Question

a b x y属于R.满足a^2+b^2=p^2,x^2+y^2=q^2,p q都大于0,则ax+by的最大值是?p^2

Answer 1

你做的似乎是错误的

a^2+b^2=p^2,x^2+y^2=q^2(p>0,q>0)

(a^2+b^2)(x^2+y^2)=a^2x^2+b^2y^2+a^2y^2+b^2x^2

=(a^2x^2+b^2y^2+2axby)+(a^2y^2+b^2x^2-2aybx)

=(ax+by)^2+(ay-bx)^2=p^2q^2

=>(ax+by)^2=p^2q^2-(ay-bx)^2 =>ax+by=根号[p^2q^2-(ay-bx)^2]

∵(ay-bx)^2>=0,

∴只有当(ay-bx)^2=0时

ax+by=根号[p^2q^2-(ay-bx)^2]才有最大值为根号（p^2q^2）=pq