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: \[ 3x + y \leq 6, \quad x \leq 1, \quad x \geq 0, \quad y \geq 0 \] These constraints are plotted on a 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 defined by $x \geq 0$ and $y \geq 0$ is the first quadrant. The feasible region is the area enclosed by these lines. The corner points are the intersections of these lines: - The intersection of $3x + y = 6$ and $x = 1$ is found by substituting $x = 1$ into the equation: \[ 3(1) + y = 6 \quad \Rightarrow \quad y = 3 \] This gives the point $(1, 3)$. - The intersection of $x = 1$ and $y = 0$ is $(1, 0)$.
- The intersection of $3x + y = 6$ and $y = 0$ is $(2, 0)$.
The corner points are therefore $(0, 6)$, $(1, 3)$, and $(2, 0)$.
The objective function $Z = 5x + 4y$ will be maximized at one of these corner points. Evaluating $Z$ at each point yields: - At $(0, 6)$, $Z = 5(0) + 4(6) = 24$.
- At $(1, 3)$, $Z = 5(1) + 4(3) = 17$.
- At $(2, 0)$, $Z = 5(2) + 4(0) = 10$.
The maximum value of $Z$ is $24$, occurring at the point $(0, 6)$.
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