# Calculus math question

So the cost per hour of running a cruiser is $\$ \left(\frac {V^2}{40} + 10\right)$, where$V $is the speed in knots. So I’ve answered the first question showing the cost would be$\$\frac DV \left(\frac {V^2}{40} + 10\right)$. And then they asked me to find the most economical speed for running the cruiser, and I have no idea how to get it

• Note: I reformatted your post pretty heavily, please check to see that no errors were introduced.
– lulu
Jan 23 '18 at 13:55
• This has now become a "minimize" problem. Find the value of $V$ so that $\frac DV \left(\frac {V^2}{40} + 10\right)$ is minimum. A standard problem from a beginning calculus course. Jan 23 '18 at 13:56
• And how would I find the value of V? I’m self-teaching myself calculus and I am finding this question challenging Jan 23 '18 at 14:04
• Nvm. I’ve solve the answer Jan 23 '18 at 14:17
• Does it now occur that you next need to find minimum of $V/40+10/V?$ Jan 23 '18 at 19:30

The most economical speed will be the one at which the cost ($C$) is minimum. So the problem boils down to \begin{align} & \dfrac{dC}{dV} = 0 \\ \implies & \dfrac{d}{dV} \left[ \dfrac{D}{V} \left( \dfrac{V^{2}}{40} + 10\right)\right] = 0 \\ \implies & D \dfrac{d}{dV} \left( \dfrac{V}{40} + \dfrac{10}{V}\right) = 0 \\ \implies & \dfrac{1}{40} - \dfrac{10}{V^{2}} = 0 \tag{assuming $D > 0$} \\ \implies & V = \sqrt{400} = 20 \tag{since $V \geq 0$} \end{align}
Here is a plot of cost vs speed for $D = 20$ (you can choose any positive value). It is clear from the plot that cost attains its minimum value at $V = 20$ knots.
Let your cost function be $C(V)$, which is dependent on $V$, the velocity of the cruise. $$C(V) = \frac{D}{V}\left( \frac{V^{2}}{40} + 10 \right)$$ with $D$ as a constant.
Since we already have the cost function, we can find this economical velocity by using first derivative of $C(V)$. Since $C(V)$ is continuous for $V>0$, $C(V)$ will attain a minima or maxima value at a certain velocity when its first derivative equals 0 : $$C(V) = \frac{DV}{40} + \frac{10D}{V}$$ $$C'(V) = \frac{D}{40} - 10DV^{-2}$$ now you may try to solve $$C'(V) = 0$$ You could get more than 1 velocities, choose the one that makes $C(V)$ as minimum as possible. Is this okay?