# Math Help - Calculus BC Problem

1. ## Calculus BC Problem

A taxi driver is going from LaGuardia Airport to Kennedy Airport at an average speed of v miles per hour. The distance is 30 miles and gas costs $1.33 per gallon. The taxi consumes gas at a rate of f(v)=2+(v^2)/(600) gallons per hour and v is between ten and 55. What average speed will minimize the total cost of fuel. So far ive taken the derrivitave of f(v) to find the minimum but when you set that to 0 v=0 nad i need to find a value for v. Please help --> 2. Originally Posted by andyaddition A taxi driver is going from LaGuardia Airport to Kennedy Airport at an average speed of v miles per hour. The distance is 30 miles and gas costs$1.33 per gallon. The taxi consumes gas at a rate of f(v)=2+(v^2)/(600) gallons per hour and v is between ten and 55. What average speed will minimize the total cost of fuel.

So far ive taken the derrivitave of f(v) to find the minimum but when you set that to 0 v=0 nad i need to find a value for v. Please help -->
The first thing we need to take note of is that $d=v\cdot t \iff t=\frac{d}{v}=\frac{30}{v}$ where t is in hours

2nd we need to construct a cost function that depends on velocity.
Dimentional analysis will be helpful

$c(v)=\left( 2+\frac{v^2}{600}\right)\left( \frac{gal}{ hour}\right)\cdot \left( t \mbox{hour} \right)\cdot \left( \frac{1.33 dollar}{gal}\right)$

but we know what t is in terms of v so we get and reducing all of our units we get

$c(v)=\left( 2+\frac{v^2}{600}\right)\cdot \left( \frac{30}{v} \right)\cdot (1.33 dollar)=\frac{(1.33 dollar)(30)}{600}\left( \frac{1200+v^2}{v}\right)=$

$c(v) = \frac{(1.33 dollar)(30)}{600}\left( 1200v^{-1}+v\right)$

taking the derivative we get

$c'(v) = \frac{(1.33 dollar)(30)}{600}\left( \frac{-1200}{v^2}+1\right)$

setting this equal to zero we get

$\left( \frac{-1200}{v^2}+1\right)=0 \iff v^2=1200 \iff v=20\sqrt{3}$

3. Originally Posted by andyaddition
A taxi driver is going from LaGuardia Airport to Kennedy Airport at an average speed of v miles per hour. The distance is 30 miles and gas costs \$1.33 per gallon. The taxi consumes gas at a rate of f(v)=2+(v^2)/(600) gallons per hour and v is between ten and 55. What average speed will minimize the total cost of fuel.

So far ive taken the derrivitave of f(v) to find the minimum but when you set that to 0 v=0 nad i need to find a value for v. Please help -->
since $v$ is an average speed, time to make the trip ...

$t = \frac{30}{v}$

total gas consumed ...

$f(v) \cdot t = \left(2 + \frac{v^2}{600}\right) \frac{30}{v}$

$g(v) = \frac{60}{v} + \frac{v}{20}$

minimum gas consumed ...

$g'(v) = -\frac{60}{v^2} + \frac{1}{20} = 0$

$v = 20\sqrt{3}$ mph