When does this function of two variables have a minimum?

When does this function of two variables have a minimum?

Question

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 and show that this value of $z$ is a minimum.

Suggestion

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…

This looks tricky as $z$ is given as a function of two variables. Can we use the equation connecting $x$ and $y$ to turn $z$ into a function of just one variable?

How do we then calculate stationary points for $z$?

… and show that this value of $z$ is a minimum.

What does the second derivative tell us about a stationary point?

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

$$\begin{equation} y=8x^{-4}. \end{equation}$$

We can substitute this for

$y$

in the equation we are given for

$z$

to get

$$\begin{equation} z=x+8x^{-4}. \end{equation}$$

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

$(1)$

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 $(2)$ 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.