CBSE 2025, Set 3

Question

For the given graph of a Linear Programming Problem, write all the constraints satisfying the given feasible region.

![Diagram LPP Feasible Region](https://www.edvaya.com/images/cbse/2025/mathematics/set3/q31-lpp-feasible-region.webp)

Edvaya · Accepted Answer

The correct answer is Subjective. Let us identify the linear equations corresponding to the boundary lines of the shaded feasible region $OABCD$.

**1. Equation of Line AB:**
 The line passes through points $A(0, 200)$ and $B(50, 250)$.
 First, find the slope ($m$):
 $$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{250 - 200}{50 - 0} = \frac{50}{50} = 1$$
 Using the point-slope form $y - y_1 = m(x - x_1)$:
 $$y - 200 = 1(x - 0) \implies y - x = 200 \implies -x + y = 200$$
 The shaded region contains the origin $(0,0)$ and lies "below" this line. Testing $(0,0)$ gives $0 \le 200$, which is True.
 Constraint 1: **$-x + y \le 200$**

**2. Equation of Line BC:**
 The line passes through points $B(50, 250)$ and $C(150, 150)$.
 Slope $m = \frac{150 - 250}{150 - 50} = \frac{-100}{100} = -1$
 Using the point-slope form:
 $$y - 150 = -1(x - 150) \implies y - 150 = -x + 150 \implies x + y = 300$$
 The shaded region contains the origin. Testing $(0,0)$ gives $0 \le 300$ (True).
 Constraint 2: **$x + y \le 300$**

**3. Equation of Line CD:**
 The line passes through points $C(150, 150)$ and $D(200, 0)$.
 Slope $m = \frac{0 - 150}{200 - 150} = \frac{-150}{50} = -3$
 Using the point-slope form:
 $$y - 0 = -3(x - 200) \implies y = -3x + 600 \implies 3x + y = 600$$
 The shaded region contains the origin. Testing $(0,0)$ gives $0 \le 600$ (True).
 Constraint 3: **$3x + y \le 600$**

**4. Non-negativity Constraints:**
 The entire feasible region is strictly bound within the first quadrant, bordered by the y-axis ($OA$) and x-axis ($OD$).
 Constraint 4: **$x \ge 0, y \ge 0$**

**Final Set of Constraints:**
 $-x + y \le 200$
 $x + y \le 300$
 $3x + y \le 600$
 $x \ge 0, y \ge 0$

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Mathematics > Vectors And Linear Programming > Linear Programming

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