1.

Refers to question 15.Determine the maximum distance that the man can travel.

Answer»


Solution :Referring to solution 15, we have

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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