Q. A factory manufactures two types of screws, $A$ and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes $4 min$ on the automatic and $6 \min$ on hand operated machines to get of screws $A$, while it takes $6\min$ on automatic and $3 min$ on the hand operated machines to manufacture a package to screws $B$. Each machine is available for atmost $4 h$ on any day. The manufacturer can sell a package of screws $A$ at a profit of $₹ 7$ and screws $B$ at a profit of $₹ 10$. Assuming that he can sell all the screws he manufactures, The maximum profit is
Linear Programming
Solution:
Let the manufacturer produces $x$ package of screws $A$ and $y$ package of screws $B$. We construct the following table
Item
Number of packages
Time on automatic machine (in min)
Time on hand machine (in min)
Prófit (in ₹)
A
x
4x
6x
7x
B
y
6y
3 y
10 y
Total
x+y
4x + 6y
6x + 3y
17 x + 10 y
Availability
$4 \times 60=240$
$4 \times 60=240$
The profit on a package of screws $A$ is $₹ 7$ and on the package of screws $B$ is $₹ 10$.
Our problem is to maximise $Z=7 x+10 y$...(i)
Subject to constraints are
$ 4 x+6 y \leq 240 \Leftrightarrow 2 x+3 y \leq 120 $...(ii)
$6 x+3 y \leq 240 \Leftrightarrow 2 x+y \leq 80$...(iii)
and $ x \geq 0, y \geq 0$...(iv)
Firstly, draw the graph of the line $2 x+3 y=120$.
x
0
60
y
40
0
Putting $(0,0)$ in the inequality $2 x+3 y \leq 120$, we have
$2 \times 0+3 \times 0 \leq 120$
$\Rightarrow 0 \leq 120 \text { (which is true) }$
So, the half plane is towards the origin.
Secondary, draw the graph of the line $2 x+y=80$
x
40
0
y
0
80
Putting $(0,0)$ in the inequality $2 x+y \leq 80$, we have
$2 \times 0+0 \leq 80$
$\Rightarrow 0 \leq 80$(which is lrue)
So, the half plane is towards the origin. Since, $x, y \geq 0$
So, the feasible region lies in the first quadrant.
On solving equations $2 x+3 y=120$ and $2 x+y=80$, we get $B(30,20)$.
$\therefore$ Feasible region is $OABCO$.
The corner points of the feasible region are $O(0,0), A(40,0)$, $B(30,20)$
and $C(0,40)$. The values of $Z$ at these points are as follows
Corner point
$z=7 x+10 y$
$O(0,0)$
0
$A(40,0)$
280
$B(30,20)$
$410 \rightarrow$ Maximum
$C(0,40)$
400
The maximum value of $Z$ is $₹ 410$ at $B(30,20)$.
Thus, the factory should produce 30 packages of screws $A$ and 20
packages of screws $B$ lo yel the maximum profil of $₹ 410$.
| Item | Number of packages | Time on automatic machine (in min) | Time on hand machine (in min) | Prófit (in ₹) |
|---|---|---|---|---|
| A | x | 4x | 6x | 7x |
| B | y | 6y | 3 y | 10 y |
| Total | x+y | 4x + 6y | 6x + 3y | 17 x + 10 y |
| Availability | $4 \times 60=240$ | $4 \times 60=240$ |
| x | 0 | 60 |
| y | 40 | 0 |
| x | 40 | 0 |
| y | 0 | 80 |
| Corner point | $z=7 x+10 y$ |
|---|---|
| $O(0,0)$ | 0 |
| $A(40,0)$ | 280 |
| $B(30,20)$ | $410 \rightarrow$ Maximum |
| $C(0,40)$ | 400 |