The variables \(x\) and \(y\) are such that \(x^4y=8\). A third variable \(z\) is defined by \(z=x+y\). Find the values of \(x\) and \(y\) that give \(z\) a stationary value…

We want to find the stationary points of \(z\), which (unfortunately) is a function of both \(x\) and \(y\).

But (luckily) we’ve been given a constraint connecting those two variables, namely, the first equation.

If \(x=0\), the constraint wouldn’t be satisfied, so we can divide both sides by \(x^4\) to give \[\begin{equation} y=8x^{-4}.\label{eq:1} \end{equation}\] We can substitute this for \(y\) in the equation we are given for \(z\) to get \[\begin{equation} z=x+8x^{-4}. \label{eq:2} \end{equation}\]

Here \(z\) is a function of \(x\) alone. From this we can calculate stationary points, by setting the first derivative equal to zero.

\[\begin{align*} z'(x) &= 1 - 32x^{-5} \\ \text{so} \qquad 1 - 32x^{-5} &= 0 \\ 1 &= 32x^{-5} \\ x^5 &= 32 \\ x &= 2. \end{align*}\]

As \(x=2\) is the only solution of \(x^5=32\), we only have one stationary point.

Substituting this value into equation \(\eqref{eq:1}\) gives us the corresponding \(y\)-value of our stationary point \[y=\frac{8}{16}=\frac{1}{2}.\]

… and show that this value of \(z\) is a minimum.

To check whether \(z\) has a minimum when \(x=2\) and \(y=\dfrac{1}{2}\), we can look at the second derivative of \(z\) with respect to \(x\). A positive second derivative corresponds to a local minimum. From \(\eqref{eq:2}\) we obtain \[z''(x) = 160x^{-6},\] which gives \(z''(2)=2.5>0\) (or we could note that \(z''(x)\) is positive for all values of \(x\)). Therefore the value of \(z\) when \(x=2\) and \(y=\dfrac{1}{2}\) is indeed a minimum.