Question

The cost of running a bus from $A$ to $B$ , is $Rs.\left(av+\frac{b}{v}\right)$ where $v$ km/h is the average speed of the bus. When the bus travels at $30\, km/h$, the cost comes out to be $Rs.\, 75$ while at $40\, km/h$, it is $Rs. \,65$. Then the most economical speed (in $km/ h$) of the bus is :

• A. $45$
• B. $50$
• C. $60$
• D. $40$

Answer 1

The Correct Option is C

To find the most economical speed of the bus, we need to minimize the cost which is given by the function:

Cost = Rs.\left(av+\frac{b}{v}\right)

where v is the speed in km/h. Given the costs at different speeds:

• v = 30 km/h, Cost = Rs. 75
• v = 40 km/h, Cost = Rs. 65

From these two conditions, we write two equations:

30a + \frac{b}{30} = 75 -- (1)

40a + \frac{b}{40} = 65 -- (2)

Multiply equation (1) by 40 and equation (2) by 30 to eliminate b:

1200a + \frac{40b}{30} = 3000

1200a + \frac{30b}{40} = 1950

Subtract the second equation from the first:

\frac{40b}{30} - \frac{30b}{40} = 1050

Simplify to find b:

\frac{40b}{30} - \frac{30b}{40} = \frac{1600b - 900b}{1200} = \frac{700b}{1200} = 1050

b = \frac{1050 \times 1200}{700} = 1800

Substitute b = 1800 back into equation (1) to find a:

30a + \frac{1800}{30} = 75

30a + 60 = 75

30a = 15

a = \frac{1}{2}

Now we have a = 0.5 and b = 1800. To find the most economical speed, differentiate the cost function:

\frac{d}{dv}(0.5v + \frac{1800}{v}) = 0.5 - \frac{1800}{v^2}

Set the derivative to zero to minimize cost:

0.5 - \frac{1800}{v^2} = 0

\frac{1800}{v^2} = 0.5

v^2 = \frac{1800}{0.5} = 3600

v = \sqrt{3600} = 60

Thus, the most economical speed is 60 km/h.