A truck gets 10 miles per gallon (mpg) of diesel fuel traveling along an interstate highway at 50 mph. This mileage decreases by 0.15 mpg for each mile per hour increase above 50 mph. If the truck driver is paid 32 dollars an hour and diesel fuel costs P = 3 dollars/gal, which speed $v$ between 50 and 70 mph will minimize the cost of a trip along the highway?
I started by forming the equation:
$C(v) = \frac{3}{17.5-.15v}+\frac{32}{v} $
To find the minimum of this function I want to do the derivative to find the zeros.
$C'(v) = \frac{.45}{(17.5-.15v)^2} -\frac{32}{v^2}$
How am I supposed to find zeros from this equation to find the minimum?
$C'(v) = \frac{.45}{(17.5-.15v)^2} -\frac{32}{v^2} = 0$
So
$0.45v^2 - 32(17.5 - 0.15v)^2 = 0$
Now expand the bracket, multiply by the $-32$ before collecting like terms