Find the speed at which the motorist will make the cost per mile a minimum. (Looking for hints) 
The cost of fuel (per hour) in running a locomotive is proportional to the square of the speed and is 25 dollars per hour for a speed of 25 m.p.h. Other costs amount to $100 per hour regardless of speed. Find the speed at which the motorist will make the cost per mile a minimum.


I've been on this question for a while now, here's a graph I produced through Geogebra, $f(x)$ being the cost per hour and $x$ being the speed...I think that's what it should look like since it said that all other costs $100 regardless of speed so I assumed it would look like that, though I wouldn't have been able to come up with that without Geogebra. If I'm wrong please hint me in the direction I should go...
I'm also stuck on the "Find the speed at which the motorist will make the cost per mile a minimum" if anyone can give me a hint on that also, I'd like to figure out the rest myself.
Apologies for the lack of research, I'm pretty stuck one where to begin, all I could come up with was that graph.
Thanks in advance.
 A: You tried to make things too complicated. 
The cost of fuel per hour $c_f$ is proportional to the square of the speed $v$.
You have thus $c_f=\alpha v^2$, with $\alpha$ the constant of proportionality. 
You know that $c_f(25 mph)=25\$$
You should be able to find easily $\alpha$...
After that, you have the total cost per hour $c_t=c_f+100$
Now what is the cost per mile $c_m$? It depends on $v$ of course. 
$c_m=\dfrac{c_t}{v}=\dfrac{\alpha v^2+100}{v}=\alpha v+\dfrac{100}{v}$
Find out the minimum and you are set. 
A: Your formula for the cost per hour seems to be wrong. Let's start from the beginning.
We are told "cost of fuel (per hour) in running a locomotive is proportional to the square of the speed." Letting $x$ be the speed in miles per hour and $f$ be the cost of fuel per hour we get
$$f=kx^2$$
for some constant $k$. Since $f$ "is $25$ dollars per hour for a speed of $25$ m.p.h." we substitute $f=25$ and $x=25$ and solve for $k$, getting $k=\frac 1{25}$. Now our formula is
$$f=\frac{x^2}{25}$$
Then since "Other costs amount to $\$100$ per hour regardless of speed" the total cost per hour, $h$, is
$$h=\frac{x^2}{25}+100$$
However, you want to minimize cost per mile, not cost per hour. The total cost $c$ for time $t$ hours is given by
$$c=ht$
Using the formula "distance = rate times time" we get, for distance $d$,
$$\frac cd=\frac c{xt}=\frac{ht}{xt}=\frac hx$$
$$=\frac{\frac{x^2}{25}+100}x=\frac x{25}+\frac{100}x$$
That last is what you want to minimize. Can you finish from here?
