Question

Cho \(x^2+4y^2=1\). Tìm \(minS,maxS\): \(S=x+y\)

Answer 1

Sửa đề: Tìm min, max của S = x + 2y

Áp dụng BĐT Bunhiacopxki ta có:

\(\left(x+2y\right)^2\) \(\le\) \(\left(1+1\right)\left(x^2+4y^2\right)\)

=> \(\left(x+2y\right)^2\) \(\le\) 2.1 = 2 ( Do \(x^2+4y^2=1\) )

=> \(-\sqrt{2}\le x+2y\le\sqrt{2}\)

=> Dấu = xảy khi \(\left\{{}\begin{matrix}\dfrac{1}{x}=\dfrac{1}{2y}\\x^2+4y^2=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=\pm\dfrac{\sqrt{2}}{2}\\y=\pm\dfrac{\sqrt{2}}{4}\end{matrix}\right.\)

=> Min S = \(-\sqrt{2}\) khi \(x=-\dfrac{\sqrt{2}}{2}\) ; y = \(\dfrac{-\sqrt{2}}{4}\)

Max S = \(\sqrt{2}\) khi \(x=\dfrac{\sqrt{2}}{2}\) ; y = \(\dfrac{\sqrt{2}}{4}\)