$EI ~ \theta_{AB} = 0$

$\frac{1}{2}(0.5L)(0.25PL) - 0.5L(M_{wall}) = 0$

$\frac{1}{2}M_{wall}L = \frac{1}{16}PL^2$

$M_{wall} = \frac{1}{8}PL$ *answer*

$\delta_{max} = t_{A/B}$

$EI ~ \delta_{max} = EI ~ t_{A/B}$

$EI ~ \delta_{max} = (Area_{AB}) \cdot \bar X_A$

$EI ~ \delta_{max} = \frac{1}{2}(0.5L)(0.25PL)[ \, \frac{2}{3}(0.5L) \, ] - 0.5L(M_{wall})[ \, \frac{1}{2}(0.5L) \, ]$

$EI ~ \delta_{max} = \frac{1}{48}PL^3 - \frac{1}{8}M_{wall}L^2$

$EI ~ \delta_{max} = \frac{1}{48}PL^3 - \frac{1}{8}(\frac{1}{8}PL)L^2$

$EI ~ \delta_{max} = \frac{1}{48}PL^3 - \frac{1}{64}PL^3$

$EI ~ \delta_{max} = \frac{1}{192}PL^3$ *answer*