Lagrange multipliers question

[Maxim]

Hi I am a student chemistry currently in my second bachelor year and i have a question regarding the Lagrange multipliers.

"find the extreme values of the function: f= xy+z² on the circle formed by the intersection of the sphere with equation: x²+y²+z² = 4 and the plane with equation x-y = 0."

I have tried solving this multiple times but can't seem to find the right answer. I have tried using both statements as seperate condition and working with 2 L. mulitpliers, but that leads me to a set of false equations, and when trying to find the circle in question, which i believe should be y²+z²?, i get concrete values for my multiplier, namely -1/2 but no concise answer for the variables.

Sorry should there be mistakes in this question as I am unfamiliar with the English terms I had to translate them from my native language.

kind regards
-maxim

 

[JeffM]

[MATH]L(x,\ y,\ z,\ \lambda ,\ \kappa) = xy + z^2 - \lambda (4 - x^2 - y^2 - z^2) - \kappa (x - y - 0).[/MATH]
[MATH]\therefore \dfrac{\delta L}{\delta x} = 0 \iff y + 2 \lambda x - \kappa = 0.[/MATH]
[MATH]\dfrac{\delta L}{\delta y} = 0 \iff x + 2 \lambda y + \kappa = 0.[/MATH]
[MATH]\dfrac{\delta L}{\delta z} = 0 \iff 2z + 2 \lambda z = 0.[/MATH]
[MATH]\dfrac{\delta L}{\delta \lambda} = 0 \iff 4 = x^2 + y^2 + z^2.[/MATH]
[MATH]\dfrac{\delta L}{\delta \kappa} = \iff x = y.[/MATH]
Using the last partial, we get

[MATH]\dfrac{\delta L}{\delta x} = 0 \iff x + 2 \lambda x - \kappa = 0.[/MATH]
[MATH]\dfrac{\delta L}{\delta y} = 0 \iff x + 2 \lambda x + \kappa = 0.[/MATH]
[MATH]\dfrac{\delta L}{\delta z} = 0 \iff 2z + 2 \lambda z = 0.[/MATH]
[MATH]\dfrac{\delta L}{\delta \lambda} = 0 \iff 4 = 2x^2 + z^2.[/MATH]
Eliminate [MATH]\kappa.[/MATH]
[MATH](x + 2 \lambda x - \kappa) + (x + 2 \lambda x + \kappa) = 0 + 0 \implies[/MATH]
[MATH]2x + 4 \lambda x =0 \implies 2x(1 + 2 \lambda ) = 0.[/MATH]
[MATH]\text {CASE I: } x = 0.[/MATH]
[MATH]\therefore y = 0 \text { and } z^2 = 4 \implies z = \pm 2.[/MATH]
[MATH]max\{f(x,\ y,\ z)\} = 4.[/MATH]
[MATH]\text {CASE II: } x \ne 0.[/MATH]
[MATH]\therefore (1 + 2 \lambda) = 0 \implies \lambda = -\ 0.5.[/MATH]
[MATH]\therefore 0 = 2z + 2 \lambda z = 2z - z = z \implies z = 0.[/MATH]
[MATH]\therefore x = \pm \sqrt{2} = y.[/MATH]
[MATH]min\{(f(x,\ y,\ z)\} = 2[/MATH]
EDIT: You can solve this without Lagrangian multipliers by substitution, simplification, and checking boundary conditions.