\(4x-3y=7\Rightarrow x=\frac{3y+7}{4}\)
\(B=-xy=-y\left(\frac{3y+7}{4}\right)=-\frac{3}{4}\left(y^2+\frac{7}{3}y\right)\)
\(B=-\frac{3}{4}\left(y^2+2.\frac{7}{6}y+\frac{49}{36}\right)+\frac{49}{48}\)
\(B=-\frac{3}{4}\left(y+\frac{7}{6}\right)^2+\frac{49}{48}\le\frac{49}{48}\)
\(B_{max}=\frac{49}{48}\)
\(B_{min}\) không tồn tại, chắc bạn chép sai đề