Question

水渠道断面为等腰梯形,如图所示,渠道深为h,梯形面积为S,为了使渠道的渗水量达到最小,

Answer 1

令下底为m,上底为n

则：n=m+2h/tanα

S = (m+n)h/2 = (m+m+2h/tan)h/2 = h(m+h/tanα)

m = S/h-h/tanα

腰长L = h/sinα

两腰及下底之和f(α) = 2h/sinα+S/h-h/tanα

= S/h + h(2/sinα-1/tanα)

= S/h + h(2/sinα-1/tanα)

= S/h + h(2-cosα)/sinα

f‘(α) = h {sinα*[-(-sinα)]-(2-cosα)*cosα}/(sin^2α) = h(1-2cosα)/sin^2α

当cosα＜1/2时,函数单调增；当cosα＞1/2时,函数单调减

当cosα=1/2,即α=60°时,两腰及下底之和达到最小.