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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 更新: Upper length is 3. Bottom length is 8 AB=5

最佳解答:

AB = 5 => BP = 5 - x The 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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