Câu 1:
A = (3 - y)(4 - x)(2y + 3x)
6A = (6 - 2y)(12 - 3x)(2y + 3x)
Ta có: \(\hept{\begin{cases}0\le x\le4\\0\le y\le3\end{cases}\Leftrightarrow\hept{\begin{cases}4-x\ge0\\3-y\ge0\\2y+3x\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}12-3x\ge0\\6-2y\ge0\\2y+3x\ge0\end{cases}}}\)
Áp dụng BĐT cô-si ta được:
\(\left(12-3x\right)+\left(6-2y\right)+\left(2y+3x\right)\ge3.\sqrt[3]{\left(12-3x\right)\left(6-2y\right)\left(2y+3x\right)} \)
\(\Leftrightarrow3.\sqrt[3]{6A}\le18\Leftrightarrow A\le36\)
Dấu = xảy ra khi:
12 - 3x = 6 - 2y = 2y + 3x
=> \(\hept{\begin{cases}3x+4y=6\\6x+2y=12\end{cases}\Rightarrow\hept{\begin{cases}x=2\left(n\right)\\y=0\left(n\right)\end{cases}}}\)
Vậy.....
