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Question:

Land Cruiser Prado, the latest SUV from Toyota Motors, consumes diesel at the rate of $\frac{1}{400}\left\{\frac{1000}{x}+x\right\}$ per km, when travelling at the speed of x km/hr. The diesel costs Rs. 65 per litre and the driver is paid Rs. 50 per hour. Find the steady speed that will minimize the total cost of a 1000 km trip?

Explanation:

Total cost incurred = total cost incurred due to petrol + total cost incurred due to driver charges

= (total litres consumed × cost per litre) + (total driver hours × cost per hour)

Total petrol consumed in 1000 km = 1000 × $\frac{1}{400}\left\{\frac{1000}{x}+x\right\}=2.5\left\{\frac{1000}{x}+x\right\}$

∴ Total petrol cost = $2.5\left\{\frac{1000}{x}+x\right\}\times 65$

$=162.5\times \left\{\frac{1000}{x}+x\right\}$

Time taken to travel 1000 km = 1000/x

∴ Total driver cost = (1000/x) × 50 = Rs. (50000/x)

∴ Total cost = 162.5 × $\left\{\frac{1000}{x}+x\right\}+\left(\frac{50000}{x}\right)$

$=\left(\frac{1000}{x}\right)(162.5+50)+162.5x$

$=\frac{212500}{x}+162.5x$

You can now substitute options and verify that the least cost is for x = 36.

Hence, option (b).