Question

Cho x + y = 15. Tìm min, max

\(B=\sqrt{x-4}+\sqrt{y-3}\)

Answer 1

Min :AD BĐT vs a,b>0
\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\)
=>\(B=\sqrt{x-4}+\sqrt{y-3}\ge\sqrt{x-4+y-3}\)
Bình phương 2 vế
=> \(B^2\ge x+y-7=8=2\sqrt{2}\)
Vậy Min B=\(2\sqrt{2}\Leftrightarrow\left(x;y\right)=\left(4;11\right);\left(12;3\right)\)
Max: AD BĐT Buhiacopski ta có:
\(B^2=\left(1.\sqrt{x-4}+1.\sqrt{y-3}\right)^2\)
=> \(B^2\le\left(1+1\right)\left(x-4+y-3\right)=2.\left(15-7\right)=16\)
=> B ≤ 4
Vậy Max B=4 ⇔\(\left\{{}\begin{matrix}x\ge4\\y\ge3\\\sqrt{x-4}=\sqrt{y-3}\\x+y=15\end{matrix}\right.\Leftrightarrow\left(x;y\right)=\left(8;7\right)\)