Thread: 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.

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.

Thanks Alex,

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

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 and vanish?

What do you mean by 'vanish'?

Originally Posted by bagels0

What do you mean by 'vanish'?

equal zero

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

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

What does this mean? Now what do I do?

Thanks.

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.

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

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.

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?

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?

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)?

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.

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?