Solve the following problem graphically: Minimise and Maximise $Z = 3x + 9y$, Subject to the constraints: $x + 3y \le 60$, $x + y \ge 10$, $x \le y$, $x \ge 0, y \ge 0$.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Min: $60$, Max: $180$ ##
Let
$x + 3y \le 60 \dots(i)$
$x + y \ge 10 \dots(ii)$
$x \le y \dots(iii)$
$x \ge 0, y \ge 0 \dots(iv)$
For $x+3y=60$
|
$x$ |
0 |
60 |
|
$y$ |
20 |
0 |
For $x+y=10$
|
$x$ |
0 |
10 |
|
$y$ |
10 |
0 |
From the graph, we see that the feasible region is $ABCDA$. The coordinates of the corner points are $A(0, 10)$, $B(5, 5)$, $C(15, 15)$, and $D(0, 20)$.
Now we evaluate $Z$ at each corner point:
|
Corner Point |
Value of $Z=3x+9y$ |
|
$A(0, 10)$ |
$90$ |
|
$B(5, 5)$ |
$60$ (Minimum) |
|
$C(15, 15)$ |
$180$ (Maximum) |
|
$D(0, 20)$ |
$180$ (Maximum) |
Hence, the maximum value of $Z$ on the feasible region occurs at the two corner points $C(15, 15)$ and $D(0, 20)$ and it is 180 in each case. The minimum value of $Z$ is 60 at the point $B(5, 5)$ of the feasible region.