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There are 101 men, some of them are drivers and some of them are fuel carrier. each car can carries enough fuel for a single race. How do we maximize the number of race?

attempt: let $x$ be the number of drivers, $y$ be number of fuel carriers. $x+y=101$, and we want to maximize $x+xy$. I would go with $x=51$ and $y=50$ this is just my feeling.

Is this right?

stopwatchingmesleep
  • 73
  • I don't understand how the word problem (first paragraph) translates into maximizing $x+xy$, but if you want to maximize $x+xy = x(y+1)$ subject to $x+(y+1)=102$ then yeah the optimal is when $x = y+1 = 51$. – antkam Dec 29 '19 at 07:14
  • This is because each fuel carrier can provide to more than one car! – stopwatchingmesleep Dec 30 '19 at 04:06

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