Có x + y = 10
=> x = 10 - y
Thay \(x=10-y\) vào bt P ta có
\(P=y\left(10-y\right)\)
\(=-\left(y^2-10y\right)\)
\(=-\left(y-5\right)^2+25\le25\forall y\)
=> MaxP = 25 \(\Leftrightarrow\left\{{}\begin{matrix}y-5=0\\x=10-y\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=5\end{matrix}\right.\)
Ta có: \(\left(x-y\right)^2\ge0\forall x,y\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\forall x,y\)
\(\Leftrightarrow x^2+y^2+2xy\ge2xy+2xy\forall x,y\)
\(\Leftrightarrow\left(x+y\right)^2\ge4xy\forall x,y\)
\(\Leftrightarrow4xy\le100\)
hay \(xy\le25\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}xy=25\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy=25\\x=10-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(10-y\right)\cdot y=25\\x=10-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}10y-y^2-25=0\\x=10-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2-10y+25=0\\x=10-y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(y-5\right)^2=0\\x=10-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-5=0\\x=10-y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=10-5=5\end{matrix}\right.\)
Vậy: Khi x+y=10 thì giá trị lớn nhất của P=xy là 25 khi x=5;y=5
