So you end up with either $10000-p$ or $11000-p$ instead of either $9000$ or $11000$. You must solve for $p$ in $$\frac 12\ln(10000-p)+\frac12\ln(11000-p) = \frac12\ln 9000+\frac12\ln 11000$$ or equivalently $$ (10000-p)(11000-p)=9000\cdot 11000$$ whihc is a quadratic in $p$ where one of the roots is between $0$ and $1000$ (and the other makes both factors on the left negative and can be ignored). It turns out that the premium will eat more than half of your hoped-for profit ...