更新1:
Upper length is 3. Bottom length is 8 AB=5
Cal Prob
2009-11-24 3:31 am
Where should the point P be chosen on line segment AB (x = AP) so as to maximize the angle θ? Picture: http://www.webassign.net/scalcet/4-7-69.gif
回答 (1)
2009-11-24 4:01 am
✔ 最佳答案
AB = 5 => BP = 5 - xThe angle θ = arctan(x/8) + arctan[(5-x)/3]
dθ/dx = {1/[1 + (x/8)^2]}(1/8) + {1/[1 + (5-x)^2/9}(-1/3)
= 8/(64+x^2) - 3/[9 + (5-x)^2]
= 8/(x^2 + 64) - 3/(x^2 - 10x + 34)
= [8(x^2 - 10x + 34) - 3(x^2 + 64)] / [(x^2 + 64)(x^2 - 10x + 34)]
= (5x^2 - 80x + 80) / [(x^2 + 64)(x^2 - 10x + 34)]
dθ/dx = 0 => 5x^2 - 80x + 80 = 0
x^2 - 16x + 16 = 0
x = [16 +/- √(256 - 64)]/2
x = 14.93 (rejected) or x = 8 - 4√3 = 1.072
when x = 1.072, θ =1.051761
when x = 1.06, θ = 1.051758
when x = 1.08, θ = 1.051759
Therefore when x = 8 - 4√3 = 1.072, θ is a maximum.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20091123000051KK01061