$P = \frac{{ - {P_0}}}{{{V_0}}}V + 3P$

$[slope = \frac{{ - {P_0}}}{{{V_0}}},c = 3{P_0}]$

              $P{V_0} + {P_0}V = 3{P_0}{V_0}\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$But\,\,\,\,\,\,\,PV = nRT\,$

$\therefore P = \frac{{nRT}}{V}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$

$From\left( i \right)\& \left( {ii} \right)\frac{{nRT}}{V}{V_0} + {P_0}V = 3{P_0}{V_0}$

$\therefore nRT{V_0} + {P_0}{V^2} = 3{P_0}{V_0}$

$...\left( {iii} \right)$

For temperature to be maximum $\frac{{dT}}{{dV}} = 0$

Differentiating $e.q.(iii)\,by\,'v'\,we\,get$

$nR{V_0}\frac{{dT}}{{dV}} + {P_0}\left( {2v} \right) = 3{P_0}{V_0}$

$\therefore nR{V_0}\frac{{dT}}{{dV}} = 3{P_0}{V_0} - 2{P_0}V$

$\frac{{dT}}{{dV}} = \frac{{3{P_0}{V_0} - 2{P_0}V}}{{nR{V_0}}} = 0$

$V = \frac{{3{V_0}}}{2}\,\,\,\,\,\,\,\,\,\therefore P = \frac{{3{P_0}}}{2}$          $[From (i)]$

$\therefore \,{T_{\max }} = \frac{{9{P_0}{V_0}}}{{4nR}}\,\,\left[ {From\,\left( {iii} \right)} \right]$