The Problem To Be Considered:
Given a function f(x,y) of two variables x and y, find the points (x0,y0), if any, such that for all (x,y),f(x,y) <= f(x0,y0)
and find the value f(x0,y0). This last
number is the maximum value of f, and
(x0,y0) is the location of a maximum.
This problem is easily solved by our strategy of building
single variable functions F(t) by combining
multivariable functions f(x,y) with parameterized paths
by defining
F(t) = f(x(t),y(t))
To do so we just have to choose the right paths. Here is how to do this:
Suppose that (x0,y0) is the location
of a maximum of f.
Suppose also that (x(t),y(t)) is any parameterized path with
(x(0),y(0)) = (x0,y0)
Then we have the following conclusion: The function F(t) = f(x(t),y(t))
has a maximum at t = 0, because when t = 0,
(x(t),y(t)) is at a maximum of f, so that
F(t) <= F(0)
for all t. But we know from the one variable theory that this means that
F'(t) = 0
Now let's get back from F to f: rewrite this
equation in terms of the gradient of f. If as before, we write
V = (x'(0),y'(0))
then
F'(0) = dot(Gradf(x0,y0),
V) = 0
Now the key point is that the path could be any path
just as long as (x(0),y(0)) = (x0,y0).
We can do this with any velocity vector
V. Thus,
dot(Gradf(x0,y0),
V) = 0
for all V. But then
Gradf(x0,y0 is orthogonal
to every vector V.
There is only one way this can be true:
Gradf(x0,y0) = 0
This analysis has led to a definition and a theorem:
Gradf(x0,y0) = 0
The term defined in every good definition occurs occurs in the first line
of a useful theorem.
Of course the case of minima is so similar to the case of maxima that it
need not be written out explicitly here.
This gives us a method for solving the problem posed towards the beginning
of the lecture:
We put emphasis on the point "if there is one" in Step 3
because it can seem a bit subtle at first. Here's is an example that
brings the point out:
Consider the function
f(x,y) = x2 + y2
for which
Gradf(x,y) = (2x,2y)
So the points we are looking for in Step 1 are the solutions of the system
2x = 0
2y = 0
and it is no problem to see that there is only one solution, namely,
(0,0). So the short list is really short here -- just one item long.
Step 2 is easy: f(0,0) = 0. So this is the minimum value if
there is one, and it is the maximum value if there is one.
Is it either?
Well, f(x,y) is a sum of sqaures, so f(x,y) >= 0
for all (x,y).
But then
f(0,0) <= f(x,y)
for all (x,y). So there is a minimum value, namely 0 and
this occurs at just one location, namely (0,0).
On the other hand there is no maximum value. To see this, consider what happens
for points (x,y) of the form, say, (x,0). At
such a point f(x,y) = x2, and by taking x
larger and larger, we can make the value of f(x,y) get as large as we like. There is no upper limit to it; i.e., no maximum.
As a second example, let's consider a variation on the first:
Consider the function
f(x,y) = x2 - y2
The difference is just the minus sign. For this function,
Gradf(x,y) = (2x,-2y)
So the points we are looking for in Step 1 are still the solutions of the system
2x = 0
2y = 0
and it is no problem to see that there is only one solution, namely,
(0,0). Again the "short list" is just one item long.
Step 2 is the same as before: f(0,0) = 0.
So this is the minimum value if
there is one, and it is the maximum value if there is one.
Is it either?
No, in this case it is neither. To see that there is no maximum,
consider points (x,y) of the form (x,0) as before.
Then to see that there is no minimum, consider points
(x,y) of the form (0,y). At
such a point f(x,y) = -y2, and by taking y
larger and larger, we can make the value of f(x,y) get as negative as we like. There is no lower limit to it; i.e., no minimum.
What these examples point out is that our three step procedure is not
really complete yet -- we need a way to decide if there is a minimum
or a maximum.
We will see soon that when we are considering maximums and minimums in
subsets of the x,y plane that are closed
and bounded, there is always both a
maximum and a minimum. The rub is that either or both may be
located on the boundary. And while there are only two boundary points to
consider for an interval in the one variable thoery -- 0
and 2 in the case discussed at the beginning of this section --
there are infinitely many boundary points of a
closed bounded region in the plane. So we need a method to deal with them
before we can treat such problems.
That method is the method of Lagrange multipliers, and we shall come to it
soon.
There is another thing that may seem to be lacking in our development so far:
What about second derivative tests?
In the context of the problem at hand, these are not so useful, even
in the single variable version. To see why, recal what the tests say:
To find the maximum of f(t) on [0,2], we
don't need to compute the values of f(t) at all
the critical points tj, just those with
f''(tj) <= 0
This fact has great theoretical importance -- and so does the multivariable analog -- but if we really want ot compute a maximum, does it help us?
That is, does it save us any computation?
Probably not. To apply it, you'd probably have to compute f''(tj)
for each critical point to see where it was positive. You'd throw out those
critical points, if any, and then go back and compute
f(tj) for the possibly shortened short list. The total
amount of computation is likely to be significantly more.
This is especially true if your are writing a program to do such a
thing. A person could look at an expression for f''(t)
and might be able to determine where it was positive or negative just by looking without computing. This is why I said "proabably" alot above.
But to write a program to recognize such things would likely be difficult --
and the result inefficient.
We will discuss topics related to the second derivatives test in a later
section of notes. But first, lets close this section by relating
critical points, tangent planes, and best linear approximation.
z = f(x0,y0)
or, in other words, tangent plane is flat. This is what you'd expect at a
local maximum or minimum: If it weren't flat there would be an "uphill"
direction, and a "downhill" direction.
Also,
the best linear appoximation h(x,y) to f(x,y) at
(x0,y0) is constant
h(x,y) = f(x0,y0)
So at the level the best linear approximation, all critical points look pretty much alike.
There is a differnce though, as we shall see, at the level of the best quadrtic
approximation.
f(x,y) = x2 + y2 - (x-1)4 -
(y-1)4 -4*x - 4*y
gradf(x,y) = (2*x - 4*(x-1)3 -4,2*y - 4*(y-1)3-4)
Thus, after simplifying, we see that
(x,y) is a critical point if and only if
-4*x3+ 12*x2-10*x = 0 and -4*y3+ 12*y2-10*y =0
These equations are each satisfied if x=0 or
y=0 respectively. To find the other solutions, we may assume that neither x nor
y is zero, and divide by 2*x and 2*y respectively.
We get
-2*x2+ 6*x-5 = 0 and -2*y2+ 6*y-5 =0
These two quadratics are irreducible, there are no roots and no more solutions.
Thus, there is just 1 critical point:
(0,0)
This is our short list. And you can't say it isn't short!
Now, to decide if there is a minimum or maximum, look at large values of the variables. At large values the highest powers in x and y dominate, and the function behaves like
- x4 -
y4
which clearly has no lower bound, but is bounded above.
Thus there is a maximum, but no minimum. To find the maximum, we just plug in
the critical points to f(x,y). We get:
f(0,0) = -2
Evidently, the maximum value is -2, and it occurs at (0,0)
and only there
This answers part (b). And we have already observed that there is no minimum, which answers part (c).
We can
check our work with the applet that draws contour curves and gradient fields: This was used to produce the following contour graphs with gradient vectors drawn in in red. (The little dots are at their tails.)
wehre the scale in the first picture is such that it shows -8 < x,y < 8, and the second is a close up on the origin. You should recognize from the pictures that they say there is one critical point, and it is a maximum.
(See the gradient vectors pointing inward? Also you can see by the color -- green denotes low values, and white high values in these applets -- think deep green valleys and snowy mountain peaks.)
If you use the applet, which reads x and y coordinates for you
as you move the cursor around, you can also find the coordinates of any
critical points you see in the picture.
So the applets can be used to check all of the homework problem! This gives you
a very large supply of answered problems.
f(x,y) = -x2*y2+x6 +y6
gradf(x,y) = (-2*x*y2 + 6*x5,
-2*y*x2 + 6*y5 )
Thus, we see that
(x,y) is a critical point if and only if
-x*y2 + 2*x3 = 0 and -y*x2 + 2*y3 =0
These equations are each satisfied if x=0 and
y=0. So (0,0> is one critical point. To find the others, we may assume that neither x nor
y is zero, and divide by 2*x and 2*y respectively.
We get
-y2+ 3*x4 = 0 and -x2+ 3*y4 =0
Eliminating y yields
27*x8 = 3*y4 = x2
or
x2 = 1/3,
We get the same results for y becasue of the symmetry in the equations.
Thus, there are 5 critical points:
(0,0) (1/sqrt(3),1/sqrt(3)) (-1/sqrt(3),1/sqrt(3)) (1/sqrt(3),-1/sqrt(3))
(-1/sqrt(3),-1/sqrt(3))
This is our short list.
Now, to decide if there is a minimum or maximum, look at large values of the variables. At large values the highes power dominates, and the function behaves like
x6 +
y6
which clearly has no upper bound, but is bounded below.
Thus there is a minimum, but no maximum. To find the minimum, we just plug in
the critical points to f(x,y). We get:
f(0,0) = 0, f(1/sqrt(3),1/sqrt(3)) = -1/27, and the same; i.e., 1/27 for the rest.
Evidently, the minimum value is -1/27, and it occurs at exactly four places
(1/sqrt(3),1/sqrt(3)) (-1/sqrt(3),1/sqrt(3)) (1/sqrt(3),-1/sqrt(3))
(-1/sqrt(3),-1/sqrt(3))
This answers part (c). And we have already observed that there is no maximum which answers part (b).
Finally, for part (d), the eqauation for the tangent plane at each of the four minima is just
z = -1/27
Here are the pictures that confirm our work is correct:
The first is a wide view, in which you see all five critical points. Notice that the graph is lighter around the edges, so the highest values are off at the edges. Is this what you would expect here? Next is
a close-up on the origin. We can see that it is higher up than the other four points -- so it is not the minimum, as we know. Finally, the third is a close up of (1/sqrt(3),1/sqrt(3)). You see the visual signature of a local minimum -- notice the gradient vectors pointing away from the center.
f(x,y) = x2*y + (x-1)*(y-1)2
This one has exactly two critical points, and both have integer coefficients.
You will probably have to solve a cubic equation in x or y,
but this is a simple case that can be done by inspection. You can aid your
powers of inspection by drawing a graph -- or using one of the applets to look for the critical points.
Here is the picture drawn by the applet:
The first shows the two critical points. If you have trouble solving the eqautions, you can use the applet to read off the coordinates of the
two
critical points that you see here -- just move the cursor over the points, and read
the result off the position panel. The second picture is a close up of one of the two critical points. Notice how regular things look up close!
f(x,y) = x*y/(1+ x2 + y2)2
Here is what the applet draws for us:
Here is the picture drawn by the applet:
From this you should be able to see how many critical points there are, and what type they are.
Definition of Critical Points
A point (x0,y0) is called a
critical point of a function f(x,y) in case
Theorem on Maxima, Minima, and Critical Points
A point (x0,y0) is the location
of a maximum of f(x,y) only if (x0,y0)
is a critical point of this function.
Similarly,
A point (x0,y0) is the location
of a minimum of f(x,y) only if (x0,y0)
is a critical point of this function.
Critical Points and Tangent Planes
Consider a function f(x,y) of two variables.
At any critical point (x0,y0) of f(x,y), the
equation of the tangent plane is
WORKED PROBLEMS
Worked Problem 1
Let
Solution to Worked Problem 1
We easily compute that
Things to Notice
Worked Problem 2
Let
Solution to Worked Problem 2
We easily compute that
POSED PROBLEMS
Posed Problem 1
Let
Posed Problem 2
Let
