Question

Using Excel Solver: We have 3,000 units of product to sell over a five-day period. From historical sales data, we have estimated the following demand curves: P = price/unit in $, Q = number of units sold. Day 1: P = 10?0.01 Q valid for prices between $3 and $8. Day 2: same as Day 1. Day 3: P = 15?0.01Q valid for prices between $6 and $10 Day 4: P = 20?0.01Q valid for prices between $6 and $12 Day 5: same as Day 1. 1) The revenue maximizing price for Day 1 is _________ (Hint: Please keep one decimal point.), and quantity sold is ________ (Hint: Please enter an integer.). 2) The revenue maximizing price for Day 2 is ___________ (Hint: Please keep one decimal point.), and quantity sold is ________ (Hint: Please enter an integer.). 3) The revenue maximizing price for Day 3 is __________ (Hint: Please keep one decimal point.), and quantity sold is _________ (Hint: Please enter an integer.). 4) The revenue maximizing price for Day 4 is __________ (Hint: Please keep one decimal point.), and quantity sold is _________ (Hint: Please enter an integer.). 5) The revenue maximizing price for Day 5 is ____________ (Hint: Please keep one decimal point.), and quantity sold is _________ (Hint: Please enter an integer.)

Answer 1

Solution:-

Price at Day 1

P= 10-0.01Q

The price ranges between $3 and $8, including the limits.

The revenue-maximizing point

The maximum revenue is found by using the principle of maxima

i.e. dR/dQ =0 AND d²R/dQ²<0

The first order differential dR/dQ (10Q-0.01Q²)= 10-2*0.01Q=0

=10-0.2Q=0

=0.2Q=10

=Q=10/0.2= 500 Unit

The Second order differential (10-0.2Q) = -0.2

Since the second order differential is negative, the two conditions of maxima is satisfied and maximum Revenue point Q=500

and Maximum revenue , when Q=500,will be

The revenue R= 10Q-0.01Q^{2 =} 10*500-0.01*500² =2500

The price of the product will be P=10-0.01Q =10-0.01*500 = $5

2) The maximum revenue for day 2 (Day 2 is same as day 1)

The maximum revenue for day 2 will be same as Day1 , since the figures are not changing.

3) The maximum revenue for day 3

Price function P = 15?0.01Q

The revenue maximizing point

The maximum revenue is found by using the principle of maxima

i.e. dR/dQ =0 AND d²R/dQ²<0

Revenue = Price *Qunatity = (15?0.01Q)*Q

Revenue R= 15Q-0.01Q²

The first order differential dR/dQ (15Q-0.01Q²)= 15-2*0.01Q=0

=15-0.2Q=0

=0.2Q=15

=Q=15/0.2= 750 Unit

The Second order differential (15-0.2Q) = -0.2,

The second order differential is negative, hence it satisfies the conditions of maxima. The maximum revenue is at the point where Q=750 Unit.

The price of the product is 15-0.01*Q , When Q=750 ,

Price = 15-0.01*750 = $7.5

Maximum Revenue = 15Q-0.01Q² = 15*750-0.01*750² = $5625

4)

The maximum revenue for day 4

Price function P = 20?0.01Q

The revenue maximizing point

The maximum revenue is found by using the principle of maxima

i.e. dR/dQ =0 AND d²R/dQ²<0

Revenue = Price *Qunatity = (20?0.01Q)*Q

Revenue R= 20Q-0.01Q²

The first order differential dR/dQ (20Q-0.01Q²)= 20-2*0.01Q=0

=20-0.2Q=0

=0.2Q=20

=Q=20/0.2= 1000 Unit

The Second order differential (20-0.2Q) = -0.2,

The second order differential is negative, hence it satisfies the conditions of maxima. The maximum revenue is at the point where Q=1000 Unit.

The price of the product is 20-0.01*Q, When Q=1000,

Price = 20-0.01*1000 = $10

Maximum Revenue = 15Q-0.01Q² = 20*1000-0.01*1000² = $10000

5) The Maximum Revenue for the day 5 will be 1000 units of an item and maximum revenue will be $10000 as it is same as day 4