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A Factory Owner Purchases Two Types of Machines, a and B, for His Factory. the Requirements and Limitations for the Machines Are as Follows: - Mathematics

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Sum

A factory owner purchases two types of machines, A and B, for his factory. The requirements and limitations for the machines are as follows:

  Area occupied by the
machine
Labour force for each
machine
Daily output in
units
Machine A
Machine B
1000 sq. m
1200 sq. m
12 men
8 men
60
40

He has an area of 7600 sq. m available and 72 skilled men who can operate the machines.
How many machines of each type should he buy to maximize the daily output?

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Solution

Let machines of type A and y machines of type B were purchased.
Number of machines cannot be negative.
Therefore,

\[x, y \geq 0\]

We are given,

  Area occupied by the
machine
Labour force for each
machine
Daily output in
units
Machine A
Machine B
1000 sq. m
1200 sq. m
12 men
8 men
60
40

The area of 7600 sq m is available and there are 72 skilled men available to operate the machines.
Therefore, the constraints are

\[1000x + 1200y \leq 7600\]
\[\text{ and } 12x + 8y \leq 72\]

Total daily output = Z =  \[60x + 40y\]\

which is to be maximised.

Thus, the mathematical formulat‚Äčion of the given linear programming problem is 

Max Z =  \[60x + 40y\]

subject to

\[1000x + 1200y \leq 7600\]
\[12x + 8y \leq 72\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows :
1000x + 1200y = 7600, 12x + 8y = 72, x = 0 and y = 0
Region represented by 1000x + 1200y ≤ 7600:
The line 1000x + 1200y = 7600 meets the coordinate axes at \[A_1 \left( \frac{38}{5}, 0 \right)\] and \[B_1 \left( 0, \frac{19}{3} \right)\] respectively. By joining these points we obtain the line
1000x + 1200y = 7600. Clearly (0,0) satisfies the 1000x + 1200y = 7600. So, the region which contains the origin represents the solution set of the inequation 1000x + 1200y ≤ 7600.
Region represented by 12x + 8y ≤ 72:
The line 12x + 8y = 72 meets the coordinate axes at C1(6, 0) and D1(0, 9) respectively. By joining these points we obtain the line 12x + 8y = 72 .Clearly (0,0) satisfies the inequation 12x + 8y ≤ 72. So,the region which contains the origin represents the solution set of the inequation 12x + 8y ≤ 72.
Region represented by x ≥ 0 and y ≥ 0:
Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequations x ≥ 0, and ≥ 0.
The feasible region determined by the system of constraints 1000x + 1200y ≤ 7600, 12x+ 8y ≤ 72, x ≥ 0, and y ≥ 0 are as follows.


The corner points are O(0, 0)

\[B_1 \left( 0, \frac{19}{3} \right)\] , E1(4, 3), C1(6, 0)
The values of Z at these corner points are as follows
 
Corner point Z= 60x + 40y
O 0
B1 253.3
E1 360
C1 360

The maximum value of Z is 360 which is attained at E1(4, 3) and C1(6, 0). 

Thus, the maximum output is Rs 360 obtained when 4 units of type A and 3 units of type B or 6 units of type A are manufactured.
 
Concept: Graphical Method of Solving Linear Programming Problems
  Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.4 | Q 8 | Page 51

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