Question

已知△ABC中,A(1,1),B(m,√m),C(4,2)(1＜m＜4),求m为何值时,△ABC的面积S最大.

Answer 1

用ac做低,再求出b点到ac直线的距离

ac直线x-3y+2=0

b点到ac直线的距离=(│mx+y√m+2│)÷√10

s=1/2×√10×(│m-3√m+2│)÷√10

=1/2×│m-3√m+2│

因为1＜m＜4

所以m-3√m+2 是负的

s=1/2(-m+3√m-2)

=1/2[-(√m-3/2)^2+1/4]

所以当m=9/4时

△ABC的面积=1/8