JEE Main 2013 (Online) 23rd AprilQuestion 1
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 :
Differentiation and its Application
Solution
Concept. Two data points give $a$ and $b$; then minimise $av+\frac bv$. Formulas used. (1) $av+\frac bv$ ($a,b>0$) is least at $v=\sqrt{\frac ba}$Step 1 - Find $a,b$. $$30a+\frac{b}{30}=75\ \Rightarrow\ 900a+b=2250,\qquad40a+\frac{b}{40}=65\ \Rightarrow\ 1600a+b=2600.$$ Subtracting: $700a=350$, so $a=\frac12$ and $b=1800$. Step 2 - Minimise. By (1), $v=\sqrt{3600}=60$ (derivative $\frac12-\frac{1800}{v^2}=0$).Answer: (C)