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Refers to question 15.Determine the maximum distance that the man can travel. |
Answer» <html><body><p><br/></p>Solution :Referring to solution 15, we have <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_MATH_XII_C12_S01_020_S01.png" width="<klux>80</klux>%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/maximise-2172432" style="font-weight:bold;" target="_blank" title="Click to know more about MAXIMISE">MAXIMISE</a> Z=x+y subject to <br/> `2x+3y le 120, 8x+5y le 400, x ge 0, y ge 0` <br/> On solving we get `8x +5y=400 and 2x+3y=120 we get ` <br/> `(30)/(7),(80)/(7)` and (0,40) <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_NCERT_EXE_MATH_XII_C12_S01_020_S02.png" width="80%"/> <br/> <a href="https://interviewquestions.tuteehub.com/tag/hence-484344" style="font-weight:bold;" target="_blank" title="Click to know more about HENCE">HENCE</a> the maximum distance that the man can <a href="https://interviewquestions.tuteehub.com/tag/travel-1426745" style="font-weight:bold;" target="_blank" title="Click to know more about TRAVEL">TRAVEL</a> is `54(2)/(7)<a href="https://interviewquestions.tuteehub.com/tag/km-1064498" style="font-weight:bold;" target="_blank" title="Click to know more about KM">KM</a>`.</body></html> | |