To determine the number of corner points for the feasible region, we analyze the given constraints: 1. \(x \geq 0\): This defines the region to the right of and including the \(y\)-axis. 2. \(y \geq 0\): This defines the region above and including the \(x\)-axis. 3. \(x + y \geq 4\): This defines the region above the line \(x + y = 4\). This line can be rewritten as \(y = 4 - x\), intersecting the axes at \(x = 4\) and \(y = 4\). The feasible region is the intersection of these conditions, located in the first quadrant (\(x \geq 0\), \(y \geq 0\)) and above the line \(x + y = 4\). Although this region is unbounded, it possesses two corner points: the intersection of \(x + y = 4\) with \(x = 0\), which is \((0, 4)\), and the intersection of \(x + y = 4\) with \(y = 0\), which is \((4, 0)\). Therefore, there are \(2\) corner points. The correct answer is (C).
Assertion (A): A line in space cannot be drawn perpendicular to \( x \), \( y \), and \( z \) axes simultaneously.
Reason (R): For any line making angles \( \alpha, \beta, \gamma \) with the positive directions of \( x \), \( y \), and \( z \) axes respectively, \[ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1. \]