Question

The solution set of the following system of inequations:
$x+2 y \leq 3,3 x+4 y \geq 12, x \geq 0, y \geq 1$, is

• A. bounded region
• B. unbounded region
• C. only one point
• D. empty set

Answer 1

Answer: (D)

The solution region is bounded by the straight lines
$x+2 y=3$...(i)
$3 x+4 y=12$...(ii)
$x =0$...(iii)
$y =1$...(iv)
The straight lines (i) and (ii) meet the $x$ -axis in $(3,0)$ and $(4,0)$ and for $(0,0), x+2 y \leq 3 \Rightarrow 0 \leq 3$ which is true.
Hence $(0,0)$ lies in the half plane $x+2 y \leq 3$, Also the lines (1) and $(2)$ meet the $y$ -axis in $(0,3 / 2)$ and $(0,3)$ and for $(0,0) 3 x+ 4 y \geq 12 \Rightarrow 0 \geq 12$ which is not true.
Hence $(0,0)$ doesn't belong to the half plane $3 x+4 y \geq 0$.
Also $x \geq 0, y \geq 1 \Rightarrow $ the solution set belongs to the first quadrant. Moreover all the boundary lines are part of the solution.
From the shaded region, We find that there is no solution of the given system. Hence the solution set is an empty set.