For a ship travelling at ${x}$ km/h the running cost in £ is ${(x^2 + {13500\over x})}$ per hour. Find the speed that minimises the cost of a 300km journey.

The speed formula is ${speed = {distance\over time}}$

=> ${speed = {300\over x}}$

The speed will be ${S(x) = (x^2 + {13500\over x})({300\over x})}$

I would then find the derivative or ${S'}$ and solve for x and plug it back into my ${S(x)}$ function.

Am I on the right track or is my function incorrect?


When asked to minimize or maximize a value, this is an optimization question which usually implies finding the relative extrema of a function. In your case, the cost function $C$ with respect to speed $x$ is

$$ C(x) = x^2 + \frac{13500}{x} $$

To find the relative extrema, you probably already know about finding the slope by differentiating the function as

$$ C'(x) = 2x - \frac{13500}{x^2} $$

Then, you probably also know that the relative extrema is where the slope is zero

$$ \begin{eqnarray} 2x - \frac{13500}{x^2} &=& 0 \\ 2x &=& \frac{13500}{x^2} \\ 2x^3 &=& 13500 \\ x^3 &=& 6750 \\ x &=& 15 \sqrt[3]{2} \end{eqnarray} $$

To find whether the extrema is a relative minimum or a relative maximum, you might then know to use the second derivative for the concavity or curvature of the graph

$$ \begin{eqnarray} C''(x) &=& 2 + \frac{27000}{x^3} \\ C''(15\sqrt[3]{2}) &=& 6 \\ \end{eqnarray} $$

When the second derivative is positive, the slope is increasing which implies a relative minimum. So, the speed that minimizes the cost of the journey is $15 \sqrt[3]{2}$ km/h or approximately $18.9$ km/h.

  • $\begingroup$ Thanks I don't understand how you go from ${ 2x - \frac{13500}{x^2} = 0}$ To ${x^3 = 6750}$ $\endgroup$
    – dagda1
    Jan 26 '16 at 6:51
  • $\begingroup$ I edited my answer with more steps and fixed the second derivative. $\endgroup$
    – cr3
    Jan 26 '16 at 14:15

Yes you are incorrect.

Your cost function is given and $x$ is the speed, if you observe.

So you just minimise the cost i.e. solve the equation $$\frac{d}{dx}\left(x^2+\frac{13500}{x}\right)=0$$

Find the second derivative of the cost function and put the above solution in it to check if the 2nd derivative is positive or not.

If it is positive, then this solution is the required speed.


The cost of fuel consumed in propelling a steamer through the water varies as the cube of her speed, and is $ \$\,25$ per hour when the speed is 10 miles per hour. The other expenses are $\$\,100$ per hour. Find the most economical speed.


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