Stationary Points: Multi-variable functions

We now move on to multi-variable functions. Again, we start off considering a function of two variables, $f=f(x, y)$ and a point, $(a, b)$ in the domain of $f$.

Let $(x, y)=(a, b)$ be a point in the domain of $f(x, y)$. Then, the point $(a, b)$ is a

local minimum if $\quad f(x, y) \geq f(a, b)$ when $(x, y)$ is near $(a, b)$

local maximum if $\quad f(x, y) \leq f(a, b)$ when $(x, y)$ is near $(a, b)$

global minimum if $f(x, y) \geq f(a, b) \forall(x, y)$ in domain of $f$

global maximum if $f(x, y) \leq f(a, b) \forall(x, y)$ in domain of $f$

The above definitions are shown graphically in Fig. 1.9:

Figure 1.9: Plot of a function $f(x, y)$ showing stationary points. The global maximum is shown at the point $(0.5,1.2)$ while a local maximum is shown at $(-1,-1.2)$. The global minimum is shown at $(0.6,-1)$ and a local minimum is shown at $(-1.2,1)$. Note that the global extrema are also local extrema near their respective points.

The function has four extrema (The term 'extrema' (singular: extremum) refers to minimum or maximum values): two minima and two maxima; these are labelled in Fig. 1.9. A local extreme value is a value $f(a, b)$ that is a maximum or minimum near the point $(a, b)$. A global extreme value represents the maximum or minimum value of a function $f$ on a given domain, $D$ i.e. these are the absolute extreme values of $f$ on $D$.

Local minima and maxima

Consider a function $f(x, y)$ defined on a domain, $D$ and the point $(a, b)$ is in the domain of $f$. Then, $f$ has a local maximum or minimum if

this means we have zero partial derivatives at $(a, b)$, i.e. $f_{x}(a, b)=0$ and $f_{y}(a, b)=0$.

It is clear to see where Theorem 1.5 comes from by considering the singlevariable case. Let us define $g(x)=f(x, b)$, where $b$ is a fixed point. Then, by Theorem 1.3, $g$ has a local maximum (or minimum) at $x=a$ if $g^{\prime}(a)=0$, i.e. $f_{x}(a, b)=0$. Similarly, by defining $h(y)=f(a, y)$ and re-using Theorem 1.3, we have $h^{\prime}(y)=0$ and hence $f_{y}(a, b)=0$.

Definition 1.5 Consider a function $f(x, y)$. The point $(a, b)$ is a critical point (or stationary point) of $f$ if

Being critical is necessary for a local maximum or local minimum. In other words, local maxima and minima must be looked for among critical points. Example 1.12 Suppose we have a function $f(x, y)$ given by $f(x, y)=x^{2}+y^{2}-2 x-6 y+14$. Determine any critical points.

Solution We need to compute the partial derivatives of $f$, i.e. $f_{x}, f_{y}$ and set them to zero to find any critical points. So, we need to solve the following 2 equations:

Equations (1.41) give $x=1$ and $y=3$. There is only one critical point and it occurs at $(a, b)=(1,3)$.

In Example 1.12, we found one critical point but at this point we did not classify it as maximum or minimum. In fact, not all critical points are maxima or minima; a stationary point can also be classified as a saddle point. To determine the nature of the critical points, we use the second derivative test given by Theorem 1.6.

Second derivative test

Consider a function $f(x, y)$ that has continuous partial derivatives. Suppose that $f_{x}(a, b)=0$ and $f_{y}(a, b)=0$. Now take the Hessian matrix [see Eq. (1.13)], evaluated at the point $(a, b)$, and define its determinant, i.e.

Then, if

• $D>0$ and $f_{x x}<0 \Rightarrow(a, b)$ is a local maximum;
• $D>0$ and $f_{x x}>0 \Rightarrow(a, b)$ is a local minimum;
• $D<0 \Rightarrow(a, b)$ is a saddle point;
• $D=0$, the test is inconclusive (we need to consider higher order). Example 1.13 Consider the function $f(x, y)=x^{4}+y^{4}-4 x y+1$. Find the stationary points and determine whether they are local maxima, minima or saddle points.

Solution The first step is to compute the partial derivatives of $f$ and set them to zero:

From Eqs. (1.42), we have the equalities $y=x^{3}$ and $y^{3}=x$. Eliminating $y$, gives:

Note that the terms in red are always $>0$ which means that Eq. (1.43) is equal to zero if $x=0$ or $x^{2}-1=0$ (i.e. $x=-1$ or $x=1$ ). This gives us the following stationary points:

To classify the critical points, we compute the Hessian determinant given in Theorem 1.6. This requires calculation of the second-order partial derivatives, $f_{x x}, f_{x y}$ and $f_{y y}$ (recall that $f_{x y}=f_{y x}$ at each one of the critical points found). The Hessian determinant at an arbitrary point $(x, y)$ is given by,

Evaluating Eq. (1.44) at the point $(0,0)$ gives,

Similarly, at the points $(1,1)$ and $(-1,-1)$ we have