# Functions of fuel consumption

• Sep 18th 2011, 07:48 PM
I-Think
Functions of fuel consumption
For a truck, its fuel consumption in $\frac{ml}{hr}$ is given by
$f(v)=(v-10)^2+9900$, where $v$ is the speed in $\frac{km}{hr}$. 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 $\frac{f(v)}{v}$, but didn't know how to find the minimum without calculus
• Sep 19th 2011, 01:44 AM
Sudharaka
Re: Functions of fuel consumption
Quote:

Originally Posted by I-Think
For a truck, its fuel consumption in $\frac{ml}{hr}$ is given by
$f(v)=(v-10)^2+9900$, where $v$ is the speed in $\frac{km}{hr}$. 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 $\frac{f(v)}{v}$, 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 $f(v)$. $f(v)$ will be minimum when $(v-10)^2$ is a minimum. It is clear that the minimum value of $(v-10)^2$ is zero. Hence the corresponding speed is,

$(v-10)^2=0$

$v=10\mbox{ kmh^{-1}}$
• Sep 19th 2011, 06:34 AM
I-Think
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 $v=10$, we have $f(v)=9900$, so that is for every $10km$ traveled in an hour, $9900ml$ of fuel is consumed, giving us a consumption rate of $990mlkm^{-1}$

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

Is there something wrong with this reasoning?
• Sep 19th 2011, 08:13 AM
Sudharaka
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 $v=10$, we have $f(v)=9900$, so that is for every $10km$ traveled in an hour, $9900ml$ of fuel is consumed, giving us a consumption rate of $990mlkm^{-1}$

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

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 $\frac{f(v)}{v}$. But alas I still didn't get an idea to tackle this problem.