# Extrema of multivariable function

1. Jun 11, 2007

### haroldholt

Hi

I'm studying for a calculus exam and I'm a little stuck on finding the extrema for multivariable functions.

For the particular question I'm trying to do now I need to find and classify the extrema for the function f(x,y) = (4x^2)(e^y) - 2x^4 - e^4y.

I can find the first derivatives, being fx = 8xe^y - 8x and fy = (4x^2)(e^y) - 4e^4y and I know I have to let them be equal to 0 to find where the extrema are located but I'm not sure how to do that. I guess it's just the exponentials that are throwing me off.

Any help would be appreciated.

Last edited: Jun 11, 2007
2. Jun 11, 2007

### D H

Staff Emeritus
You have done well so far. The next step is to simultaneously set those partial derivatives to zero. So try that and show us the results.

3. Jun 11, 2007

### haroldholt

Thanks

Well it's 0 = 8xe^y - 8x and 0 = (4x^2)(e^y) - 4e^4y. But I'm not sure where to go from there. I'm not sure what to do with the exponentials, I know they can never equal zero, but I'm not sure what that means for my equations.

4. Jun 11, 2007

### D H

Staff Emeritus
Simplify the fx=0 equation.

5. Jun 11, 2007

### haroldholt

0 = 8x(e^y - 1). But I still don't know what to do with the exponential.

6. Jun 11, 2007

### D H

Staff Emeritus
Any equation of the form a*b=0 tells you that either a=0 or b=0 (or both). This is obviously of that form. x=0 doesn't lead to any extrema. (Why not?) What happens when you set e^y-1 to zero?

7. Jun 11, 2007

### haroldholt

Thanks for that. I now realise that y can be set to zero in the equation 8x(e^y - 1). And then if you sub y = 0 into the other equation you get 1 and -1 which yields the points (-1,0) and (1,0) (which is the answer in the back of the book ). And if x = 0 then the other equation doesn't work. Thanks heaps. I now see how stupid I was initially lol.

8. Jun 13, 2007

### uiulic

I wish when I learned calculus (or any other courses) I could have such instruction.