question_answer

A resourceful home decorator manufactures two types of lamps say A and B. Both lamps go through two technicians, first a cutter, second a finisher. Lamp A requires 2 h of the cutter's time and 1 h of the finisher's' time. Lamp B requires 1 h of cutter's and 2 h of finisher's time. The cutter has 104 h and finisher has 76 h of time available each month. Profit on one lamp A is Rs. 6.00 and on one lamp B is Rs. 11.00. Assuming that he can sell all that he produces, how many of each type of lamps should he manufacture to obtain the best return.

Answer:

The information given in the question can be put in the following tabular form            

Lump	Cutter?s Time	Finisher?s Time	Profit (in Rs.)
A	2	1	6
B	1	2	11
Maximum time available	104	76

Let the decorator manufacture x lamps of type A and y lamps of type B. \[\therefore \] Total profit \[=Rs.(6x+11y)\] Total time taken by the cutter in preparing x lamps of type A and y lamps of type B is \[(2x+y)\] hours. But the cutter has 104 hours only for each month.  \[\therefore \]     \[2x+y\le 104\] Similarly, the total time taken by the finisher in preparing x lamps of type A and y lamps of type B is \[(x+2y)\] hours. But the cutter has 76 h only for each month. \[\therefore \]      \[x+2y\le 76\] Since, the number of lamps cannot be negative. \[\therefore \]      \[x\ge 0\,\,\,\text{and}\,\,\,y\ge 0\] Let Z denotes the total profit. Then, \[Z=6x+11y.\] Since, the profit is to be maximised. So, the mathematical formulation of the given LPP is as follows Maximise \[Z=6x+11y.\] Subject to the constraints             \[2x+y\le 104,\] \[x+2y\le 76\] and       \[x\ge 0,\] \[y\ge 0.\]