Solve the following linear programming problem graphically: Minimise: $Z = 5x + 10y$, subject to the constraints: $x + 2y \le 120$, $x + y \ge 60$, $x - 2y \ge 0$, $x \ge 0, y \ge 0$.
Answer & explanation
Correct answer: option 3
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 $x + 2y = 120$:
$\Rightarrow \frac{x}{120} + \frac{y}{60} = 1$
|
$x$ |
120 |
0 |
60 |
|
$y$ |
0 |
60 |
30 |
(ii) For $x + y = 60$:
$\Rightarrow \frac{x}{60} + \frac{y}{60} = 1$
|
$x$ |
60 |
0 |
30 |
|
$y$ |
0 |
60 |
30 |
(iii) For $x - 2y = 0$:
$\Rightarrow x = 2y$
|
$x$ |
0 |
60 |
120 |
|
$y$ |
0 |
30 |
60 |
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 $Z$:
|
Point |
$Z=5x+10y$ |
|
$A(40, 20)$ |
$400$ |
|
$B(60, 30)$ |
$600$ |
|
$C(120, 0)$ |
$600$ |
|
$D(60, 0)$ |
$300 \text{ (Min)}$ |
$∴$ The value of $Z$ is minimum at $x = 60$ and $y = 0$, and the minimum value = 300.