7. Given positive real numbers $a, b$ satisfy $9a+b=36$, then $\frac{1}{a}+\frac{1}{b}$ reaches its minimum value when $ab=$ $\qquad$ -
$\frac{1}{a}+\frac{1}{b}=\frac{9}{9 a}+\frac{1}{b} \geqslant \frac{(3+1)^{2}}{9 a+b}=\frac{4}{9}$, equality holds when $3 a=b \Rightarrow a=3, b=9 \Rightarrow a b=27$.