The cost of running a bus from A to B, is operatornameRs(a v+(b/v)), where v k ⋅ m / hr. is the average speed of the bus. when the bus travels at 30 k . m / hr., the cost comes out to be Rs. 75 while at 40 k . m / hr., it is Rs. 65 . The most economical speed (in k.m/hr.) of the bus is

Question

The cost of running a bus from A to B, is operatornameRs(a v+(b/v)), where v k ⋅ m / hr. is the average speed of the bus. when the bus travels at 30 k . m / hr., the cost comes out to be Rs. 75 while at 40 k . m / hr., it is Rs. 65 . The most economical speed (in k.m/hr.) of the bus is (A) 30 (B) 40 (C) 45 (D) 60

Tardigrade Admin · Accepted Answer

Here a =(1/2), b =1800. Now, c ( v )=( v /2)+(1800/ v ) ⇒ c prime( v )=0 ⇒ v =60 k ⋅ m / hr. Also, .( d 2 c / dv 2)] v =60= positive quantity. ∴ c text min. (v=60)=60

Q. The cost of running a bus from A to B, is Rs(av+vb​), where vk⋅m/hr. is the average speed of the bus. when the bus travels at 30k.m/hr., the cost comes out to be Rs. 75 while at 40k.m/hr., it is Rs. 65 . The most economical speed (in k.m/hr.) of the bus is

30

40

45

60

Here a=21​,b=1800. Now, c(v)=2v​+v1800​⇒c′(v)=0⇒v=60k⋅m/hr. Also, dv2d2c​]v=60​= positive quantity. ∴cmin. ​(v=60)=60

Here $a = \frac{1}{2}, b = 1800$ . Now, $c (v) = \frac{v}{2} + \frac{1800}{v} \Rightarrow c^{'} (v) = 0 \Rightarrow v = 60 k \cdot m / h r$ .
Also, $\frac{d ^{2} c}{d v ^{2}}]_{v = 60} =$ positive quantity.
$∴ c_{min.} (v = 60) = 60$