Question

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

Which of the following are corner points of the convex polygon region of the solution?

• A. $(0,0)$
• B. $(2,3)$
• C. $(0,4)$
• D. $(\frac{3}{2},\frac{5}{2})$

Answer 1

The correct option is D $(\frac{3}{2},\frac{5}{2})$

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.

solving $3x-y-2=0$ and $x+y=4$ we get the intersecting point.

adding the equations

$⟹3x+x=2+4⟹4x=6⟹x=\frac{3}{2}$

substituting $x=\frac{3}{2}⟹y=4-\frac{3}{2}⟹y=\frac{5}{2}$

As shown in the above figure, the blue shaded region is the feasible region with the three corner points $(\frac{3}{2},\frac{5}{2}),(\frac{2}{3},0),(4,0)$