Maths. Section ALinear Programming

Linear inequalities corresponding to the shaded feasible region OABCO…

Linear inequalities corresponding to the shaded feasible region OABCO in the given figure are

$Graph of the feasible region OABCO$

Answer & explanation

Correct answer: option 1

The correct answer is Option (1) → $4x + 3y \le 12,\ 3x+4y \le 12,\ x \ge 0, y \ge 0$

The line through D(0, 4) and A(3, 0), in intercept form:

$ \displaystyle \frac{x}{3} + \frac{y}{4} = 1 $

$4x + 3y = 12$

The line through C(0, 3) and E(4, 0):

$ \displaystyle \frac{x}{4} + \frac{y}{3} = 1 $

$3x + 4y = 12$

The shaded region contains the origin O, and O gives $0 \le 12$ in both, so the region lies on the side ≤ of each line:

$4x + 3y \le 12$

$3x + 4y \le 12$

The region is in the first quadrant, so $x \ge 0$ and $y \ge 0$.