You have Lagrangian
with partial derivatives (set equal to 0)
Solving the first four equations for
, respectively, we can substitute these solutions in terms of
into the fifth equation to find
, which in turn will lead to
. Denoting by
, we have
and substituting into the fifth equation yields
From either choice of
we arrive at
, i.e. exactly two critical points at
, for which we get a maximum value of 14 and minimum value of -14, respectively.