Finding the extrema of a $f(x,y,z) = xy+yz$ with $x+y+z =1$ and $x^2+y^2+z^2 = 6$

Question

Find the extreme values of $$f(x,y,z) =xy+yz$$ under the constraints that $$I:x+y+z =1$$ and $$II:x^2+y^2+z^2 = 6$$

by parametrisation of the curve given by the side conditions.

I think that an obvious starting point would be using $I$ to see that $y = 1-x-z$ and plugging this into $f(x,y,z)$ to obtain:

$$f(x,z) = (1-x-z)(x+y)$$

under the constraint

$$II: x^2+ (1-x-z)^2+z^2 = 6.$$

I would normally try to get antoher substitution out of $II$, plug it into $f$ and then calculate the zereos of the gradient and check the Hessian matrix. But here $II$ seems very complicated to me. Do I get anything wrong?

Answer 1

Notice that $f(x,y,z)=(x+z)y$. Thus, by condition I, $x+z=1-y$, and

$f(x,y,z)=y-y^2$

Now condition II simply states $-\sqrt{6}\leq y\leq \sqrt{6}$ and the minimum and maximum of your function on this interval are easily found.

Answer 2

Hint: With $z=1-x-y$ we get $$f=xy+y-xy-y^2=xy+y-xy+y^2=y-y^2$$ and you will get a function in only one variable.