Solve the following linear programming problem graphically: Minimise: $Z = 5x + 10y$, subject to the constraints: $x + 2y \le 120$, $x + y \ge 60$, $x - 2y \ge 0$, $x \ge 0, y \ge 0$.

$Z = 300$ at $(60, 0)$

The correct answer is Option (3) → $Z = 300$ at $(60, 0)$ ##

Objective Function:

$Z = 5x + 10y$

Constraints:

$x + 2y \le 120, \quad x + y \ge 60, \quad x - 2y \ge 0, \quad x \ge 0, \quad y \ge 0$

Points to plot the boundary lines:

(i) For $x + 2y = 120$:

$\Rightarrow \frac{x}{120} + \frac{y}{60} = 1$

(ii) For $x + y = 60$:

$\Rightarrow \frac{x}{60} + \frac{y}{60} = 1$

(iii) For $x - 2y = 0$:

$\Rightarrow x = 2y$

The corner points of the feasible region $ABCD$ are:

$A(40, 20), \quad B(60, 30), \quad C(120, 0), \quad D(60, 0)$

Evaluation of $Z$:

$∴$ The value of $Z$ is minimum at $x = 60$ and $y = 0$, and the minimum value = 300.