Answer:
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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 \[x\ge 0,\] \[y\ge 0.\] Lump Cutter?s Time Finisher?s Time Profit (in Rs.) A 2 1 6 B 1 2 11 Maximum time available 104 76
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