# Thread: Multivariable calculus problem (Global Max/Min)

1. ## Multivariable calculus problem (Global Max/Min)

Find the global maximum and minimum of f(x,y) = (x^2) -2xy + 2y on the domain x ≥ 0, y ≥ 0 and y ≤ 2-x

Thanks.

2. ## Re: Multivariable calculus problem (Global Max/Min)

Originally Posted by bagels0
Find the global maximum and minimum of f(x,y) = (x^2) -2xy + 2y on the domain x ≥ 0, y ≥ 0 and y ≤ 2-x

Thanks.
Use the second partial derivatives test.

3. ## Re: Multivariable calculus problem (Global Max/Min)

Thanks Alex,

but could you please give a little more detail? I get very confused with the shading and all that...

4. ## Re: Multivariable calculus problem (Global Max/Min)

Originally Posted by bagels0
Thanks Alex,

but could you please give a little more detail? I get very confused with the shading and all that...
Where do $f_x$ and $f_y$ vanish?

5. ## Re: Multivariable calculus problem (Global Max/Min)

What do you mean by 'vanish'?

6. ## Re: Multivariable calculus problem (Global Max/Min)

Originally Posted by bagels0
What do you mean by 'vanish'?
equal zero

7. ## Re: Multivariable calculus problem (Global Max/Min)

2x - 2y = 0 [fx]
-2x + 2 = 0 [fy]

So, x = 1, and y = -1?

What does this mean? Now what do I do?

Thanks.

8. ## Re: Multivariable calculus problem (Global Max/Min)

Originally Posted by bagels0
2x - 2y = 0 [fx]
-2x + 2 = 0 [fy]

So, x = 1, and y = -1?

What does this mean? Now what do I do?

Thanks.
It means that there is a local extremum at (1, -1). Determine whether it is a maximum or minimum by using the second derivative test.

9. ## Re: Multivariable calculus problem (Global Max/Min)

Cool. But what about the Domain? How do I take care of that?

10. ## Re: Multivariable calculus problem (Global Max/Min)

Originally Posted by bagels0
Cool. But what about the Domain? How do I take care of that?
In the end, we will have to evaluate the function at the end-points of the domain.

11. ## Re: Multivariable calculus problem (Global Max/Min)

Ok, so lets do the second derivative test.

fxx = 2

fyy = 0

Also, my previous points were wrong. It was (1,1), not (1,-1).

So, what do I do now? Now that I've taken the second derivatives? What do they show?

12. ## Re: Multivariable calculus problem (Global Max/Min)

Originally Posted by bagels0
Ok, so lets do the second derivative test.

fxx = 2

fyy = 0

Also, my previous points were wrong. It was (1,1), not (1,-1).

So, what do I do now? Now that I've taken the second derivatives? What do they show?
Don't you know what the Second partial derivatives test states?

13. ## Re: Multivariable calculus problem (Global Max/Min)

Oh, so find the D? D = fxx*fyy - fxy^2

D = 0-4 = -4

So, that means...there is a saddle point at (1,1)?

14. ## Re: Multivariable calculus problem (Global Max/Min)

Originally Posted by bagels0
Oh, so find the D? D = fxx*fyy - fxy^2

D = 0-4 = -4

So, that means...there is a saddle point at (1,1)?
Yes. So there is no local extremum in the interior of the domain (the triangle). So you only need to evaluate the function at the endpoints of the domain (the sides of the triangle) to find the extrema.

15. ## Re: Multivariable calculus problem (Global Max/Min)

Ok, so here is what I did.

Found the critical point (1,1). f(1,1) = 1

The Discriminant showed it is a saddle point.

Found the edge points to be (2,0), (0,2) and (0,0) (Is this correct?)

Evaluating the f(edge points), I got 4, 4 and 0 respectively.

So.....global maximum is 4 at (2,0) and (0,2)? Global minimum is 0 at (0,0)?

Saddle point isn't max or min right? Also, is it ok if the maximum happens at 2 points? Is all this correct?

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