The production cost per kilogram, \(C\) (in thousands of pounds) when \(x\) kilograms of a chemical are made is given by \[ C=3x+\frac{100}{x},\; x>0. \] Find the value of \(x\) for which the cost is a minimum, and the minimum cost.

We find the stationary points of \(C\) by differentiating: \[ \frac{dC}{dx}=3-\frac{100}{x^2}. \] Setting the derivative to zero, we find that the stationary points of \(C\) are given by the solutions of \[ 3=\frac{100}{x^2}, \] which means that \[ x^2=\frac{100}{3}. \] We only need to take the positive square root, since \(x>0\), and so \[ x=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}. \] The cost at this value of \(x\) is \[\begin{align*} C&=3\times\frac{10\sqrt{3}}{3}+100\times\frac{\sqrt{3}}{10} \\ &=10\sqrt{3}+10\sqrt{3} \\ &=20\sqrt{3}. \end{align*}\]

Show, also, that this cost is a minimum rather than a maximum.

We could take the second derivative, \[ \frac{d^2C}{dx^2}=\frac{200}{x^3}. \] Since \(x>0\), the second derivative is positive. Therefore the stationary point at \(x=\dfrac{10\sqrt{3}}{3}\) is certainly a local minimum.

As there is only one stationary point for \(x>0\), the local minimum at \(x=\dfrac{10\sqrt{3}}{3}\) gives the smallest value of \(C\) over all \(x>0\), so we have found the minimum cost of producing the chemical.

We can also decide whether a stationary point is a maximum or a minimum by looking at how the function or the first derivative behaves near the stationary point. This is particularly helpful if it would be difficult to compute the second derivative, or if the second derivative is \(0\).