Question

Tìm GTLN của

A = y - \(2y^2\) + 4040

Answer 1

\(A=y-2y^2+4040=-2\left(y^2-\dfrac{y}{2}+\dfrac{1}{16}\right)+\dfrac{32321}{8}\)

\(=-2\left(y-\dfrac{1}{4}\right)^2+\dfrac{32321}{8}\le\dfrac{32321}{8}\)

\(maxA=\dfrac{32321}{8}\Leftrightarrow y=\dfrac{1}{4}\)