Question

已知3a^2+4b^2=4,求b*根号下（1+a^2)的最大值

Answer 1

3a^2+4b^2=4

b^2=1-3a^2/4

b=√(1-3a^2/4)

b√(1+a^2)

=√(1-3a^2/4)√(1+a^2)

=√[-3/4(a^4-a^2/3-4/3)]

=√[49/48-3/4(a^2-1/6)^2]

∴当a^2=1/6和b^2=7/8时,

b√(1+a^2)的值最大,

最大值=7√3/12.