The running cost of a truck travelling at a steady speed of v km/h is Rs. \(\left(\displaystyle\frac{v^{2}}{10} + 250\right)\) per hour. For a journey of 400 km, the speed at which the total running cost of the journey is least is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 50 km/h
At v km/h the journey of 400 km takes \(\frac{400}{v}\) hours. So the total cost of the journey is
\( \displaystyle C(v) = \frac{400}{v}\left(\frac{v^{2}}{10} + 250\right) \)
\( \displaystyle = 40v + \frac{100000}{v}, \quad v \gt 0 \)
Setting the derivative equal to zero:
\( \displaystyle C'(v) = 40 - \frac{100000}{v^{2}} = 0 \)
\( v^{2} = 2500 \), so v = 50
Since \( \displaystyle C''(v) = \frac{200000}{v^{3}} \gt 0 \), this is a minimum:
\( C(50) = 2000 + 2000 = 4000 \)
For comparison, C(25) = 5000, C(40) = 4100 and C(100) = 5000, all more than Rs. 4000.