We have added the asterisks to differentiate the
optimal values from general values of the variables.
In order to complete the proof we need to
confirm that these are valid values. From (4) and
the condition of this case (pDo
), there is
no problem with A being real and positive. In this
case (7) always produces a real positive value. (We
should note parenthetically here that we may have
a situation in which (7) gives a positive value but
that is not the solution! The discussion of that is
given in the next paragraph.) We also need to
check if the presumed optimal value of S is
between 0 and Q: The nonnegativity of S can
be established if we show hAXpD: Actually S can
be shown to be strictly positive, i.e. h2A24p2D2:
This can be confirmed from (5) and the condition
of this case