Question

已知m,n,x,y属于R,且m^2+n^2=a,x^2+y^2=b,求mx+ny的最大值?

Answer 1

其实不诡异,观察一下已知条件象什么公式,配一配就行了

m^2+n^2+x^2+y^2=a+b

(x^2-2mx+m^2)+(y^2-2ny+n^2)+2(mx+ny)=a+b

(mx+ny)={a+b-[(x^2-2mx+m^2)+(y^2-2ny+n^2)]}/2

(mx+ny)={a+b-[(x-m)^2+(y-n)^2]}/2

(x-m)^2+(y-n)^2>=0

所以 mx+ny