# Homework Help: Absolute Extrema: 2 variables

1. Mar 9, 2013

### Reefy

1. The problem statement, all variables and given/known data

Find the absolute extrema of the function on the set D.

f(x,y)= x^2 + 4y^2 + 3x -1

D= {(x,y) l x^2 + y^2 ≤ 4}

2. Relevant equations

3. The attempt at a solution

The only thing I've done so far was find the critical point. I found f-sub x = 2x+3 and f-suby= 8y giving me a critical point of (-3/2,0), correct?

Using that critical point, I found f(-3/2,0) = -1

After that, I don't know how to proceed. I know -2 ≤ x ≤ 2 and -2 ≤ y ≤ 2 but how do I find the other critical numbers with that info? I'm unsure because the closed boundary graph is a circle and not a square, box, or triangular region.

2. Mar 10, 2013

### ehild

f(-3/2) is not equal to -1.

The absolute extreme can be on the perimeter. Find the critical point with the condition that x2+y2=4.

ehild

3. Mar 10, 2013

### Reefy

Ah, you're right, thank you. For that, I got -13/4.

So set y=√(4-x^2) ? and find f(x,√(4-x^2)). Then find values of x and use the result to find y. Then find critical numbers? Because that's what I did but it didn't seem right.

Last edited: Mar 10, 2013
4. Mar 10, 2013

### Ray Vickson

You have found the global minimum of f, subject to g = x^2 + y^2 <= 4. (although your f is wrong while your (x,y) is correct). However, an extremum can also be a maximum, and you have not yet found that.

5. Mar 10, 2013

### ehild

That is right.

ehild

6. Mar 10, 2013

### Reefy

Thanks, guys! I got it now. I forgot to take the derivative of f(x,√(4-x^2)).

Instead of finding the derivative, I found x and y of f(x,√(4-x^2)) and plugged it into the original function which was wrong.