Question

Determine the minimum value of f(x, y, z) = 2x 2 + y 2 + 2z 2 + 5 subject to the constraint 3x + 2y + 3z = 4. +

Answer 1

We want to optimize f(x,y,z)=x^2 y^2 z^2, subject to g(x,y,z) = x^2 + y^2 + z^2 = 289. 

Then, ∇f = λ∇g ==> <2xy^2 z^2, 2x^2 yz^2, 2x^2 y^2 z> = λ<2x, 2y, 2z>. 

Equating like entries: 
xy^2 z^2 = λx 
x^2 yz^2 = λy 
x^2 y^2 z = λz. 

Hence, x^2 y^2 z^2 = λx^2 = λy^2 = λz^2. 

(i) If λ = 0, then at least one of x, y, z is 0, and thus f(x,y,z) = 0 <---Minimum 
(Note that there are infinitely many such points.) 
(f being a perfect square implies that this has to be the minimum.) 

(ii) Otherwise, we have x^2 = y^2 = z^2. 
Substituting this into g yields 3x^2 = 289 ==> x = ±17/√3. 

This yields eight critical points (all signage possibilities) 
(x, y, z) = (±17/√3, ±17/√3, ±17/√3), and  
f(±17/√3, ±17/√3, ±17/√3) = (289/3)^3 <----Maximum 

I hope this helps!