Thread: 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?

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!

Hey Aschu.

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

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

Thank you very much

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.

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