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Refers to question 15.Determine the maximum distance that the man can travel. |
Answer» MAXIMISE Z=x+y subject to `2x+3y le 120, 8x+5y le 400, x ge 0, y ge 0` On solving we get `8x +5y=400 and 2x+3y=120 we get ` `(30)/(7),(80)/(7)` and (0,40) HENCE the maximum distance that the man can TRAVEL is `54(2)/(7)KM`. |
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