Question

Tìm Max của: \(C=\sqrt{x-2}+\sqrt{y-3}\) với x+y=6

Answer 1

Áp dụng BĐT Bunhiacopxki , ta có :

\(\left(x-2+y-3\right)\left(1^2+1^2\right)\) ≥ \(\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\)

⇔ C² ≤ 2

⇔ C ≤ \(\sqrt{2}\)

⇒ C_MAX = \(\sqrt{2}\)

Dấu " = " xảy ra khi và chỉ khi : \(\sqrt{x-2}=\sqrt{y-3}vàx+y=6\)

⇔ y = \(\dfrac{7}{2}\) ; x = \(\dfrac{5}{2}\)

Answer 2

x-2≥0=>x≥2

y-3≥0=>y≥3

x+y=6=>y-3=3-x

C>0;

C^2=x-2+(3-x)+2√(x-2)(3-x)

C^2=2√[1/4-(x-5/2)^2]+1≤2/2+1=2

C≤√2

(x,y)=(5/2;7/2) tm dk

maxC=√2

Answer 3

\(C=\sqrt{x-2}+\sqrt{y-3}\\ \Rightarrow C^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\\ =x-2+2\sqrt{\left(x-2\right)\left(y-3\right)}+y-3\\ =\left(x+y\right)-5+2\sqrt{\left(x-2\right)\left(y-3\right)}\\ =6-5+2\sqrt{\left(x-2\right)\left(y-3\right)}\\ =1+2\sqrt{\left(x-2\right)\left(y-3\right)}\)

Áp dụng \(BDT:Cô-si\)

\(\Rightarrow1+2\sqrt{\left(x-2\right)\left(y-3\right)}\le1+\left(x-2+y-3\right)\\ =1+\left(x+y-5\right)\\ =1+\left(6-5\right) =2\\ \Rightarrow C^2\ge2\\ \Rightarrow C\ge\sqrt{2}\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}x-2=y-3\\x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=-1\\x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{6+\left(-1\right)}{2}=\dfrac{5}{2}\\y=\dfrac{6-\left(-1\right)}{2}=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(C_{Max}=\sqrt{2}\) khi \(\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\)