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How many corner points are in the solution region of \(x+4y\leq 16\), \(3x+2y\leq 18\), \(x\geq 0\), \(y\geq 0\)?

Answer and explanation

Correct answer: 4

The feasible region lies in the first quadrant and below both boundary lines. Its intercept-based vertices are \((0,0)\), \((6,0)\), and \((0,4)\). Solving \(x+4y=16\) and \(3x+2y=18\) gives their intersection as \((4,3)\). Therefore, the region has 4 corner points. Although \((5,1)\) satisfies both inequalities, it is not an intersection of active boundary lines, so it is not a corner. Exam tip: check axis intersections and intersections of all active boundary lines to identify vertices.

Related tags

Linear InequalitiesGraphical SolutionCorner PointsFeasible RegionFirst QuadrantGraphical Solution Of Linear Inequalities In Two VariablesMathematicsClass 11 Mcq

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