In this section we are going to extend one of the more
important ideas from Calculus I into functions of two variables. We are going to start looking at trying to
find minimums and maximums of functions.
This in fact will be the topic of the following two sections as well.
In this section we are going to be looking at identifying
relative minimums and relative maximums.
Recall as well that we will often use the word extrema to refer to both
minimums and maximums.
The definition of relative extrema for functions of two
variables is identical to that for functions of one variable we just need to
remember now that we are working with functions of two variables. So, for the sake of completeness here is the
definition of relative minimums and relative maximums for functions of two
variables.
Definition
Note that this definition does not say that a relative
minimum is the smallest value that the function will ever take. It only says that in some region around the
point 
the function will always be larger than . Outside of that region it is completely
possible for the function to be smaller.
Likewise, a relative maximum only says that around the function will always be smaller than . Again, outside of the region it is completely
possible that the function will be larger.
Next we need to extend the idea of critical points up to functions of two variables. Recall that a critical point of the function was a number so that either or doesn’t exist.
We have a similar definition for critical points of functions of two
variables.
Definition
To see the equivalence in the first part let’s start off
with and put in the definition of each part.
The only way that these two vectors can be equal is to
have and . In fact, we will use this definition
of the critical point more than the gradient definition since it will be easier
to find the critical points if we start with the partial derivative definition.
Note as well that BOTH of the first order partial
derivatives must be zero at . If only one of the first order partial
derivatives are zero at the point then the point will NOT be a critical point.
We now have the following fact that, at least partially,
relates critical points to relative extrema.
Fact
Proof
Note that this does NOT say that all critical points are
relative extrema. It only says that
relative extrema will be critical points of the function. To see this let’s consider the function
The two first order partial derivatives are,
The only point that will make both of these derivatives zero
at the same time is and so is a critical point for the function. Here is a graph of the function.
Note that the axes are not in the standard orientation here
so that we can see more clearly what is happening at the origin, i.e. at . If we start at the origin and move into
either of the quadrants where both x
and y are the same sign the function
increases. However, if we start at the
origin and move into either of the quadrants where x and y have the opposite
sign then the function decreases. In
other words, no matter what region you take about the origin there will be
points larger than and points smaller than . Therefore, there is no way that can be a relative extrema.
Critical points that exhibit this kind of behavior are
called saddle points.
While we have to be careful to not misinterpret the results
of this fact it is very useful in helping us to identify relative extrema. Because of this fact we know that if we have
all the critical points of a function then we also have every possible relative
extrema for the function. The fact tells
us that all relative extrema must be critical points so we know that if the
function does have relative extrema then they must be in the collection of all
the critical points. Remember however,
that it will be completely possible that at least one of the critical points
won’t be a relative extrema.
So, once we have all the critical points in hand all we will
need to do is test these points to see if they are relative extrema or
not. To determine if a critical point is
a relative extrema (and in fact to determine if it is a minimum or a maximum)
we can use the following fact.
Fact
Note that if then both and will have the same sign and so in the first
two cases above we could just as easily replace with . Also note that we aren’t going to be seeing
any cases in this class where . We will be able to classify all the critical
points that we find.
Let’s see a couple of examples.
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Example 1 Find
and classify all the critical points of .
Solution
We first need all the first order (to find the critical
points) and second order (to classify the critical points) partial
derivatives so let’s get those.
Let’s first find the critical points. Critical points will be solutions to the
system of equations,
This is a non-linear system of equations and these can, on
occasion, be difficult to solve.
However, in this case it’s not too bad. We can solve the first equation for y as follows,
Plugging this into the second equation gives,
From this we can see that we must have or . Now use the fact that to get the critical points.
So, we get two critical points. All we need to do now is classify
them. To do this we will need D.
Here is the general formula for D.
To classify the critical points all that we need to do is
plug in the critical points and use the fact above to classify them.
:
So, for D
is negative and so this must be a saddle point.
:
For D
is positive and is positive and so we must have a relative
minimum.
For the sake of completeness here is a graph of this
function.
 
Notice that in order to get a better visual we used a
somewhat nonstandard orientation. We
can see that there is a relative minimum at and (hopefully) it’s clear that at we do get a saddle point.
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Example 2 Find
and classify all the critical points for
Solution
As with the first example we will first need to get all
the first and second order derivatives.
We’ll first need the critical points. The equations that we’ll need to solve this
time are,
These equations are a little trickier to solve than the
first set, but once you see what to do they really aren’t terribly bad.
First, let’s notice that we can factor out a 6x from the first equation to get,
So, we can see that the first equation will be zero if or . Be careful to not just cancel the x from both sides. If we had done that we would have missed .
To find the critical points we can plug these
(individually) into the second equation and solve for the remaining variable.
:
:
So, if we have the following critical points,
and if the critical points are,
Now all we need to do is classify the critical
points. To do this we’ll need the
general formula for D.
:
:
:
:
So, it looks like we have the following classification of
each of these critical points.
Here is a graph of the surface for the sake of
completeness.
 
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Let’s do one more example that is a little different from
the first two.
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Example 3 Determine
the point on the plane that is closest to the point .
Solution
Note that we are NOT asking for the critical points of the
plane. In order to do this example we
are going to need to first come up with the equation that we are going to
have to work with.
First, let’s suppose that is any point on the plane. The distance between this point and the
point in question, ,
is given by the formula,
What we are then asked to find is the minimum value of this
equation. The point that gives the minimum value of this
equation will be the point on the plane that is closest to .
There are a couple of issues with this equation. First, it is a function of x, y
and z and we can only deal with
functions of x and y at this point. However, this is easy to fix. We can solve the equation of the plane to
see that,
Plugging this into the distance formula gives,
Now, the next issue is that there is a square root in this
formula and we know that we’re going to be differentiating this
eventually. So, in order to make our
life a little easier let’s notice that finding the minimum value of d will be equivalent to finding the
minimum value of .
So, let’s instead find the minimum value of
Now, we need to be a little careful here. We are being asked to find the closest
point on the plane to and that is not really the same thing as
what we’ve been doing in this section.
In this section we’ve been finding and classifying critical points as
relative minimums or maximums and what we are really asking is to find the smallest
value the function will take, or the absolute minimum. Hopefully, it does make sense from a
physical standpoint that there will be a closest point on the plane to . Also, this point should be a relative
minimum.
So, let’s go through the process from the first and second
example and see what we get as far as relative minimums go. If we only get a single relative minimum
then we will be done since that point will also need to be the absolute
minimum of the function and hence the point on the plane that is closest to .
We’ll need the derivatives first.
Now, before we get into finding the critical point let’s
compute D quickly.
So, in this case D
will always be positive and also notice that is always positive and so any critical
points that we get will be guaranteed to be relative minimums.
Now let’s find the critical point(s). This will mean solving the system.
To do this we can solve the first equation for x.
Now, plug this into the second equation and solve for y.
Back substituting this into the equation for x gives .
So, it looks like we get a single critical point : . Also, since we know this will be a relative
minimum and it is the only critical point we know that this is also the x and y coordinates of the point on the plane that we’re after. We can find the z coordinate by plugging into the equation of the plane as
follows,
So, the point on the plane that is closest to is .
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