Question

When travelling x km/hour, a truck burns diesel at the rate of $\frac{1}{300}(\frac{900}{x}+x)$ litres per km. If the diesel oil costs $40$p.per litre and driver is paid $Rs.1.50$ per hour, find the steady speed that will minimize the total cost of the trip of $500$ km.

• A. $15$ km. p.h.
• B. $27$ km.p.h.
• C. $42$ km.p.h.
• D. $45$ km.p.h.

Answer 1

Answer: (D)

$45$ km.p.h.

For $500km$

diesel cost $=\frac{1}{300}(\frac{900}{x}+x)\frac{litre}{km}\times 0.4\frac{Rs.}{litre}\times 500km$

$=\frac{2}{3}(\frac{900}{x}+x)Rs.$

Driver Cost $=\frac{500}{x}\times 1.50Rs=\frac{750}{x}Rs$

Total Cost $=\frac{750}{x}+\frac{2}{3}(\frac{900}{x}+x)$

C $=\frac{1350}{x}+\frac{2}{3}x$

For minimization

$\frac{dC}{dx}=0$

$-\frac{1350}{x^2}+\frac{2}{3}=0$

$x=\sqrt{\frac{1350\times 3}{2}}=\sqrt{2025}$

$x=45km/hour$