

Correct answer: (\frac{25}{2}) square units

Direct answer: Option C, \(\frac{25}{2}\) square units. The inequalities \(x\geq2\) and \(y\geq1\) place the region to the right of \(x=2\) and above \(y=1\). The inequality \(x+y\leq8\) places it below the line \(x+y=8\). The boundary intersections are: \(x=2,y=1\), giving \((2,1)\); with \(y=1\), \(x+1=8\), giving \((7,1)\); with \(x=2\), \(2+y=8\), giving \((2,6)\). These three points form a right triangle. Its horizontal base is \(7-2=5\), and its vertical height is \(6-1=5\). Therefore area \(=\frac12\times5\times5=\frac{25}{2}\) square units. Option C is correct. Option A, 10, would require a smaller triangle. Option B, 12, is not obtained from the base-height calculation. Option D, 15, is also too large; it ignores the one-half factor used for a triangle or uses an incorrect dimension. The boundary lines are included because all inequalities are non-strict, but boundaries do not change area. Exam cue: shift the origin to \((2,1)\); the remaining triangle has legs 5 and 5.