To make more than one assumption I would use And
:
$Assumptions = And[a > 0, \[Beta] > 0, d > 0, 1 > b > 0,
c > 0, a > \[Tau]*\[Theta]A, \[Tau] >
0, \[Theta]A > \[Theta]B > \[Theta]C, a > \[Tau]*\[Theta]B,
a > \[Tau]*\[Theta]C, Fb > 0, Fc > 0, 1 > \[Alpha] > 0]
Your constraint is then
myConstraint = Simplify[\[Tau] < 1/(2 (-\[Theta]A^2 + \[Theta]B^2))*
(-2 a \[Theta]A + 2 a \[Theta]B -
Sqrt[((2 a \[Theta]A - 2 a \[Theta]B)^2 -
4 (-\[Theta]A^2 + \[Theta]B^2)*
(-((a^2 \[Theta]A)/(\[Theta]A + \[Theta]B)) - (
a^2 \[Beta] \[Theta]A)/(\[Theta]A + \[Theta]B) + (
a^2 p \[Beta] \[Theta]A)/(\[Theta]A + \[Theta]B) - (
a^2 p \[Alpha] \[Beta] \[Theta]A)/(\[Theta]A + \[Theta]B) \
+ (a^2 \[Theta]B)/(\[Theta]A + \[Theta]B) +
(a^2 \[Beta] \[Theta]B)/(\[Theta]A + \[Theta]B) - (
a^2 p \[Beta] \[Theta]B)/(\[Theta]A + \[Theta]B) + (
a^2 p \[Alpha] \[Beta] \[Theta]B)/(\[Theta]A + \
\[Theta]B)))])]
I retract what I said about incompatibility: I had miswritten a symbol. Your constraint is indeed an interval, but its endpoints have different expressions, depending on the parameters:
intrvl =Reduce[myConstraint && $Assumptions, \[Tau], Reals]
To maximize a parametric expression, FindMaxValue
is not suitable, because it uses numerical methods. MaxValue
can handle symbolic expressions:
fnct = -((d (a - \[Theta]A \[Tau])^2)/(2 (b + c)^2)) + (
b (1 + \[Beta]) (a - \[Theta]A \[Tau])^2)/(8 (b + c)^2) +
((1 + \[Beta]) (a - \[Theta]A \[Tau])^2)/(4 (b + c));
MaxValue[{fnct, intrvl}, \[Tau]]// Simplify
The interpretation of the symbolic result is tricky.