I was practising calculus where I got this question:

The operating cost of a truck is $12 + \frac{x}{6}$ per km, when the truck travels x km/hour. If the driver earns 6 Rs. per hour, what is the most economical speed to operate the truck on a 400 km road? Also due to construction, the truck can travel only between 35 and 60 km/hour.

My approach:

$OP = \left(12 + \frac{x}{6}\right)400$

$DE = \frac{400}{x}*6 = \left(\frac{6}{x}\right)400$

Net Profit = Net Earnings - Operating Cost of Truck


$P = DE - OP$

$P = 400\left(\frac{6}{x} - 12 - \frac{x}{6}\right)$

Differentiating w.r.t x,

$P' = 400\left(\frac{-6}{x^2} - \frac{1}{6}\right)$

Equating this to $0$, x has no solution(s). Where am I going wrong?

P.S = I think the question might be slightly wrong as the operating cost should be in paisa (1 Rupee = 100 paisa). But in the book, there is no mention of that.

  • $\begingroup$ Is it an optimization problem ? If so, take derivatives. $\endgroup$
    – Spectre
    Aug 12 '20 at 3:18
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    $\begingroup$ I have taken the derivatives and equated it to $0$ as well to get maxima - minima but there is no real value of x satisfying the equation, check my approach. Maybe my approach is wrong here? $\endgroup$ Aug 12 '20 at 3:25
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    $\begingroup$ @SamRubenAbraham no your guess is completely correct here, the driver must have to go at a lower speed only but I wanted the correct approach for this question. $\endgroup$ Aug 12 '20 at 3:47
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    $\begingroup$ All your equations seem to be right. Since there is a limit on the speed, I would say that going at $35$ km/hour is the best for the driver. Maybe the question is asking about the company paying the driver? $\endgroup$ Aug 12 '20 at 3:57
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    $\begingroup$ @VarunVejalla yes you are right the correct answer is 35 km/hour only. My question is how? $\endgroup$ Aug 12 '20 at 4:03

All your equations are correct: you ended up with a net profit of $$P(x) = 400\left( \frac{6}{x} - 12 - \frac{x}{6} \right)$$

We now want to find the absolute maximum of this in the range $35 \le x \le 60$.

You also found that $$\frac{d}{dx}P = 400\left( -\frac{6}{x^2} - \frac{1}{6} \right)$$

This is always negative, so $P$ is decreasing everywhere it is defined. In general, when finding absolute extrema in an interval, you must check the critical points and end points of the interval.

Because there are no real critical points (as you realized), all that remains is to check $x = 35$ km/hour and $x = 60$ km/hour. It is clear that $x = 35$ km/hour maximizes the profit because $P$ is decreasing (or because $P(35)>P(60)$).

Specifically, with $x = 35$ km/hour, the profit will be $$400\left( \frac{6}{35} - 12 - \frac{35}{6} \right) \approx −7064.762$$

which is a loss, in fact.

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    $\begingroup$ Yes, I understand now. Thank you for explaining it so clearly. $\endgroup$ Aug 12 '20 at 4:31

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