Find the linear inequations for which the solution set is the shaded region given in Fig. 15.42
Find the linear iAlso the shaded region is in the first quadrant. Therefore, we must have $x \geq 0$ and $y \geq 0$nequations for which the solution set is the shaded region given in Fig. 15.42
Considering the line $x+y=4$, we find that the shaded region and the origin $(0,0)$ are on the same side of this line and $(0,0)$ does not satisfy the inequation $x+y \leq 4$ So, the first inequation is $x+y \leq 4$
Considering the line $y=3$, we find that the shaded region and the origin $(0,0)$ are on the same side of this line and $(0,0)$ satisfies the inequation $y \leq 3$ So, the corresponding inequation is $y \leq 3$
Considering the line $x=3$, we find that the shaded region and the origin $(0,0)$ are on the same side of this line and $(0,0)$ satisfies the inequation $x \leq 3$ So, the corresponding inequation is $x \leq 3$
Considering the line $x+5 y=4$, we find that the shaded region and the origin $(0,0)$ are on the opposite side of this line and $(0,0)$ does not satisfy the inequation $x+5 y \geq 4$ So, the corresponding inequation is $x+5 y \geq 4$
Considering the line $6 x+2 y=8$, we find that the shaded region and the origin $(0,0)$ are on the opposite side of this line and $(0,0)$ does not satisfy the inequation $6 x+2 y \geq 8$ So, the corresponding inequation is $6 x+2 y \geq 8$
Thus, the linear inequations comprising the given solution set are given below:
$x+y \leq 4, y \leq 3, x \leq 3, x+5 y \geq 4,6 x+2 y \geq 8, x \geq 0$ and $y \geq 0$