Extrema of functions of Two Variables

1. Aug 2, 2008

CalleighMay

Hello, my name is Calleigh and i am new to the forum! I am in Calculus II and have a few questions on some problems. I am using the textbook Calculus 8th edition by Larson, Hostetler and Edwards. Could someone please help me?

The problem is on pg 959 in chapter 13.8 in the text, number 56. It reads:

a. Find the absolute extrema of the function.
b. From the form of he function, determine whether a relative maximum of a relative minimum occurs at each point.

and it gives:
f(x,y)=2x-2xy+y^2
and R:The region in the xy-plane bounded by the graphs of y=x^2 and y=1

What does it mean when it states f(x,y)??? I've already heard of f(x). I don't know what it means when it has both x and y in the parenthesis. My professor gave these problems to us even though we haven't covered it yet, but he expects us to know how to do it?? Please help me, i'm so lost...

Any help would be greatly appreciated! Thanks guyssss ;)

2. Aug 2, 2008

nicksauce

f(x,y) means that f is a function of both x and y. Example, let's say M is how much money I have (let's say I only have 5 dollar bills - F and twenty dollar bills - T). Then M is a function of both F and T, ie: M(F,T) = 5F + 20T. Make sense?

3. Aug 3, 2008

schroder

Hi! As nicksauce has already pointed out, this is a function of more two independent variables, x and y. The resultant is usually labeled as z and the function is written z = f(x,y). In general, if z is a function of several independent variables, then z = f(x,y,w,s…) in this fashion. In this case, you only have two independent variables, so it is a fairly simple matter to differentiate and find the extrema. But in order to do this, you must take the partial derivatives of the function, which hopefully you have already studied. For the extremum, both first derivatives (partial derivatives) will be zero, as usual. But for a multi-variable function there is no simple second derivative test for the extremum, although in most situations, the second partial derivatives of z with respect to x and y are both negative at a maximum point and positive at a minimum point. But in order to be sure, you need to take the mixed second derivative also. In order for the point to be an extremum, the product of the partial derivatives must be equal or greater than the mixed second derivative. This is really not as difficult as it sounds. Once you study a bit more on partial derivatives it will all make sense!

4. Aug 3, 2008

Epsilon36819

Hi Calleigh,

It may help to visualize the function as a surface in 3D, where every point (x,y) of the x-y plane is mapped to a function z=f(x,y). If you fix y, say to 3, you can "cut the plane" at that particular y and obtain a function z=f(x,3), ie a function of one variable which can be graphed in the x-z plane. You can do this for every y (or every x) and think of the resulting surface in 3D as all these curves put side by side.

5. Aug 3, 2008

CalleighMay

Thanks for the replies guys! ;)

I tried taking the derivative (partial derivative?) of the function and got 2. I used my calculator to find the derivative. Something tells me this is wrong though because i will have a y' somewhere, no? I don't know how to do it out by hand.

Also what's a "mixed second derivative"? I cant find the second deriv until i do the first, which i cant do lol =(

6. Aug 3, 2008

schroder

OK, I will go ahead and walk you through this one because it seems you need the help. Hopefully you will get more out of this than just the answer. As Epsilon36819 pointed out, with two independent variables the graph of the function is represented by a surface in three dimensions. The way to analyze the total effect of the x and y variables is to differentiate the function against one variable while treating the other as a constant. That is what is meant by a partial derivative. Then you do the same thing with the second variable while treating the first as a constant. In the problem set you have given us: f(x,y)=2x-2xy+y^2 differentiating it with respect to x, while treating y as a constant yields 2 – 2y = 0. Now hold x constant and differentiate the function against y. This yields: -2X + 2y = 0. These are your two partial derivatives in x and y. You need to solve these equations simultaneously but in this case the first one is obviously y = 1 so substituting that into the second gives us x =1 also. The point which we are evaluating as an extremum is (1,1). Now differentiate them again in the same manner to get the second partial derivatives which gives us zero for the first (differentiating against x) and 2 for the second, differentiating against y. So the point does appear to be an extremum and because of the positive 2 it would appear to be a minimum. As I said before, with multi-valued functions such as this you can be fooled by something called a “saddle point” which is neither a maximum nor a minimum! To be sure this is a true extremum we need to take the mixed second derivative, which will be the same for both x and y. Differentiating 2 – 2y against y yields -2 and differentiating -2x + 2y against X also yields -2.
The final test is to multiply the second partial derivatives with each other and compare the product with the mixed second derivative as follows: (0) x (2) = 0 which is greater value than the mixed second derivative of -2 so this would seem to confirm that the point is indeed an extremum but because of the zero result it can still be a saddle point! It is best to graph it to be absolutely sure in this case. To find the absolute value of the function at (1,1) simply substitute these values into the Original Equation and you get a value of +1.
Note: Somebody else please verify this result to make sure I am not giving Calleigh a bum steer here!

Last edited: Aug 3, 2008
7. Aug 3, 2008

schroder

Whoops! I need to edit my post but it won't let me. In any case, I have confirmed that the point (1,1) is indeed a saddle point. The mixed second derivative should be squared before it is compared with the product of the second partial derivatives. So we have a +4 which indicates a saddle point. Sorry for any confusion I may have caused.

8. Aug 4, 2008

CalleighMay

thanks schroder, that helps! I tried it your way and it makes sense, but my friends and i worked on it tonight and they don't understand what you're talking about lol.

We got:

f(x,y)=2x-2xy+y^2
derivative with respect to x is f sub x =6x=0, x=0
with respect to y, 4y-4=0, y=1
then f(0,1)=-2
on line line of y=4, -2<=x<=2
so, f(x,y)=f(x)=3x^2+32-16=3x^2-16
and the max would be at 28 with a minimum at 16
on the curve of y=x^2, -2<=x<=2
f(x,y)=f(x)=3x^2+2(x^2)^2-4x^2=2x^4-x^2=x^2[2(x^2)-1]
and the max would be at 28 and the minimum at -1/8

absolute max 28 at (+ or -2, 4)
absolute min -2 at (0,1)

Does this seem about right? ;) thanks

9. Aug 4, 2008

schroder

With math there can only be one correct solution (in most cases) so either I am wrong or your friends are wrong, and I am fairly certain it is the latter!
I don’t see how you get the partial derivatives of the function to come out to 6x = 0 with respect to x,
and 4y – 4 = 0 with respect to y? Remember, when differentiating with respect to one variable, your treat the second variable the same as any constant. So the partial derivative of 2x -2xy +y^2 with respect to x is clearly 2 -2y = 0 and the partial derivative of 2x -2xy +y^2 with respect to y is clearly -2x +2y = 0. This solves for x =1 and y = 1. Plugging those values into the original equation gives an absolute value of z = 1. So the “extremum” point in 3D is (1,1,1). The only tricky part of the solution is in testing whether the extremum is either a minimum or a maximum. In this case it is neither as it is a saddle point of a hyperbolic paraboloid. The best way to see this is with a 3D graph. It is a very interesting function to graph and the saddle point of (1,1,1) becomes very clear. Give it another try and I think you will find my solution is correct.
Cheers!

10. Aug 4, 2008

CalleighMay

so you're saying that the extrema of the function is at (1,1,1) which isn't a max or min, but a "saddle point of a hyperbolic paraboloid" ? And what about part b? lol