Cost for a Trucking Company Imagine you are the manager of a trucking company. One employee is to drive a truck down a $400$ mile road that has a minimum speed limit of $40$ mph and a maximum speed limit of $70$ mph. You have to pay him $\$6$ per hour, and the cost of operating the truck (in cents, not including the wages) is $12+\dfrac{x}{6}$, where $x$ is the speed of the truck in mph. What driving speed costs the least for this road?
The equations I have are $C_1(x)=12+\dfrac{x}{6}$ and $C_2(x)=600\times\dfrac{400}{x}$. These equations have different units, so how can I combine them to optimize?
 A: I realize this is over three years late, but hopefully someone else will stumble across this answer and find it helpful!

The equations do in fact have the same units! They look something like this (dividing everything by $100$ to work with dollars instead of cents):
\begin{align}
C_1(x)&=\frac{1}{100}(\text{cents}+\text{cents}) \\[0.2ex]
&=\text{dollars}+\text{dollars} \\
&=\text{dollars} \\
C_2(x)&=\frac{1}{100}\left(\frac{\text{cents}}{\text{hour}}\times\frac{\text{miles}}{\frac{\text{miles}}{\text{hour}}}\right) \\[0.3ex]
&=\frac{1}{100}\left(\frac{\text{cents}}{\text{hour}}\times\text{hours}\right) \\
&=\frac{1}{100}(\text{cents}) \\[0.5ex]
&=\text{dollars}
\end{align}
Armed with this information, we can now add the two cost functions to find the total cost function:
\begin{align}
C(x)&=\frac{1}{100}\Big(C_1(x)+C_2(x)\Big) \\[0.4ex]
&=\frac{1}{100}\left(12+\frac{x}{6}+\frac{240000}{x}\right) \\[0.3ex]
&\bbox[5px,border:1px solid red]{=\frac{2400}{x}+\frac{x}{600}+\frac{3}{25}}
\end{align}
As always in single-variable optimization problems, we can now set $C'(x)$ equal to $0$ and find the points at which cost is minimized:
\begin{align}
-\frac{2400}{x^2}+\frac{1}{600}&=0 \\[0.2ex]
\frac{2400}{x^2}&=\frac{1}{600} \\
x^2&=1440000 \\
x&=1200
\end{align}
Wait, what? Weren't we supposed to get an answer in miles per hour that was between $40$ and $70$?
Not necessarily. All this answer means is that cost would be minimized if the truck were to drive at $1200$ mph, but that's a little difficult to achieve for a number of reasons. The best we can do in this case is to pick the value in our range of allowed values that's closest to $1200$, so we can conclude that you should have the truck driver drive at $70$ mph.
