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 :

Updated On: Aug 15, 2024
  • $45$
  • $50$
  • $60$
  • $40$
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The Correct Option is C

Solution and Explanation

Let cost $C=av+\frac{b}{v}$ According to given question, $30a+\frac{b}{30}=75\,...\left(i\right)$ $40a+\frac{b}{40}=65\,...\left(ii\right)$ On solving $\left(i\right)$ and $\left(ii\right)$, we get $a=\frac{1}{2} and b=1800$ Now, $C=av+\frac{b}{v}$ $\Rightarrow \frac{dC}{dv}=0 = a-\frac{b}{v^{2}}$ $\frac{dC}{dv}=0 \Rightarrow a-\frac{b}{v^{2}}=0$ $\Rightarrow v=\sqrt{\frac{b}{a}}=\sqrt{3600}$ $\Rightarrow v=60\,kmph$
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Concepts Used:

Rate of Change of Quantities

The rate of change of quantities can be expressed in the form of derivatives. Rate of change of one quantity with respect to another is one of the major applications of derivatives. The rate of change of a function with respect to another quantity can also be done using chain rule.

If some other quantity ‘y’ causes some change in a quantity of certain ‘x’, in view of the fact that an equation of the form y = f(x) gets always satisfied, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by 

This is also called the Average Rate of Change.

If the rate of change of a function is to be defined at a specific point i.e. a specific value of ‘x’, it is known as the Instantaneous Rate of Change of the function at that point. From the definition of the derivative of a function at a point, we have

From this, it is to be concluded that the instantaneous Rate of Change of the function is represented by the derivative of a function. From the rate of change formula, it represents the case when Δx → 0. Thus, the rate of change of ‘y’ with respect to ‘x’ at x = x0 = (dy/dx)x = x0