Lagrange Multiplier stuff!
Find the max and min values
f(x,y,z)= x^2-y-z, x^2-y^2+z=0

Question

Lagrange Multiplier stuff!
Find the max and min values
f(x,y,z)= x^2-y-z, x^2-y^2+z=0

anonymous · Answer

Gradient of f= <2x,-1,-1> Gradient of g (constraint)= <2x, -2y, 1> Now im stuck

anonymous · Answer

I did
$$2x=\lambda(2x) $$
$$-1=\lambda (-2y)$$
$$-1=\lambda $$
which doesnt make any sense cuz I have different values for lambda

anonymous · Answer

What are you assuming your second value for lambda is?

anonymous · Answer

1 because 2x=lambda (2x)
So i have -1 and 1 for lambda...

anonymous · Answer

Well, my best guess is that lambda is -1 only. If we look at that 2x = lambda(2x) part:
$$2x = \lambda (2x) \implies 2x-\lambda (2x) = 0 \implies 2x( 1 - \lambda) = 0$$ So lambda = 1 OR x = 0. Clearly lambda must be -1, so this forces x to be 0 for this to be valid. From there, since we know lambda is -1, we can get y to be -1/2 and from there ue the constraint to see what x must be.

anonymous · Answer

To see what z must be, my bad

anonymous · Answer

oh ok. so then lambda=-1, y= 1/2, x=0, then z=1/4
f(0,1/2,1/4)=-3/4 Right?
Is this the min or max?

anonymous · Answer

Right. You can use the second derivative test to see if its a min or a max.

anonymous · Answer

AWESOME:) thank you!!

anonymous · Answer

No problem :)