The number of corner points of the feasible region determined by the constraints \( 2x-y\ge0,\ x-3y+3\ge0,\ x\ge0,\ y\ge0 \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 2
The constraints say \( y\le2x \), \( x-3y+3\ge0 \), \( x\ge0 \) and \( y\ge0 \).
The lines \( y=2x \) and \( x-3y+3=0 \) meet where
\( x-6x+3=0 \)
\( \displaystyle x=\frac{3}{5},\ y=\frac{6}{5} \)
This point satisfies every constraint. The line \( y=2x \) also passes through the origin, which satisfies every constraint too.
The line \( x-3y+3=0 \) meets the axes at (0, 1), which fails \( 2x-y\ge0 \), and at (−3, 0), which fails \( x\ge0 \), so neither of these is a corner.
The region runs along the x-axis without end (it is unbounded), and its only corner points are (0, 0) and \( \left(\frac{3}{5},\frac{6}{5}\right) \). So there are 2 corner points.