Solve the following LPP graphically: Maximize $Z = x + 3y$, subject to the constraints: $x + 2y \le 200$, $x + y \le 150$, $y \le 75$, $x, y \ge 0$.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $Z = 275$ at $(50, 75)$ ##
Maximize $Z = x + 3y$ subject to the constraints:
- $x + 2y \le 200$
- $x + y \le 150$
- $y \le 75$
- $x, y \ge 0$
Now, $x + 2y = 200$
|
$x$ |
0 |
200 |
100 |
50 |
|
$y$ |
100 |
0 |
50 |
75 |
$x + y = 150$
|
$x$ |
0 |
150 |
75 |
100 |
|
$y$ |
150 |
0 |
75 |
50 |
$y = 75$
Corner points and evaluation of $Z$:
|
Corner point |
$Z=x+3y$ |
|
$(0, 0)$ |
$0 + 3(0) = 0$ |
|
$(0, 75)$ |
$0 + 3 \times 75 = 225$ |
|
$(50, 75)$ |
$50 + 3 \times 75 = 275 ⇒\text{(Maximise)}$ |
|
$(100, 50)$ |
$100 + 3 \times 50 = 250$ |
|
$(150, 0)$ |
$150 + 3 \times 0 = 150$ |
Maximum value of $Z = 275$ at $x = 50, y = 75$