d
\(Q=\left(\alpha t-\beta t^2+\gamma t^3\right)\)

\(i=\frac{d Q}{d t}=\left(\alpha-2 \beta t+3 \gamma t^2\right)\)

\(\frac{d i}{d t}=(3 \gamma t-2 \beta)=0\)

\(\Rightarrow t=\frac{\beta}{3 \gamma}\)

\(i=\left(\alpha-2 \beta t+3 \gamma t^2\right)=\left(\alpha-\frac{\beta^2}{3 \gamma}\right)\)