Maximise $Z = x + y$ subject to $x + 4y \leq 8, 2x + 3y \leq 12, 3x + y \leq 9, x \geq 0, y \geq 0$.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3.9
We are given that $Z = x + y$ subject to the constraints
$x + 4y \leq 8 \dots(i)$
|
x |
0 |
8 |
|
y |
2 |
0 |
$2x + 3y \leq 12 \dots(ii)$
|
x |
0 |
6 |
|
y |
4 |
0 |
$3x + y \leq 9 \dots(iii)$
|
x |
0 |
3 |
|
y |
9 |
0 |
$x \geq 0, y \geq 0$
On solving eq. $(i)$ and $(iii)$ we get
$x = \frac{28}{11} \text{ and } y = \frac{15}{11}$
Here, $OABC$ is the feasible region whose corner points are
$O(0, 0), A(3, 0), B\left(\frac{28}{11}, \frac{15}{11}\right), C(0, 2)$
Let us evaluate the value of $Z$
|
Corner points |
Value of $Z=x+y$ |
|
$O(0, 0)$ |
$Z=0+0=0$ |
|
$A(3,0)$ |
$Z=3+0=3$ |
|
$B(\frac{28}{11},\frac{15}{11})$ |
$Z=\frac{28}{11}+\frac{15}{11}=\frac{43}{11}=3.9$ Maximum |
|
$C(0, 2)$ |
$Z=0+2=2$ |
Hence, the maximum value of Z is 3.9 at $B(\frac{28}{11},\frac{15}{11})$