Functions of fuel consumption

For a truck, its fuel consumption in $\displaystyle \frac{ml}{hr}$ is given by

$\displaystyle f(v)=(v-10)^2+9900$, where $\displaystyle v$ is the speed in $\displaystyle \frac{km}{hr}$. Find the speed that gives the best distance to fuel ratio

No calculus is to be used to solve this question

Any pointers? Was thinking of trying $\displaystyle \frac{f(v)}{v}$, but didn't know how to find the minimum without calculus

Re: Functions of fuel consumption

Quote:

Originally Posted by

**I-Think** For a truck, its fuel consumption in $\displaystyle \frac{ml}{hr}$ is given by

$\displaystyle f(v)=(v-10)^2+9900$, where $\displaystyle v$ is the speed in $\displaystyle \frac{km}{hr}$. Find the speed that gives the best distance to fuel ratio

No calculus is to be used to solve this question

Any pointers? Was thinking of trying $\displaystyle \frac{f(v)}{v}$, but didn't know how to find the minimum without calculus

Hi I-Think,

To find the best distance to fuel ratio we should find the speed that minimizes $\displaystyle f(v)$. $\displaystyle f(v)$ will be minimum when $\displaystyle (v-10)^2$ is a minimum. It is clear that the minimum value of $\displaystyle (v-10)^2$ is zero. Hence the corresponding speed is,

$\displaystyle (v-10)^2=0$

$\displaystyle v=10\mbox{ kmh^{-1}}$

Re: Functions of fuel consumption

Are you sure? Talked to a friend and he said the answer was 100. He didn't give reasons, just hints.

Checking at $\displaystyle v=10$, we have $\displaystyle f(v)=9900$, so that is for every $\displaystyle 10km$ traveled in an hour, $\displaystyle 9900ml$ of fuel is consumed, giving us a consumption rate of $\displaystyle 990mlkm^{-1}$

But at $\displaystyle v=100$, we have $\displaystyle f(v)=8100+9900=18000$, that is for every $\displaystyle 100km$ traveled in an hour, $\displaystyle 18000ml$ of fuel is consumed, giving us a consumption rate of $\displaystyle 180mlkm^{-1}$

Is there something wrong with this reasoning?

Re: Functions of fuel consumption

Quote:

Originally Posted by

**I-Think** Are you sure? Talked to a friend and he said the answer was 100. He didn't give reasons, just hints.

Checking at $\displaystyle v=10$, we have $\displaystyle f(v)=9900$, so that is for every $\displaystyle 10km$ traveled in an hour, $\displaystyle 9900ml$ of fuel is consumed, giving us a consumption rate of $\displaystyle 990mlkm^{-1}$

But at $\displaystyle v=100$, we have $\displaystyle f(v)=8100+9900=18000$, that is for every $\displaystyle 100km$ traveled in an hour, $\displaystyle 18000ml$ of fuel is consumed, giving us a consumption rate of $\displaystyle 180mlkm^{-1}$

Is there something wrong with this reasoning?

No, there isn't. My method is incorrect. I had found the speed for which the fuel consumption is minimum, not the speed corresponding to the best distance to fuel ratio. As you have correctly stated we have to find the minimum of $\displaystyle \frac{f(v)}{v}$. But alas I still didn't get an idea to tackle this problem.