\(3a+5b=12\Rightarrow b=\frac{12-3a}{5}\)
\(\Rightarrow B=a\left(\frac{12-3a}{5}\right)=\frac{-3a^2+12a}{5}=\frac{-3\left(a^2-4a+4\right)+12}{5}=-\frac{3}{5}\left(a-2\right)^2+\frac{12}{5}\le\frac{12}{5}\)
\(B_{max}=\frac{12}{5}\) khi \(\left\{{}\begin{matrix}a=2\\b=\frac{6}{5}\end{matrix}\right.\)