Question

In a Linear Programming Problem, the objective function $Z = 5x + 4y$ needs to be maximised under constraints $3x + y \leq 6$, $x \leq 1$, $x \geq 0$, $y \geq 0$. Express the LPP on the graph and shade the feasible region, and mark the corner points.

Hint:

In Linear Programming Problems, always identify the feasible region by graphing the constraints and checking the corner points for the optimal solution.

Answer 1

The constraints are: \[ 3x + y \leq 6, \quad x \leq 1, \quad x \geq 0, \quad y \geq 0 \] We plot these constraints on the graph: - The line $3x + y = 6$ intersects the axes at $(0, 6)$ and $(2, 0)$.
- The line $x = 1$ is a vertical line.
- The region $x \geq 0$ and $y \geq 0$ represents the first quadrant.
The feasible region is the area enclosed by these lines, and the corner points are where the lines intersect: - Intersection of $3x + y = 6$ and $x = 1$: Substitute $x = 1$ into $3x + y = 6$: \[ 3(1) + y = 6 \quad \Rightarrow \quad y = 3 \] So the point is $(1, 3)$. - Intersection of $x = 1$ and $y = 0$: The point is $(1, 0)$.
- Intersection of $3x + y = 6$ and $y = 0$: The point is $(2, 0)$.
Thus, the corner points are $(0, 6)$, $(1, 3)$, and $(2, 0)$.
The objective function $Z = 5x + 4y$ will be maximised at one of these corner points. Calculate $Z$ at each point:
- At $(0, 6)$, $Z = 5(0) + 4(6) = 24$.
- At $(1, 3)$, $Z = 5(1) + 4(3) = 17$.
- At $(2, 0)$, $Z = 5(2) + 4(0) = 10$.
Thus, the maximum value of $Z$ occurs at $(0, 6)$, and the maximum value is $24$.
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