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 4 minutes on the automatic and 6 minutes on hand opereated machines to manufacture a package of screw A, while it takes 6 minutes on automatic and 3 minutes on hand operated machines to manufacture a package of screw B. Each, machine is available for almost 4 hours on any day. The manufacturer can sell a package of screws A at a profit of 70 paise and screws B at Rs 1. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit ?

Answer 1

Let manufacturer should produce x packets of screw of type A and y packets of screw of type B.

∴ Profit on x packets = Rs 0.70x
and profit on y packets = Rs y

∴ Objective function to earn maximum profit

Z = 0.70 x + y

Time to produce x screws of type A = 4x minute
and time to produce y screws of type B = 6y minute

But automatic machine is available for 4 hours only.

∴ According to question.

4x + 6y ≤ 4 hour

⇒ 4x + 6y ≤ 240 minute

Similarly time taken to prepare screws of type A on handmade machine = 6x minutes and time taken to prepare screws of type B on handmade machine = 3y minutes

But handmade machine is available for 4 hours only a day.

∴ 6x + 3y ≤ 4 hour

⇒ 6x + 3y ≤ 240 minute

∵ x and y are the number of screws.

∴ x ≥ 0 and y ≥ 0

Mathematically formulation of this Linear

Programming Problem is as the following :

Maximize Z = 0.70x + y

Subject to the constraints

4x + 6y ≤ 240

6x + 3y ≤ 240

x ≥ 0, y ≥ 0

Converting the given inequations into equations

4x + 6y = 240 …..(1)

6x + 3y = 240 …..(2)

Region represented by 4x + 6y ≤ 240 : 

The line 4x + 6y = 240 meets the coordinate axis at A(60, 0) and B(0, 40).

4x + 6y = 240

x	60	0
y	0	40

A(60, 0); B(0, 40)

Join the points A and 5 to obtain the line. 

Clearly (0, 0) satisfies the in equations 4 × 0 + 6 × 0 = 0 ≤ 240. 

So, the region containing the origin represents the solution set of the in equation.

Region represented by 6x + 3y ≤ 240 : 

The line 6x + 3y = 240 meets the coordinate axis at points C(40, 0) and D(0, 80).

6x + 3y = 240

x	40	0
y	0	80

C(40,0);D(0,80)

Join C and D to obtain the line.

Clearly (0,0) satisfies the given in equation 6 × 0 + 3 × 0 = 0 ≤ 240. 

So the region containing the origin represents the solution set of the in equation.

Region represented x ≥ 0 and y ≥ 0 : 

Since every point in the first quadrant satisfies these in equations.

So the first quadrant in the region represented by the in equation x ≥ 0 and y ≥ 0.

The coordinate of point of intersection of lines 4x + 6y = 240 and 6x + 3y = 240 are x = 30 and y = 20.

The shaded region OAED represent the common region of the in equations.

This region is the feasible solution region of the in equations.

The comer points of this solution region are O(0,0), A(40, 0), E(30, 20) and D(0, 40).

The value of objective function on these points is as following table:

Point	x-coordinate	y-coordinate	Objective function Z = 0.70 x + y
O	0	0	Z0 = 0.70 x 0 + 0 = 0
A	40	0	ZA = 0.70 x 40 + 0 = 28
B	30	20	ZE = 0.70 x 30 + 20 = 41
D	0	40	ZD = 0.70 x 0 + 40 = 40

From the table the value of objective function is maximum at point E(30, 20) = Rs 41.

Hence the manufacturer should produce 30 packets of screw of type A and 20 packets of screws of type B to get the maximum profit Rs 41.