Question

Section C
Q15
Solve graphically: x+2y3,3x + 4y < 12, y > 1, x > 0, y = 0​

Answer 1

Answer:

We have x + 2y ≤ 3, 3x + 4y > 12, x > 0, y ≥ 1

Now let’s plot lines x + 2y = 3, 3x + 4y = 12, x = 0 and y = 1 in coordinate plane.

Line x + 2y = 3 passes through the points (0, 3/2) and (3, 0).

Line 3jc + 4y = 12 passes through points (4, 0) and (0, 3).

For (0, 0), 0 + 2(0) – 3 < 0.

Therefore, the region satisfying the inequality x + 2y ≤ 3 and (0,0) lie on the same side of the line x + 2y = 3.

For (0, 0), 3(0) + 4(0)- 12 ≤0.

Therefore, the region satisfying the inequality 3x + 4y ≥ 12 and (0, 0) lie on the opposite side of the line 3x + 4y = 12.

The region satisfying x > 0 lies to the right hand side of the y-axis.

The region satisfying y > 1 lies above the line y = 0