回復 18# money 的帖子
計算2.
設 \(x,y\in R\) 且滿足 \(x^2+(y-1)^2=1\),試求 \(\displaystyle \frac{x+y+1}{x-y+3}\) 的最大最小值?
向量方法 :
設 \(\displaystyle \frac{x+y+1}{x-y+3}=k\),整理得 \((k-1)x+(-k-1)y=1-3k\)
令 \(\vec{a}=(x,y-1)\) , \(\vec{b}=(k-1,-k-1)\)
則 \(\vec{a}\cdot\vec{b}=x(k-1)+(y-1)(-k-1)=1\times\sqrt{(k-1)^2+(-k-1)^2}\cos{\theta}\)
而有 \(1-3k+k+1=\sqrt{(k-1)^2+(-k-1)^2}\cos{\theta}\)
\(\displaystyle \Rightarrow \cos{\theta}=\frac{2-2k}{\sqrt{2k^2+2}}\),
故 \(\displaystyle -1\leq\frac{2-2k}{\sqrt{2k^2+2}}\leq 1\)
\(\Rightarrow 2-\sqrt{3}\leq k \leq 2+\sqrt{3}\)
[ 本帖最後由 Joy091 於 2011-8-9 10:52 PM 編輯 ]