Consider the demand equation x+1/3p=20 0< or equal to p < or equal to 60, where
ID: 3289085 • Letter: C
Question
Consider the demand equation x+1/3p=20 0< or equal to p < or equal to 60, where x = # of units, p is the unit price in $.
I need for you to show the work in detail for each problem please so I can understand it. I appreciate it.
a)Find the elasticity of demand function. Then use it to determine if at a price of $20 whether the demand is elastic, unitary, or inelastic. If the price was increased slightly from $20 would the revenue increase or decrease.
b)Find the revenue as a function of p: R(p)
c) Determine the intervals where the revenue is increasing and the intervals where it is decreasing. Show your sign graph to justify your answer.
d) At what price will the Revenue be a maximum?
e) What is the maximum Revenue?
f) What is the marginal revenue function? Also, find the marginal revenue when the price is $15 and interpret your answer.
A friend and I have posted this several times and asked for more assistance but no one has shown us the work in detail so we can understand it.
Thanks.
Explanation / Answer
a)elasticity= (-dx/dp)*(p/x)= -(-1/3)*(1/2)=1/6
b)revenue=p*x=p*(20-p/3)=20p-p^2/3
d)for maximum revenue
d(20p-p^2/3)/dp=0
20-2p/3=0 ==>p=$30
e) maximum revenue=20*30-30^2/3 =$300
f)marginal revenue=d(20p-p^2/3)/dp=20-2p/3
at p=$15
marginal revenue=20-2/3*15=$10