求解数学题
必须得出,否则就瞬间散架
OP:x=2y, 设OP交AB于D(2m,m),m≠0,A(2m+h,m+k),B(2m-h,m-k),h>0,则 m^+m^/3<=1,m^<=3/4,-√3/2<=m<=√3/2, (2m+h)^/4+(m+k)^/3=1,① (2m-h)^/4+(m-k)^/3=1,② ①-②,2mh+4mk/3=0, AB的斜率k/h=-3/2,③ |AB|=2h√(1+9/4)=h√13, AB:3x+2y-8m=0, P(2,1)到AB的距离d=|8-8m|/√13, 把③代入①,3(4m^+4hm+h^)+4(m^-3hm+9h^/4)=12, 16m^+12h^=12,h=√(1-4m^/3) ∴△APB的面积S=(1/2)d|AB|=|4-4m|h=4√[(1-m)^*(1-4m^/3)], 设f(m)=(1-m)^*(3-4m^), 则f'(m)=-2(1-m)*(3-4m^)-8m(1-m)^ =-2(1-m)[3-4m^+4m(1-m)] =-2(1-m)(3+4m-8m^) =-16(m-1)[m-(1+√7)/4][m-(1-√7)/4], 当m1=(1-√7)/4时f'(m1)=0,f(x)|max=f(m1)=(37+14√7)/16,S最大, 这时l的方程是3x+2y-2(1-√7)=0。
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