※ 引述《Stoudemi (Stoudemi)》之銘言：
: 請教版上的高手
: 這一題應該如何解會比較簡單
: 小弟一開始是利用R和r的關係來解
: http://i.imgur.com/IwBw2hz.jpg
2R + r + r + rsqrt(2) + rsqrt(2) = 2sqrt(2)
=> R + [1 + sqrt(2)]r = sqrt(2)
=> R = sqrt(2) - [1 + sqrt(2)]r
因為4r <= 2
=>r <= 1/2
=> R >= sqrt(2) - [1 + sqrt(2)]/2
= [sqrt(2) - 1]/2
(2)對
又2R <= 2
=> R <= 1
(1)對
r = [sqrt(2) - R]/[1 + sqrt(2)]
>= [sqrt(2) - 1]/[1 + sqrt(2)]
= 3 - 2sqrt(2)
[sqrt(2) - 1]/2 <= R <= 1
3 - 2sqrt(2) <= r <= 1/2
A = Pi 4 r^2 + Pi R^2
= Pi[4r^2 + R^2]
= Pi{4r^2 + 2 - 2sqrt(2)[1 + sqrt(2)]r + [3 + 2sqrt(2)]r^2}
= Pi{[7 + 2sqrt(2)]r^2 - 2sqrt(2)[1 + sqrt(2)]r + 2}
當r = sqrt(2)[1 + sqrt(2)] / [7 + 2sqrt(2)]
= [2 + sqrt(2)] / [7 + 2sqrt(2)]
A有極小值
(4)錯
當r = 3 - 2sqrt(2)時
A有極大值
此時R = sqrt(2) - [sqrt(2) + 1][3 - 2sqrt(2)]
= 1
(3)對
(5)錯
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