A factory manufactures two types of screws $A$ and $B$ , each type requiring the use of two machines, an automatic and a hand-operated. It takes $4$ minutes on the automatic and $6$ minutes on the hand-operated machines to manufacture a packet of screws $^{'} A^{'}$ while it takes $6$ minutes on the automatic and $3$ minutes on the hand-operated machine to manufacture a packet of screws 'B'. Each machine is available for at most $4$ hours on any day. The manufacturer can sell a packet of screws $^{'} A^{'}$ at a profit of $70$ paise and screws $^{'} B^{'}$ at a profit of $R s . 1$ . Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit? Formulate the given LLP and solve it graphically and find the maximum profit.

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Solution

Let $x$ be the number of packages of screws A and $y$ be the number of packages of screws B that we can make.
Clearly, $x, y \geq 0$ .
Total time on machine is $4 h r = 240 m i n$
$∴$ for automatic machine , time = $4 x + 6 y \leq 240 ⟹ 2 x + 3 y \leq 120$
$∴$ for hand machine, time = $6 x + 3 y \leq 240 ⟹ 2 x + y \leq 80$
The shaded region is the feasible region.
The profits on Screw A is Rs. 0.7 and on Screw B is Rs. 1. We need to maximize the profits, i.e. maximize $z = 0.7 x + y$ , given the above constraints.
Now, at A, $z = 28$
at O, $z = 0$
at E, $z = 41$
at D, $z = 40$
Hence, maximum profit is at point $E (30, 20)$ .

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