The cost of fuel to run a locomotive is propotional to the 3/2 power of the speed.At a speed of 23 miles per hour, the cost of fuel is $50 per hour. Other costs amount to $100 per hour.
a) find the speed that will minimize the cost per mile.
b) what is the cost of a 300-mile trip at the optimal speed?

Use the given info to find k in C=k\cdot s^{\frac{3}{2}}

Where C=cost and s=speed.

50=k\cdot 23^{\frac{3}{2}}

k=\frac{50\sqrt{23}}{529}\approx 0.4533

C=.4533s^{\frac{3}{2}}

To find the cost per mile, divide by s. Don't forget the $100 additional cost.

\frac{Cost}{Mile}=\frac{\frac{Cost}{Hour}}{\frac{M iles}{Hour}}

C(s)=\frac{.4533s^{\frac{3}{2}}+100}{s}

Differentiate:

\frac{dC}{ds}=\frac{25\sqrt{23}}{529\sqrt{s}}-\frac{100}{s^{2}}

Set to 0 and solve for s and we get:

s=23\cdot 2^{\frac{4}{3}}\approx 57.96 \;\ \frac{m}{hr}

That is the optimal speed.

Now, can you finish the second part?