Review question

# When does this function of two variables have a minimum? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9671

## Solution

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 $$$y=8x^{-4}.\label{eq:1}$$$ We can substitute this for $y$ in the equation we are given for $z$ to get $$$z=x+8x^{-4}. \label{eq:2}$$$

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.