Solve the following linear programming problem graphically: Maximize $Z = 2x + 3y$, subject to the constraints: $x + y \le 6$, $x \ge 2$, $y \le 3$, $x, y \ge 0$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Max $Z = 15$ at $(3, 3)$ ##
Constraints: $x + y \le 6, x \ge 2, y \le 3, x, y \ge 0$
For the line $x + y = 6$:
|
x |
0 |
6 |
1 |
|
$y$ |
6 |
0 |
5 |
Evaluation of Corner Points:
|
Corner points |
$Z=2x+3y$ |
|
$A(2, 0)$ |
$2 \times 2 + 3 \times 0 = 4$ |
|
$B(6, 0)$ |
$2 \times 6 + 3 \times 0 = 12$ |
|
$C(3, 3)$ |
$2 \times 3 + 3 \times 3 = 15$ |
|
$D(2, 3)$ |
$2 \times 2 + 3 \times 3 = 13$ |
Maximum value of $Z$ is 15 at $D(3, 3)$.