finding relative Max and mins and saddle points functions of two variables is a
little bit more complicated than finding them for a function of a single variable
but so still certainly very doable so how do we find a maximum minimum or a
saddle point of a function of two variables so I've got a couple graphs
here and we're gonna kind of attack that in the same way and that we're going to
look for some first derivative some first partials and find out where
they're zero and then kind of use something equivalent to the second
derivative test to look for kind of a concavity to determine then if they're a
minimum a maximum or a saddle point so for example this sort of parabolic cone
shape here we notice that if we take the derivative with respect to X we keep Y
constant basically you end up this point down here at the bottom is a minimum and
the tangent lines in both the X and the y direction in the plane are zero now on
the other hand we have a graph over here that kind of has a weird point kind of
right about there that here in part of the graph you can see if we took a cross
section it would be a minimum but if we look at it from this perspective it
would actually be a maximum so we get what's called a saddle point in that
case it kind of looks like something you would put on a horse and ride right it
has a saddle a place where you can sort of sit in it so how do we tell which one
of these we have because in this case we have write a derivative going one way
for the X variable and a derivative going the other way for the Y variable
again that we kind of have this horizontal tangent thing going on so how
do you tell which one that you have so there's essentially a test there are
some steps that we're going to take so the steps for finding the relative
maximum or relative minimum and saddle points of a function of two
variables is the following so the first step is we need to find the first
partials and then three of the four second partials so we need to find the
first partials of sub X F sub y so the partial with respect to X the partial
with respect to Y and then we need to find the three these three partials so
the second partial with respect to X the second partial with respect to Y and
then the mixed second partials so partial with respect to X and then Y and
what we're going to do with these is kind of like a combination of a first
derivative test and a second derivative test combined so the first part is very
similar to what we did in performing the first derivative test and finding
critical values there we were just finding the critical values of the
derivative but here we have to first partials right so you have to kind of
treat this as a system of equations and find a solution there so you might have
to review a little bit of your algebra there and talk about the finding the
solution to a system of equations by either the elimination method or the
substitution method and so then let's assume that we have a solution we have
the point a B is a solution and there may be more than one we it may actually
find that there are a couple of points that are that that satisfy that system
of equations but let's say that a B is one of them then we're going to compute
what we call D so D of a B and D of a B is all about the second partials so the
second partial with respect to x times the second partial with respect to Y
minus the mixed second partial squared right and so there are basically three
possibilities we're gonna look at for D first D is negative if D is negative
then we have a saddle point at a B okay so if the result of D and plugging in
our solution into D gives us a negative number we have a saddle point if D is
positive then we either have a max or a min so we have a maximum
if the second partial with respect to X is negative and we have a minimum if the
second partial is positive okay so D being greater than zero doesn't tell me
whether I have a max or a min I have to then look at the second partial with
respect to X and if that is negative so kind of think of second derivative
negative means concave down then that means I have a maximum and if the second
partial with respect to X is positive then think concave up and then we have a
relative minimum and if we end up with zero as our output then it turns out
that this test that we're doing is not applicable for the function that we have
and pretty much that's not going to be the case in the problems that we have
but it is a possibility okay so let's do some examples so the first one is we're
going to look for the relative maximum and minimum points and if there are any
saddle points we'll find those along the way to because we you know it's obvious
from the result of D if de is negative then we have a saddle point right okay
see you let's go ahead and find our partials so first one is the partial
with respect to X okay so X is our variable of interest y is treated as a
constant so the derivative of x squared with respect to X is 2x the derivative
of X times y with respect to X is going to be 1 times y y is treated as a
constant so that's just Y the derivative of Y squared with respect to X is 0
because it doesn't have X in it it doesn't vary with respect to X so it's
derivative is 0 and then minus the derivative of 3x which is just going to
be 3 so that's the partial with respect to X the partial with respect to Y is
going to be let's see derivative of x squared with respect to Y is 0 plus the
derivative of x times y so X is the constant multiple times the derivative Y
which is 1 so x times 1 is X plus the derivative of y squared is 2y
and then minus the derivative of 3x well it doesn't have a Y in it so it doesn't
vary with respect to Y so that means its constant means its derivative is 0
all right so then we also need the second partials with respect to x and y
so let's calculate up those so FXX derivative of 2x with respect to X is 2
the derivative of Y with respect to X is 0 the derivative of 3 with respect to X
is 0 so that second partial is just 2 alright the next one is FY y so the
derivative of X with respect to Y is 0 plus the derivative to I with respect to
Y is 2 and then don't forget we also need the mixed partial F X Y so go back
to F sub X the first partial with respect to X and take the derivative
with respect to Y so the derivative of 2x with respect to Y is 0 plus the
derivative of Y is 1 minus the derivative of 3 which is 0 so the second
partial mixed partial is 1 alright step 2 is to find the solution to the system
of equations partial with respect to X equals 0 and partial with respect to y
equals 0 so like we have to find a point that makes both of those true at the
same time so let's set that up 2x plus y minus 3 equals 0 and X plus 2y is equal
to 0 so our two strategies here are either elimination or substitution I'm
going to do substitution so I'm going to solve this second equation for x so
subtract 2y from both sides and I get x equals 2y now that I'm going to take and
substitute in here for X so when I do that I get two x replace X with negative
2y plus y minus 3 is equal to zero negative 4y + y + negative 3y minus 3 is
0 and add 3 to both sides divide by negative 3y would have to be negative 1
to make that true right quick check and I get it 3 times negative 1 3 minus 3 is
0 so that works okay so we've got part of our solution so what we're going to
do is we're going to plug that into here so X then would have to be negative 2
times the Y value that we got so negative 1 so that means 2 so we have a
solution is X is 2 and Y is negative 1 so step number 3 calculate D of 2 comma
negative 1 and let's recall that that's the second partial with respect to X at
that point times the second partial with respect to Y at that point - the mixed
partials mixed partial at that point and then squared in this case though if we
go back up to our all of our second partials notice they're all constant
functions so FXX no matter what x and y values we have
this is always 2 and fyy is always 2 and - the mixed partials squared so minus 1
squared so this is 4 minus 1 is 3 which is positive so D being positive tells me
we either have a max or a min and we have a max or a min by looking at the
second partial all right and if that is positive that means concave up we have a
minimum or if that is negative then we have concave down essentially and then
what's a maximum but f of X X the second partial with respect to X is 2 so that's
positive so kind of think write positive means concave up so that tells me we
have a minimum at the point 2 comma negative 1
okay and if you're interested in seeing what this looks like it's pretty easy
you can graph this in like Wolfram Alpha or geogebra org just you can Google 3d
graphing and lots of options come up and you can graph that up and you'll see
that basically this looks kind of a kind of a strange function kind of looks like
this and so here is our minimum point that we've got here so kind of kind of
sort of looks like a cone shape sort of thing it looks kind of like that all
right let's try another all right find the relative extreme points and of
course there may be a saddle point you can find that along the way for G oops
G
yiii of XY equals x cubed plus y squared minus 3x minus 4y plus three all right
so first steps we need our partials so the partial with respect to X is three x
squared plus zero minus three minus zero plus zero partial with respect to Y is
going to be zero plus two y minus zero minus four plus zero then we need the
second partials so the second partial with respect to X is going to be 6x
minus zero second partial with respect to Y is going to be two minus zero and
then our mixed partial G X Y so the derivative of G sub X the partial with
respect to X with respect to Y well there are no Y's in there so that's
basically treated all as a constant so then it is actually zero alright so step
two is we need to solve the system of equations G sub X is zero G sub y is
zero right so let's set that up 3 x squared minus 3 equals 0 and 2y minus 4
equals 0 okay well notice these are like the first one is independent of Y and
the second one is independent of X so basically what we're going to do is
solve each of them separately right so this says three x squared must be three
so that means X is either going to be one or negative one right because you
get x squared equals one so X is negative one and one and then this one
we solve let's see two Y would equal 4 so that means Y is equal to two
so this basically means we have two solutions here because we have the case
when X is 1 and you have case when X is negative 1 but no matter what the X
values are the Y value they get only one Y value so we actually end up with two
solutions here so we have two points to consider so we need to calculate D for
each of them so give myself a little bit of room here so D of 1 comma 2 is second
partial with respect to X a 1 comma 2 I'm second partial with respect to Y 1
comma 2 minus the mixed oops these are all you should be G's right not FS this
is the function G we're not using function f all right let me fix that up
for you all right so g XX g YY and g x y at 1 comma 2 quantity squared ok so G X
X is 6 X so this is going to be 6 times the x-coordinate so that's 1 times G YY
is just plain old 2 and then minus the mixed partial which is 0 squared so this
is 6 times 2 is 12 and so this is positive and since it's
positive then we also have to check FX or G I keep doing that G X X so G X X at
1 comma 2 is 6 times X so 6 times 1 which is 6 which is positive so again
think positive second derivative positive concave up so we have a
relative minimum at 1 comma 2 now we need to check D at negative 1 comma 2 so
let's see that's G of X X negative 1 comma 2 times G Y y and
negative 1 comma 2 minus G of X Y and negative 1 comma 2 squared so G of X X
that's the 6 times X so 6 times negative 1 times G YY is always 2 so times 2
minus our mixed partials squared so 0 squared so this is negative 6 so that's
less than 0 so if we go if we go back to our conditions there
remember if D is negative then we have a saddle point at negative 1 comma 2 ok so
we have a relative min at 1 comma 2 and a saddle point at negative 1 comma 2
okay all right one last example and this is an applications problem and we're
going to be looking at how many types of each widget do we need to sell in order
to maximize our profits so a company sells a widget a and widget B and they
sell X thousand a widget a so X is going to represent the number of items of
widget a I'm a number of units of widget a that we are going to produce and sell
and then we have Y thousand of widget B so Y represents the number of widget B
that we're going to produce and sell and we have a revenue function and we have a
cost function but we're asked to maximize profit so we have to remember
that profit equals revenue minus cost so in this case our profit function is
going to be 17 X plus 21y and then we subtract off all of our terms in C so
it'll be minus four x squared plus four X Y minus 2 y
squared plus 11x minus 25 y plus 3 right so if I'm going to subtract off the
constant or the cross function that means every term in here gets subtracted
off so it means every term gets an opposite sign so before I move on I want
to collect my like terms so let's see I have a term that has just an X in it
there's another one that has just an X do I have just a Y sure do just Y and
just Y and then I have an x squared no other x Squared's Y squared no other Y
Squared's and a mixed X Y so let's see so P X Y is 17 X plus 11x that's 28 X +
21 Y minus 25 Y is minus 4y and then minus 4 x squared + 4 X Y minus 2 y
squared and then plus 3 right so this is the function that we're looking to find
the maximum of so our profit function is the that function there and we are going
to need to find the maximum so it's the same process so first step is I need the
partials so P sub X it's a derivative of 28 X with respect to X is 28 minus 4 y
that's going to be a derivative is going to be 0 because it doesn't have an X and
then minus 4 x squared is going to be minus 8x plus 4 X Y is going to be
derivative of 4 X is 4 times the constant Y the derivative of minus 2 y
squared is 0 derivative of 3 is 0 so there's our partial with respect to X
now we need our partial with respect to Y derivative of 28 X with respect to Y
is 0 derivative of negative 4y is negative 4
derivative of negative 4 x squared is 0 plus
constant 4x times the derivative Y which is 1 so just for X and then minus the
derivative 2y squared which is for Y and then plus the derivative of 3 which is 0
right and then we need second partials so derivative of that with respect to X
is just going to be 0 minus 8 plus 0 and second partial with respect to Y it's
going to be 0 plus 0 minus 4 and then we need our mixed second partial so we go
back to here and take the derivative with respect to Y so 0 minus 0 plus for
me at that ok step 2 is solve the system with both partials equal to 0 okay so
let's write that down so we've got 28 minus 8x plus 4y equals 0 and negative 4
plus 4x minus 4y equals 0 so now on previous problems we used substitution
or in the case of the other problem like each equation was only with one variable
so we could solve them separately so notice here both equations have both
variables but instead of doing substitution I want to notice that I
have plus 4y here and minus 4y here so what I'm actually going to do is what's
called the elimination method is I'm going to add these two equations
together and I'll end up with an equation in just one variable so let's
see add the constant terms together to 28 and negative 4 is 24 negative 8x plus
4x is minus 4 X 4 y minus 4y is 0 and then equals 0 plus 0 is 0 so we end up
with that equation which we can then solve for X so 24 equals 4x so X would
have to be 6 okay so that's one solution 6 comma there that's half of the
solution and then we just need to find a y-value well I can pick
either of the two equations and solve for y so I'm going to plug this into 28
minus 8x so 8 times 6 plus 4y is equal to 0 so this is 28 6 times 8 is 48 plus
4y is equal to 0 so negative 20 plus 4y equals 0 so Y would have to be at 20 to
both sides divide by 4 is 5 so we get the solution 6 comma 5 all right
now just like before you can't just assume this is a maximum you have to
prove that this provides a maximum so that means we need to find D of 6 comma
5 which means second partials at 6 comma five times the other second partial at 6
comma 5 minus the mixed partial at 6 comma 5 squared okay alright second
partial with respect to X is negative 8 times second partial with respect to Y
is negative 4 minus the mixed partial which is 4 quantity squared
so negative 8 times negative 4 is 32 minus 16 which is 16 which is positive
okay good I would be really upset if this was a
saddle point if that came out negative but what we need to check now is the
second partial with respect to X which in this case is negative 8 so again
think second derivative negative means concave down so that does indeed we need
a maximum so then you would need to sell 6000 a widget a and 5,000 widget B to
maximize profit
and that's pretty much it they all kind of work the same you just have to do the
interpretation here at Step four right which we didn't really write in step
four here is if D is positive and the second derivative with respect to X is
positive or negative then you have a maximum in
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