For the following shaded region, the linear constraints are
- $x-y \leq 0,-x+3 y \leq 3, x \geq 0, y \geq 0$
- $x-y \geq 0,-\dot{x}+3 y \geq 3, x \geq 0, y \geq 0$
- $x-y \geq 0,-x+3 y \leq 3, x \geq 0, y \geq 0$
- $x-y \leq 0,-x+3 y=3, x \geq 0, y \geq 0$
Solution
Looking at the graph, we can see: - The region is bounded by the line $y=x$ - The line $-x+3 y=3$ - The $x$-axis $(y=0)$ - The $y$-axis $(x=0)$ 2.
The shaded region shows: - It's above the line $y=x($ so $x-y \leq 0)$ - It's below the line $-x+3 y=3$ - It's in the first quadrant $(x \geq 0, y \geq 0)$ 3.
Looking at the options given: (A) $x-y \leq 0,-x+3 y \leq 3, x \geq 0, y \geq 0$ (B) $x-y \geq 0,-x+3 y \geq 3, x \geq 0, y \geq 0$ (C) $x-y \geq 0,-x+3 y \leq 3, x \geq 0, y \geq 0$ (D) $x-y \leq 0,-x+3 y=3, x \geq 0, y \geq 0$ The correct answer is $(A)$ because: The region is above $y=x$, so $x-y \leq 0$ The region is below $-x+3 y=3$, so $-x+3 y \leq 3$ The region is in the first quadrant, so $x \geq 0$ and $y \geq 0$ Therefore, the linear constraints defining the shaded region are: $x-y \leq 0,-x+3 y \leq 3, x \geq 0, y \geq 0$
The shaded region is defined by the linear constraints from option (A): $\begin{aligned} & x-y \leq 0 \\ & -x+3 y \leq 3 \\ & x \geq 0 \\ & y \geq 0 \end{aligned}$
Asked in: MHT CET 2024 (11 May Shift 1)