1. ## Calculus 2

f(x,y) = x2 + y2 subject to the constraint xy = 1, find all extrema, if they exist. Where x is in [-10, 10]
The only (x,y) that satisfy the constraint are =+_1 and y=+_1. In both cases f(x) =2. Here is where I stuck, which one is the max/min?

2. ## Re: Calculus 2

Do you not understand the meaning of the words "maximum" and "minimum"? If they both have the same value for f, then you can't have one the maximum and the other a minimum. Although it is quite possible that neither is a maximum or minimum.

There is a theorem saying that a continuous function will achieve both max and min on a compact set (closed and bounded interval will work). Further there is a theorem saying that they must lie either inside the interval, where the derivative is 0, or on the boundary. So it is possible that (1, 1) and (-1, -1) both give a maximum value while the minimum occurs at x= -10 or 10. Or that both (1, 1) and (-1, -1) both give a minimum value while the maximum occurs at x= -10 or 10. Or that the maximum occurs at either x= -10 or x= 10 or the other way around. To tell, evaluate f at those points and see which is the largest and which is the smallest!

3. ## Re: Calculus 2

Hey Aschu.

What is the Hessian matrix evaluated at those points and its determinant?

4. ## Re: Calculus 2

it's not true that the only x and y that satisfy the constraint are x=y +_1 .... for example the end points x=10 y=0.1 or x=-10 y=0.1 which give f(x,y)= 10^2 +0.1^2 that's higher than 2.

This is a multivariate question i believe not calculus 2

5. ## Re: Calculus 2

Thank you very much

6. ## Re: Calculus 2

Originally Posted by Ahasueros
it's not true that the only x and y that satisfy the constraint are x=y +_1 .... for example the end points x=10 y=0.1 or x=-10 y=0.1 which give f(x,y)= 10^2 +0.1^2 that's higher than 2.

This is a multivariate question i believe not calculus 2
But after applying Lagrange Multiplier, I arrived to x=y, and the constraint xy=1. Which contradict to ur counter examples.

7. ## Re: Calculus 2

Lagrange multiplier gives you the minimum in this case, but you have to also check the end points, on the interval x from -10 to 10 for example if x were to be from -100 to 100, and one in the case where x=-100 then from the constraint y=1/-100 and if you plug those values into your f(x,y) gives a greater output compared to 2.

x=10 and y=0.1 is closed under the constraint 10*0.1=1