Calculus help needed? Thanks :)

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1 Answer
Apr 24, 2018

(a) Elasticity of demand is #-x^2/(420-x^2)#; (b) Demand is elastic at #x=$10# and (c) Revenue is maximized at #x=$14.49#

Explanation:

(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.