Maxima and minima

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We see how to find extrema of functions of several variables.

Given a function $z=F(x,y)$ , we are often interested in points where $z$ takes on the largest or smallest values. For instance, if $z$ represents a cost function, we would likely want to know what $(x,y)$ values minimize the cost. If $z$ represents the ratio of a volume to surface area, we would likely want to know where $z$ is greatest. This leads to the following definition that we will state rather generally, but use mostly in the case of a function $F:\R ^2\to \R$ .

Let $F:\R ^n\to \R$ be defined on a set $S\subset \R ^n$ containing the point $\vec {c}$ in $\R ^n$ .

If there is an open ball $B$ containing $\vec {c}$ such that $F(\vec {c}) \geq F(\vec {x})$ for all $\vec {x}$ in $B$ , then $F$ has a local maximum at $\vec {c}$ ; if $F(\vec {c}) \leq F(\vec {x})$ for all $\vec {x}$ in $B$ , then $F$ has a local minimum at $\vec {c}$ .
If $F(\vec {c})\geq F(\vec {x})$ for all $\vec {x}$ in $S$ , then $F$ has an absolute maximum at $\vec {c}$ ; if $F(\vec {c})\leq F(\vec {x})$ for all $\vec {x}$ in $S$ , then $F$ has an absolute minimum at $\vec {c}$ .
If $F$ has a local maximum or minimum at $\vec {c}$ , then $F$ has a local extremum at $\vec {c}$ ; if $F$ has an absolute maximum or minimum at $\vec {c}$ , then $F$ has a absolute extremum at $\vec {c}$ .

Critical points

If $F$ has a local maximum at $\vec {c}$ , it means the gradient will point “nowhere” since the gradient points in the initial direction of greatest increase. This means it is pointing in a “direction” whose components are zero or ”direction” is undefined. In an entirely similar way, the gradient will be a vector whose components are either zero or undefined at local minimums as well.

Let $z = F(x,y)$ be continuous on an open set $S$ . A critical point $\vec {c}=\vector {c_1,c_2}$ of $F$ is a point in $S$ such that

$\grad F(\vec {c}) = \vec {0}$ or
$\grad F(\vec {c})$ is undefined.

Let $F:\R ^n\to \R$ be defined on an open set $S$ containing $\vec {c}$ . If $F$ has a local extremum at $\vec {c}$ , then $\vec {c}$ is a critical point of $F$ .

Therefore, to find local extrema, we find the critical points of $F$ and determine which correspond to local maxima, local minima, or neither. We’ll use examples to demonstrate this process.

Let $F(x,y) = x^2+y^2-xy-x-2$ . Find the local extrema of $F$ .

We start by computing $\grad F$ : $\grad F(x,y) = \vector {\answer [given]{2x-y-1},\answer [given]{2y-x}}$ Each component of $\grad F$ is never undefined. A critical point occurs when $\grad F(x,y) = \vec {0}$ , leading us to solve the following system of linear equations: $2x-y-1 =\answer [given]{0}\quad \text {and}\quad -x+2y = \answer [given]{0}.$ This solution to this system is $x=\answer [given]{2/3}$ , $y=\answer [given]{1/3}$ . So the critical point is $\vec {c}=\vector {\answer [given]{2/3},\answer [given]{1/3}}$ . When possible, it is good to confirm your answer with a graph:

The graph above shows $F$ along with this critical point. It is clear from the graph that this is a local minimum.

Let $F(x,y) = -\sqrt {x^2+y^2}+2$ . Find the local extrema of $F$ .

We start by computing $\grad F$ : $\grad F(x,y) = \vector {\frac {-x}{\sqrt {x^2+y^2}},\frac {-y}{\sqrt {x^2+y^2}}}$ Here the only critical point is at $\vector {0,0}$ because $\grad F$ is undefined at $\vector {0,0}$ , and because $\grad F\ne \vec {0}$ at all other points.

The graph of $F$ is shown above along with the point $(0,0,2)$ . It is clear from the graph that this point is the absolute maximum of $F$ .

In each of the previous two examples, we found a critical point of $F$ and then determined whether or not it was a local (or absolute) maximum or minimum by graphing. It would be nice to be able to determine whether a critical point corresponded to a max or a min without a graph. Before we develop such a test, we do one more example that sheds more light on the issues our test needs to consider.

Let $F(x,y) = x^3-3x-y^2+4y$ . Find the local extrema of $F$ .

Once again we start by computing $\grad F$ : $\grad F(x,y) = \vector {\answer [given]{3x^2-3},\answer [given]{-2y+4}}$ Each component is always defined. Setting $\grad F(x,y) = \vec {0}$ and solving for $x$ and $y$ , we find $\begin{align*} x &=\pm \answer[given]{1}\\ y &= \answer[given]{2}. \end{align*}$

We have two critical points: $\vector {-1,2}$ and $\vector {1,2}$ . In order to determine if they correspond to a local maximum or minimum or neither, we consider the graph of $F$ below:

The critical point $\vector {-1,2}$ clearly corresponds to a local maximum. However, the critical point at $\vector {1,2}$ is neither a maximum nor a minimum, displaying a different, interesting characteristic.

If one walks parallel to the $y$ -axis towards this critical point, then this point becomes a local maximum along this path. But if one walks towards this point parallel to the $x$ -axis, this point becomes a local minimum along this path. A point that seems to act in some dircetion as a max and in another as a min is a saddle point. A formal definition follows.

Let $F:\R ^n\to \R$ and $\vec {c}$ be in the domain of $F$ where $\grad F(\vec {c})=\vec {0}$ . The function $F$ has a saddle point at $\vec {c}$ if, for every open ball $B$ containing $\vec {c}$ , there are points $\vec {a}$ and $\vec {b}$ in $B$ such that $F(\vec {c})>F(\vec {a})$ and $F(\vec {c})<F(\vec {b})$ .

The most obvious example of a saddle point is a the point determined by $\vector {0,0}$ on a hyperbolic paraboloid of the form $z = \pm x^2 \mp y^2$ .

When thinking about a graph of $z= F(x,y)$ at a saddle point, the instantaneous rate of change in all directions is $0$ and there are points nearby with $z$ -values both less than and greater than the $z$ -value of the saddle point.

The second derivative test

In theory to identify local extrema verses saddle points, we could compute the Taylor polynomial of degree $2$ at the critical point in question, and then identify the Taylor polynomial as either:

Elliptic paraboloid: Indicating we have found local extrema.
Hyperbolic paraboloid: Indicating that we are at a saddle point.

Fortunately, as we have seen, there is a second derivative test that does exactly this for us. We will now restate this test in the context of identifying local extrema.

Second derivative test Given a function $F:\R ^2\to \R$ , and a critical point $\vec {c}$ where $D(\vec {c}) = F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})-\left [F^{(1,1)}(\vec {c})\right ]^2.$

If $D(\vec {c})>0$ and $F^{(2,0)}(\vec {c})<0$ then $F$ has a local maximum at $\vec {c}$ .
If $D(\vec {c})>0$ and $F^{(2,0)}(\vec {c})>0$ then $F$ has a local minimum at $\vec {c}$ .
If $D(\vec {c})<0$ , then $F$ has a saddle point at $\vec {c}$ .
If $D(\vec {c})=0$ , the test is inconclusive.

We first practice using this test with the function in the previous example, where we visually determined we had a local maximum and a saddle point.

Let $F(x,y) = x^3-3x-y^2+4y$ . Determine whether the function has a local minimum, maximum, or saddle point at each critical point.

We determined previously that the critical points of $F$ are $\vector {-1,2}$ and $\vector {1,2}$ . To use the second derivative test, we must find the second partial derivatives of $F$ : $\begin{align*} F^{(2,0)}(x,y) &= \answer[given]{6x}\\ F^{(0,2)}(x,y) &= \answer[given]{-2}\\ F^{(1,1)}(x,y) &= \answer[given]{0} \end{align*}$

Thus: $D(x,y) = \answer [given]{-12x}$ Since $D(-1,2) = \answer [given]{12}>0$ , and $F^{(2,0)}(-1,2) = \answer [given]{-6}$ , by the second derivative test, locally looks like at $\vector {-1,2}$ , meaning $F$ has a local maximum at $\vector {-1,2}$ .

Since $D(1,2) = \answer [given]{-12} <0$ , by the second derivative test, $F$ locally looks like at $\vector {1,2}$ meaning $F$ has a saddle point at $\vector {1,2}$ .

The second derivative test has confirmed the visual evidence we found before.

Find the local extrema of $F(x,y) = x^2y+y^2+xy$ .

We start by finding the first and second partial derivatives of $F$ : $\begin{align*} F^{(1,0)}(x,y) &= 2xy+y\\ F^{(0,1)}(x,y) &= x^2+2y+x \\ F^{(2,0)}(x,y) &= 2y\\ F^{(0,2)}(x,y) &= 2\\ F^{(1,1)}(x,y) &= 2x+1 \end{align*}$

We find the critical points by finding where $\grad F(x,y) = \vec {0}$ , since the partial derivatives are defined everywhere on $\R ^2$ . With a bit of algebra we can show that there are three critical points

$\vector {0,0}$ ,
$\vector {\answer [given]{-1},0}$ , and
$\vector {\answer [given]{-1/2},\answer [given]{1/8}}$ .

Now for each critical point $\vec {c}$ , we compute compute $D(\vec {c}) = F^{(2,0)}(\vec {c})F^{(0,2)}(\vec {c})-F^{(1,1)}(\vec {c})^2$ . We find $D(x,y) = \answer [given]{4y-(2x+1)^2}$

Since $D(0,0)=\answer [given]{-1}<0$ , we see that $F$ has a saddle point at $\vector {0,0}$ .
Since $D(-1,0)=\answer [given]{-1} <0$ , we see that $F$ has a saddle point at $\vector {-1,0}$ .
Since $D(-1/2,1/8)=\answer [given]{1/2}>0$ and $F^{(2,0)}(-1/2,1/8) = \answer [given]{1/4} > 0$ , we see that $F$ has a local at $\vector {-1/2,1/8}$ .

Below we see a graph of $F$ and the three critical points.

Note how this function does not vary much near the critical points. Visually it is difficult to determine whether a point is a saddle point or local minimum (or even a critical point at all!). This is one reason why the second derivative test is so important to have.

For some interesting extra reading check out: