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A Manufacturer Makes Two Types a and B of Tea-cups. Three Machines Are Needed for the Manufacture and the Time in Minutes Required for Each Cup on the Machines is Given Below: - Mathematics

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Sum

A manufacturer makes two types A and B of tea-cups. Three machines are needed for the manufacture and the time in minutes required for each cup on the machines is given below:

  Machines
I II III
A
B
12
6
18
0
6
9

Each machine is available for a maximum of 6 hours per day. If the profit on each cup A is 75 paise and that on each cup B is 50 paise, show that 15 tea-cups of type A and 30 of type B should be manufactured in a day to get the maximum profit.

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Solution

Let units of type A and y units of type cups were made.
Quantities cannot be negative.Therefore, \[x, y \geq 0\]

As we are given,

  Machines
I II III
A
B
12
6
18
0
6
9
Every machine is available for a maximum of 6 hours per day i.e. 360 minutes per day.
Therefore, the constraints are

\[12x + 6y \leq 360\]
\[18x + 0y \leq 360\]
\[6x + 9y \leq 360\]

If the profit on each cup A is 75 paise and that on each cup B is 50 paise.

Total profit = Z = \[0 . 75x + 0 . 50y\]  which is to be maximised.

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

Max Z =  \[0 . 75x + 0 . 50y\] subject to

\[12x + 6y \leq 360\]
\[18x + 0y \leq 360\]
\[6x + 9y \leq 360\]

\[x, y \geq 0\]

First we will convert inequations into equations as follows :
12x + 6y = 360, 18x = 360,

\[6x + 9y = 360\] x = 0 and y = 0

Region represented by 12x + 6y ≤ 360:
The line 12x + 6y = 360 meets the coordinate axes at A1(30, 0) and B1(0, 60) respectively. By joining these points we obtain the line 12x + 6y = 360.Clearly (0,0) satisfies the 12x + 6y = 360. So,the region which contains the origin represents the solution set of the inequation 12x + 6y ≤ 360.

Region represented by 18x + 0y ≤ 360:
The line 18x + 0y = 360 meets the coordinate axes at C1(20, 0) . We obtain the line 18x + 0y = 360.Clearly (0,0) satisfies the inequation 18x + 0y ≤ 360. So,the region which contains the origin represents the solution set of the inequation 18x + 0y ≤ 360.

Region represented by
\[6x + 9y \leq 360\] The line
\[6x + 9y = 360\]  meets the coordinate axes at E1(60, 0) and F1(0, 40) respectively. By joining these points we obtain the line \[6x + 9y = 360\] Clearly (0,0) satisfies the inequation \[6x + 9y \leq 360\]   So,the region which contains the origin represents the solution set of the inequation  \[6x + 9y \leq 360\]
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 12x + 6y ≤ 360, 18x + 0y ≤ 360,
\[6x + 9y \leq 360\] , x ≥ 0, and y ≥ 0 are as follows.
The corner points are O(0, 0) F1(0, 40), G1(15, 30), H1(20, 20) and C1(20, 0). 

The values of Z at these corner points are as follows
 
Corner point Z= 0.75x + 0.50y
O 0
F1 20
G1 26.25
H1 25
C1 15

Thus, the maximum profit is Rs 26.25 obtained when 15 units of type A and 30 units of type B cups were made.
Concept: Graphical Method of Solving Linear Programming Problems
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APPEARS IN

RD Sharma Class 12 Maths
Chapter 30 Linear programming
Exercise 30.4 | Q 7 | Page 51

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