certainly,but we have to know if x+y+z=n ,when x=y=z,

=n/3,we get x*y*z is the Max.
V=4x*y^2=4xb^2/a^2(a^2-x^2)=4b^2/a^2(a^2x-x^3)
only a^2x-x^3 useful
a^2x-x^3=x(a-x)(a+x)=[(1-c)x(a-x)(ca+cx)]/c(1-c)=[
(1-c)x+(a-x)+(ca+cx)=a+ca=n
so,x=(a+ca)/3(1-c),a-x=(a+c)/3,cancel c we get x=(squat3/3)a.
Thank your question.