Find the linear inequalities for which the shaded area is the solution

1.	Find the linear inequalities for which the shaded area is the solution set in the figure given below.
Answer» We have seen that the shaded region and origin are on the same side of the line 3x + 4y = 12. For (0,0) we have 0 + 0 - 12 < 0. So the shaded region satisfies the inequality 3x + 4y \(\le\)12. We have seen that the shaded region and origin are on the same side of the line 4x + 3y =12 For (0,0) we have 0 + 0 -12 < 0. So the shaded region satisfies the inequality 4x + 3y \(\le\)12. Also , the region lies in the first quadrent Therefore x \(\ge\) 0 and y \(\ge\)0 Thus the linear inequation comprising the given solution set are +4y \(\le\)12, 4x + 3y \(\le\) 12, x \(\ge\) 0, y \(\ge\)0

Find the linear inequalities for which the shaded area is the solution set in the figure given below.

Answer»

We have seen that the shaded region and origin are on the same side of the line 3x + 4y = 12.

For (0,0) we have 0 + 0 - 12 < 0. So the shaded region satisfies the inequality 3x + 4y \(\le\)12.

We have seen that the shaded region and origin are on the same side of the line 4x + 3y =12

For (0,0) we have 0 + 0 -12 < 0. So the shaded region satisfies the inequality 4x + 3y \(\le\)12.

Also , the region lies in the first quadrent Therefore x \(\ge\) 0 and y \(\ge\)0

Thus the linear inequation comprising the given solution set are +4y \(\le\)12, 4x + 3y \(\le\) 12, x \(\ge\) 0, y \(\ge\)0