Solve the following linear programming problem graphically: Maximise $Z = 4x + y$, Subject to the constraints: $x + y \le 50$, $3x + y \le 90$, $x \ge 0, y \ge 0$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Max $Z = 120$ at $(30, 0)$ ##
Let
$ x + y \le 50 \dots(i)$
$3x + y \le 90 \dots(ii)$
$x \ge 0, y \ge 0 \dots(iii)$
For $x + y = 50$:
|
X |
0 |
50 |
|
Y |
50 |
0 |
For $3x + y = 90$:
|
X |
0 |
30 |
|
Y |
90 |
0 |
From the graph, the feasible region is $OABCO$. The coordinates of the corner points are $O(0, 0), A(30, 0), B(20, 30)$ and $C(0, 50)$.
|
Corner Point |
Value of Z=4x+y |
|
$O(0, 0)$ |
0 |
|
$A(30, 0)$ |
120 (Maximum) |
|
$B(20, 30)$ |
110 |
|
$C(0, 50)$ |
50 |
Hence, maximum value of $Z$ is 120 at the point $A(30, 0)$.