The cost of running a bus from A to B is Rs. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the cost function be $C(v) = av + \frac{b}{v}$.Given that at $v = 30 	ext{ km/h}$,$C = 75$,so $30a + \frac{b}{30} = 75 \implies 900a + b = 22

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) Let the cost function be $C(v) = av + \frac{b}{v}$.Given that at $v = 30 	ext{ km/h}$,$C = 75$,so $30a + \frac{b}{30} = 75 \implies 900a + b = 2250 \quad (i)$.Given that at $v = 40 	ext{ km/h}$,$C = 65$,so $40a + \frac{b}{40} = 65 \implies 1600a + b = 2600 \quad (ii)$.Subtracting $(i)$ from $(ii)$,we get $700a = 350$,which implies $a = 0.5$.Substituting $a = 0.5$ into $(i)$,we get $900(0.5) + b = 2250 \implies 450 + b = 2250 \implies b = 1800$.To find the most economical speed,we minimize $C(v)$ by setting $\frac{dC}{dv} = 0$.$\frac{dC}{dv} = a - \frac{b}{v^2} = 0 \implies v^2 = \frac{b}{a}$.$v^2 = \frac{1800}{0.5} = 3600 \implies v = 60 	ext{ km/h}$.

Similar Questions

If $x+y=6, x \geqslant 0, y \geqslant 0$,then the maximum value of $x^2 y$ is

If the function $f(x) = x^3 - 3(a - 2)x^2 + 3ax + 7$,for some $a \in R$,is increasing in $(0, 1]$ and decreasing in $[1, 5)$,then a root of the equation $\frac{f(x) - 14}{(x - 1)^2} = 0$ $(x \neq 1)$ is

The absolute maximum value of the function $f(x)=2 x^3-9 x^2+12 x+1$ on the interval $[0,2]$ is:

The absolute minimum value of the function $f(x) = x^3 - 18x^2 + 96x$ for $x \in [0, 9]$ is:

The maximum value of the function $f(x)=e^x+x \ln x$ on the interval $1 \leq x \leq 2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module