Use the method of lagrange multipliers to determine the maximum and minimum of f(x, y, z) = x + y + z subject to the two conditions g(x, y, z) = x 2 + y 2 − 2 = 0 and h(x, y, z) = x + z − 1 = 0.

Equating like coefficients yields 
y = 12λx and x = 2λy. 
==> y = 12λ(2λy) = 24λ^2y. 

(i) If y = 0, then 6x^2 + 0 - 8 = 0 ==> x = ± 2/√3. 

(ii) Otherwise, 1 = 24λ^2 ==> λ = ± 1/√24. 
This yields y = ±12x/√24 = ± x√6. 
Plug this into g: 6x^2 + 6x^2 - 8 = 0 ==> x = ±√(2/3). 
This yields (x, y) = (±√(2/3), ±2), (±√(2/3), ∓2). 

Testing the critical points: 
f(± 2/√3, 0) = 0  
f(±√(2/3), ±2) = 2√(2/3) <Maximum  
f(±√(2/3), ∓2) = -2√(2/3) <Minimum 

The answer will be y=-5+0 because the slope is -5 and the it has to go through y-axis 0