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A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and a sprayer. It takes 2 hours on grinding/cutting machine and 3 hours in sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/ cutting machine for at the most 12 hours. The profit fro the sale of the lamp is ₹5 and that from the shade is ₹3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit? |
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Answer» Let the cottage industry manufacture x pedestal LAMPS & y wooden shades.therefore, x > 0 & y > 0the given information can written as in table form as given below: PEDESTAL WOODEN time LAMPS (x) SHADES (y) available grinding/ 2 1 < 12cutting SPRAYS 3 2 < 20profit 5 3 then, required LPP is Maximise , Z = 5x + 3y therefore, the constraints 2x + y < 12 (1) 3x + 2y < 20 (2) x > 0, y > 0 USING (1) 2x + y = 12 PUT x = 0 put y = 0 2(0) + y = 12 2x + 0 = 12 0 + y = 12 2x = 12 y = 12 x = 6 x 0 6 y 12 0 now, using (2)3x + 2y = 20 put x = 0 put y = 0 3(0) + 2y = 20 3x + 2(0) = 20 2y = 20 3x = 20 y = 10 x = 6.6 x 0 6.6 y 10 0 solving (1) and (2) 2x + y = 12 (× 3) we get, 6x + 3y = 36 (3) 3x + 2y = 20 (× 2) we get, 6x + 4y = 40 (4)now , solving (3) & (4) 6x + 3y = 36 6x + 4y = 40 (-) (-) (-) -y = -4 y = 4now, putting the value of y in eqn (1) 2x + y = 12 2x + 4 = 12 2x = 12 - 4 2x = 8 x = 4 so, points of intersection are (4,4).... check for origin ;for (1) 0+0 < 12 for (2) 0+0 < 20 0 < 12 0 < 20 true... true... ( graph is as shown in attachment )Maximise Z = 5x + 3y therefore points are,O (0,0) ----- 5(0) + 3(0) = 0 + 0 = 0A (0,10) ----- 5(0) + 3(10) = 0 + 30 = 30B (6,0) ----- 5(6) + 3(0) = 30 + 0 = 30C (4,4) ----- 5(4) + 3(4) = 20 + 12 = 32 (maximum) |
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