Question

2.   Copy and paste the following data into Excel:

P	Q
$15.25	125
$14.79	133
$14.33	140
$13.57	141
$12.96	147

a.   Run OLS to determine the demand function as P = f(Q); how much confidence do you have in this estimated equation? Use algebra to invert the demand function to Q = f(P).

b.   Using calculus to determine dQ/dP, construct a column which calculates the point-price elasticity for each (P,Q) combination.

c. What is the point price elasticity of demand when P=$15.25? What is the point price elasticity of demand when P=$14.10?

d.   To maximize total revenue, what would you recommend if the company was currently charging P=$14.79? If it was charging P=$14.10?

e.   Use your first demand function to determine an equation for TR and MR as a function of Q, and create a graph of P and MR on the vertical and Q on the horizontal axis.

f.    What is the total-revenue maximizing price and quantity, and how much revenue is earned there? Compare that to the TR when P = $15.25 and P = $14.10.

Answer 1

(a)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.9470
R Square	0.8969
Adjusted R Square	0.8625
Standard Error	0.3418
Observations	5
ANOVA
	df	SS	MS	F
Regression	1	3.049441362	3.049441	26.09642
Residual	3	0.350558638	0.116853
Total	4	3.4
	Coefficients	Standard Error	t Stat	P-value
Intercept	28.3769	2.7833	10.1954	0.0020
Q	-0.1035	0.0203	-5.1085	0.0145

Estimated demand equation: P = 28.3769 - 0.1035Q

Since R² = 0.8969, this means that 89.69% of the variation can be explained by the model, indicating a good degree of goodness of fit.

P = 28.3769 - 0.1035Q

0.1035Q = 28.3769 - P

Q = (28.3769 - P) / 0.1035 = 274.17 - 9.66P

(b) and (c)

Elasticity = (dQ/dP) x (P/Q) = - 9.66 x (P/Q)

Elasticity data table as follows.

When P = 15.25, Elasticity = - 1.18

When P = 14.1, Q = 137.96 (Using inverse demand function obtained above) and Elasticity = - 0.99

P	Q	Elasticity
15.25	125.00	-1.18
14.79	133.00	-1.07
14.33	140.00	-0.99
13.57	141.00	-0.93
12.96	147.00	-0.85
14.10	137.96	-0.99

(d)

Total revenue = P x Q = 28.3769Q - 0.1035Q²

Total revenue is maximized when dTR/dQ = 0

dTR/dQ = 28.3769 - 207Q = 0

0.207Q = 28.3769

Q = 137.09

P = 28.3769 - (0.1035 x 137.09) = 28.3769 - 14.1888 = 14.1881

Therefore, if P = 14.79, price has to be decreased to maximized total revenue, and if P = 14.1, price has to be increased to maximized total revenue.

NOTE: As per Answering Policy, 1st 4 parts are answered.