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 ≥ 0. Let us construct the following table from the given data

Screw A (x) 
Screw B (x) 
# minutes 
Machine A (h) 
4 
6 
4h x 60 = 240 min 
Machine B (h) 
6 
3 
4h x 60 = 240 min 
Profit (Rs.) 
7 
10 

Since we have only 240m of machine time, we have the following constraints:
4x + 6y $\leq$ 240 $\to$2x + 3y $\leq$ 120
6x + 3y $\leq$ 240 $\to$2x + y $\leq$ 80
The profits on Screw A is Rs. 7 and on Screw B is Rs. 10. We need to maximize the profits, i.e. maximize 7x + 10y, given the above constraints.
$\textbf{Plotting the constraints}$:
Plot the straight lines 2x + 3y = 120 and 2x + y = 80
First draw the graph of the line 2x + 3y = 120
If x = 0, y = 4 and if y = 0, x = 12. So, this is a straight line between (0,4) and (12,0).
If x = 0, y = 40 and if y = 0, x = 60. So, this is a straight line between (0,40) and (60,0). At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 0. So the area associated with this inequality is bounded towards the origin.
Similarly, draw the graph of the line 2x + y = 80.
If x = 0, y = 80 and if y =0, x = 40. So, this is a straight line between (0,80) and (40,0).
At (0,0), in the inequality, we have 0 + 0 = 0 which is $\leq$ 0. So the area associated with this inequality is bounded towards the origin.
$\textbf{Finding the feasible region}$:
We can see that the feasible region is bounded and in the first quadrant.
On solving the equations 2x + 3y = 120 and 2x + y = 80, we get,
2x – 2x + 3y – y = 120 – 80 $\to$2y = 40 $\to$y = 20
If y = 20,2 x = 120 – 2x20 $\to$x = 30.
$\Rightarrow x = 30, y = 20 $
Therefore the feasible region has the corner points (0,0), (0,40), (30,20), (40,0) as shown in the figure.
$\textbf{Solving the objective function using the corner point method}$
The values of Z at the corner points are calculated as follows:
Corner Point

Z = 7x + 10y

O (0,0)

0

A (0,40)

280

E (30,20)

410 (Max value)

D (40,0)

400

$\textbf{ The maximum profit we can make is Rs. 410, which involves making 30 packages of Screw A and 20 packages of Screw B.}$