Question

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 $4min$ on the automatic and $6\min$ on hand operated machines to get of screws $A$, while it takes $6\min$ on automatic and $3min$ on the hand operated machines to manufacture a package to screws $B$. Each machine is available for atmost $4h$ 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

• A. ₹ 400
• B. ₹ 510
• C. ₹ 410
• D. $₹310$

Answer 1

Answer: (C)

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 $Z=7x+10y$...(i)
Subject to constraints are
$4x+6y\le 240\iff 2x+3y\le 120$...(ii)
$6x+3y\le 240\iff 2x+y\le 80$...(iii)
and $x\ge 0,y\ge 0$...(iv)
Firstly, draw the graph of the line $2x+3y=120$.

x	0	60
y	40	0

Putting $(0,0)$ in the inequality $2x+3y\le 120$, we have
$2\times 0+3\times 0\le 120$
$\Rightarrow 0\le 120\text{ (which is true) }$
So, the half plane is towards the origin.
Secondary, draw the graph of the line $2x+y=80$

x	40	0
y	0	80

Putting $(0,0)$ in the inequality $2x+y\le 80$, we have
$2\times 0+0\le 80$
$\Rightarrow 0\le 80$(which is lrue)
So, the half plane is towards the origin. Since, $x,y\ge 0$
So, the feasible region lies in the first quadrant.
On solving equations $2x+3y=120$ and $2x+y=80$, 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	$z=7x+10y$
$O(0,0)$	0
$A(40,0)$	280
$B(30,20)$	$410\to$ 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 $₹410$.