Question

Consider the linear inequations and solve them graphically:
$3x-y-2>0;x+y\le 4;x>0;y\ge 0$.

The solution region of these inequations is a convex polygon with _____ sides.

• A. $3$
• B. $4$
• C. $5$
• D. $7$

Answer 1

The correct option is A $3$

Given, $3x-y-2>0⟹y<3x-2$

$x+y\le 4$

$x>0$ and $y\ge 0$

First, draw the graph for equations $3x-y-2=0$

$x+y=4$

$x=0$ and $y=0$

$x=0$ is the Y-axis.

Hence, $x>0$ includes the right side region of the line.

$y=0$ is the X-axis.

Hence, $y\ge 0$ includes the upper side region of the line.

Similarly, for $y=3x-2$

Substitute y=0 we get, $3x=2⟹x=\frac{2}{3}$

Substitute x=0 we get, $y=-2$

Therefore, $y=3x-2$ line passes through (2/3,0) and (0,-2).

Hence, $y<3x-2$ includes the region below the line.

Similarly, for $x+y=4$

Substitute y=0 we get, $x=4$

Substitute x=0 we get, $y=4$

Therefore, $x+y=4$ line passes through (2/3,0) and (0,-2).

Hence, $x+y\le 4$ includes the region below the line.

As shown in the above figure, the blue shaded region is the feasible region which is the intersection of all 4 solution sets. Feasible region is the polygon with $3sides$ as shown in the figure.