Where x(measured in units of a hundred) is the quantity demanded per week and p

Question

Where x(measured in units of a hundred) is the quantity demanded per week and p is the unit price in dollars.
(a) Evaluate the elasticity at 10. E(10)=

(b) When is the demand unitary. p=

(c) What is the maximum revenue?

Suppose that 10000 people take city buses each day and pay for a ticket. A regression model suggests that the number of people taking city buses at price p dollars per ticket is given by

x=10000*sqrt(4-p)

(a) Evaluate the price elasticity of demand at $2 per ticket

E(2)=?

(b)  For what value of p is the demand unitary?

  p=?

(c)What is the maximum revenue?

Explanation / Answer

a) The formula for the elasticity of demand E (which you should know) is

E = (p/x) (dx/dp)

x = .25(225-p^2)

dx/dp = -.25*2*p= -.5p


Thus,
so E(10)=-0.5*10*(10/(0.25*(225-10^2))=-1.6

b)

-0.5*p*p/(0.25*(225-p^2))=-1

0.5*p^2=56.25-0.25*p^2

p^2=75

p=8.66

c)revenue=R=p*(0.25*(225-p^2))

dR/dp=0 will yield p for maximum revenue.

dR/dp=0

==> 0.25*225-0.25*3*p^2=0

p=8.66

so maximum revenue=R(8.66)=324.75

Q 2.

x=10000*sqrt(4-p)

price elasticity=(dx/dp)*(p/x)

dx/dp=10000*(-1)/(2*sqrt(4-p))

so price elasticity at p=2 is equal to

(-10000/(2*sqrt(4-p))*p/(10000*sqrt(4-p))

=-p/(2*(4-p))

=-0.5

b)unitary demand is when elasticity=-1

so -p/(2*(4-p))=-1

p=8-2*p

p=8/3

c)revenue=R=p*10^4*sqrt(4-p)

dR/dp=0

sqrt(4-p) - (p/(2*sqrt(4-p))=0

4-p=p

p=2

maximum revenue=10^4*sqrt(2)=14142.13

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