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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?

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  • $\begingroup$ set equal zero, rearrange for quadratic (reciprocal needed / clear denominator by multiplying by $v^2(17.5 -.15v)^2)$ in $v$ and solve $\endgroup$ – Andrew Allen Aug 25 '18 at 22:11
  • $\begingroup$ @Andrew Allen Where do I get the single v^1 term of the equation? All I got was C'(v) = .45v^2 - 9800- .675v^2 $\endgroup$ – TheGuyWhoCodes Aug 25 '18 at 22:15
  • $\begingroup$ see below answer, please note I've not checked if other aspect of forming the equation is correct or not $\endgroup$ – Andrew Allen Aug 25 '18 at 22:20
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$C'(v) = \frac{.45}{(17.5-.15v)^2} -\frac{32}{v^2} = 0$

So

$0.45v^2 - 32(17.5 - 0.15v)^2 = 0$

Now expand the bracket, multiply by the $-32$ before collecting like terms

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