Question

麻烦大神帮我解下第二问

麻烦大神帮我解下第二问2a-c

Answer 1

令x=2a-c，则c=2a-x

根据余弦定理，b^2=a^2+c^2-2ac*cosB
4=a^2+(2a-x)^2-2a(2a-x)*(1/2)
=a^2+4a^2-4ax+x^2-2a^2+ax
=3a^2-3ax+x^2
3a^2-3ax+x^2-4=0
判别式=9x^2-12(x^2-4)=48-3x^2
所以a=[3x+√(48-3x^2)]/6
S△ABC=(1/2)*ac*sinB
=(1/2)*a(2a-x)*√3/2
=(√3/4)*(2a^2-ax)
=(√3/4)*[2(3ax-x^2+4)-3ax]/3
=(√3/12)*(3ax-2x^2+8)
=(√3/12)*{[3x^2+x√(48-3x^2)]/2-2x^2+8}
=(√3/24)*[x√(48-3x^2)-x^2+16]
令S△ABC=y，t=16-x^2，则x=√(16-t)
y=(√3/24)*[√(16-t)*√(3t)+t]
=(√3/24)*[√(48t-3t^2)+t]
y'=(√3/24)*[(24-3t)/√(48t-3t^2)+1]
y''=(√3/24)*[-3(48t-3t^2)-(24-3t)^2]/(48t-3t^2)^(3/2)
=-24√3/(48t-3t^2)^(3/2)
令y'=0
√(48t-3t^2)=3t-24
48t-3t^2=9t^2-144t+576
12t^2-192t+576=0
t^2-16t+48=0
(t-12)(t-4)=0
t=12或4
因为y''(12)=y''(4)=-√3/72<0
所以t=12或4都是y的极大值点
y(12)=√3，y(4)=2√3/3，y(12)>y(4)
所以当2a-c=12时，S△ABC取到最大值√3