Linear inequalities for which the shaded region for the given figure is the solution set, are

Question

Linear inequalities for which the shaded region for the given figure is the solution set, are (A) x+y ≤ 8, x+y ≤ 4, x ≤ 5, y ≤ 5, x ≥ 0, y ≥ 0 (B) x+y ≤ 8, x+y ≥ 4, x ≤ 5, y ≤ 5, x ≥ 0, y ≥ 0 (C) x+y ≥ 8, x+y ≥ 4, x ≥ 5, y ≥ 5, x ≥ 0, y ≥ 0 (D) None of the above

Tardigrade Admin · Accepted Answer

(i) Consider the line x+y=8. We observe that the shaded region and origin lie on the same side of this line and (0,0) satisfies x+y ≤ 8. Therefore, x+y ≤ 8 is the linear inequality corresponding to the line x+y=8. (ii) Consider x+y=4. We observe that shaded region and origin are on the opposite side of this line and (0,0) satisfies x+y ≤ 4. Therefore, we must have x+y ≥ 4 as linear inequalities corresponding to the line x+y=4 (iii) Shaded portion lie below the line y=5. So, y ≤ 5 is the linear inequality corresponding to y=5. (iv) Shaded portion lie on the left side of the line x=5. So, x ≤ 5 is the linear inequality corresponding to x=5. (v) Also, the shaded region lies in the first quadrant only. Therefore, x ≥ 0, y ≥ 0. In view of (i), (ii), (iii), (iv) and (v) above the linear inequalities corresponding to the given solutions are x+y ≤ 8, x+y ≥ 4, y ≤ 5, x ≤ 5, x ≥ 0 and y ≥ 0

Q. Linear inequalities for which the shaded region for the given figure is the solution set, are

$x + y \leq 8, x + y \leq 4, x \leq 5, y \leq 5, x \geq 0, y \geq 0$

$x + y \leq 8, x + y \geq 4, x \leq 5, y \leq 5, x \geq 0, y \geq 0$

$x + y \geq 8, x + y \geq 4, x \geq 5, y \geq 5, x \geq 0, y \geq 0$

None of the above

Q. Linear inequalities for which the shaded region for the given figure is the solution set, are

x+y≤8,x+y≤4,x≤5,y≤5,x≥0,y≥0

x+y≤8,x+y≥4,x≤5,y≤5,x≥0,y≥0

x+y≥8,x+y≥4,x≥5,y≥5,x≥0,y≥0

None of the above

$x + y \leq 8, x + y \leq 4, x \leq 5, y \leq 5, x \geq 0, y \geq 0$

$x + y \leq 8, x + y \geq 4, x \leq 5, y \leq 5, x \geq 0, y \geq 0$

$x + y \geq 8, x + y \geq 4, x \geq 5, y \geq 5, x \geq 0, y \geq 0$