Question

Tìm GTLN

m,P=x+y+xy với 2 số thực x,y t/m \(2x+3y\le7\)

n,P=xy với x,y không âm và t/m x^2 + 2y =12

Answer 1

\(2x+3y\le7\Rightarrow y\le\frac{7-2x}{3}\)

\(\Rightarrow P\le x+\frac{7-2x}{3}+\frac{x\left(7-2x\right)}{3}=\frac{-2x^2+8x+7}{3}=\frac{-2\left(x-2\right)^2+15}{3}\le\frac{15}{3}=5\)

\(\Rightarrow P_{max}=5\) khi \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

b/ \(x^2+2y=12\Rightarrow y=\frac{12-x^2}{2}\)

\(P=\frac{x\left(12-x^2\right)}{2}=\frac{12x-x^3}{2}=\frac{-\left(x+4\right)\left(x-2\right)^2+16}{2}\)

Do \(x\ge0\Rightarrow\left(x+4\right)\left(x-2\right)^2\ge0\Rightarrow16-\left(x+4\right)\left(x-2\right)^2\le16\)

\(\Rightarrow P\le\frac{16}{2}=8\Rightarrow P_{max}=8\) khi \(\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)