(a) Elasticity of demand is given by #(dq)/(dx)*x/q#, where #x# is the price function and #q# is the demand.
Here we have #q=sqrt(420-x^2)#
hence #(dq)/(dx)=1/(2sqrt(420-x^2))*(-2x)=-x/sqrt(420-x^2)#
and elasticity of demand is #-x/sqrt(420-x^2)*x/sqrt(420-x^2)#
= #-x^2/(420-x^2)#
(b) At a price of #$10# per game, elasticity of demand is #-100/(420-100)=-100/320=-5/16#
as the elasticity is close to being flat, demand is elastic.
(c) Revenue #R# is given by #R=qx=xsqrt(420-x^2)#
and #(dR)/(dx)=sqrt(420-x^2)-(2x^2)/(2sqrt(420-x^2)#
and when it is #0#, revenue is maximum, it is#0#
i.e. #(2x^2)/(2sqrt(420-x^2))=sqrt(420-x^2)#
or #2x^2=2(420-x^2)# or #4x^2=840#
i.e #x=sqrt210=14.49#
Hence at price of #$14.49#, revenue is maximum.