Question

[đoàn thi hiền]

\(x^2+y^2=2\)

TÌM MAX \(P=2\left(x^3+y^3\right)-3xy\)

Answer 1

[Trà My]

\(P=2\left(x^3+y^3\right)-3xy\)

\(P=2\left(x+y\right)\left(x^2-2xy+y^2\right)-3xy\)

\(P=2\left(x+y\right)\left(2-2xy\right)-3xy\)

Mặt khác: \(x^2+y^2=2\Leftrightarrow\left(x+y\right)^2-2xy=2\Leftrightarrow xy=\frac{\left(x+y\right)^2-2}{2}\)

Thay \(xy=\frac{\left(x+y\right)^2-2}{2}\) vào P ta có:  \(P=2\left(x+y\right)\left(2-2.\frac{\left(x+y\right)^2-2}{2}\right)-3.\frac{\left(x+y\right)^2-2}{2}\)

Đặt x+y=t <=> \(\left(x+y\right)^2=t^2\le2\left(x^2+y^2\right)=2.2=4\) 

=> \(\left|t\right|\le2\) và \(P=2t\left(2-2.\frac{t^2-2}{2}\right)-3.\frac{t^2-2}{2}=-t^3-\frac{3}{2}.t^2+6t+3\) với \(\left|t\right|\le2\)

Xét \(g\left(t\right)=-t^3-\frac{3}{2}.t^2+6t+3\) trên đoạn [-2;2]

\(g'\left(t\right)=-3t^2-3t+6\)

             \(g'\left(t\right)=0\Leftrightarrow-3t^2-3t+6=\left(-3\right)\left(t^2+t-2\right)=0\)

\(\Leftrightarrow t^2+t-2=t^2-t+2t-2=t\left(t-1\right)+2\left(t-1\right)=\left(t+2\right)\left(t-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+2=0\\t-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}t=-2\in\left[-2;2\right]\\t=1\in\left[-2;2\right]\end{cases}}\)

             \(g\left(-2\right)=-7;g\left(2\right)=1;g\left(1\right)=\frac{13}{2}\)

=>\(P_{max}=\frac{13}{2}\) khi \(x=\frac{1+\sqrt{3}}{2}\) và \(y=\frac{1-\sqrt{3}}{2}\) hoặc \(x=\frac{1-\sqrt{3}}{2}\) và \(y=\frac{1+\sqrt{3}}{2}\)

Vậy .............