Question

Copy and paste the following data into Excel:

P	Q
$210.00	4280
$201.60	4335
$199.50	4513
$195.30	4655
$191.10	4696
$182.70	4949
$172.20	5142
$163.80	5313

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=$210.00? What is the point price elasticity of demand when P=$185.50?

d.   To maximize total revenue, what would you recommend if the company was currently charging P=$201.60? If it was charging P=$185.50?

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? (Round your price to the nearest cent, your quantity to the nearest whole unit, and your TR to the nearest dollar.) Compare that to the TR when P = $210.00 and P = $185.50.

Answer 1

(a) Based on the data, the regression result from Microsoft Excel is as follows:

Hence, the estimated equation is P = 386.140 - 0.042 Q (up to 3 decimal places).

Since the coefficients are highly significant (low p-value), confidence is high on this estimated regression. Also, R-square is very high.

From the above inverse demand function, using algbra, we can derive the demand function as follows:

0.042 Q = 386.140 - P => Q = (386.140/0.042) - (P/0.042) = 9299.983 - 24.084 P

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(b) Since Q = 9299.983 - 24.084 P, we get dQ/dP = -24.084. Using the formula for point price elasticity of demand [elasticity = (dQ/dP)(P/Q)], for each price and quantity combinations, elasticity has been calculated as follows.

(c) At price $210, Q = 9299.983 - 24.084 (210) = 4242.246 and elasticity = -24.084 * 210 / 4242.246 = -1.192

At price $185.5, Q = 9299.983 - 24.084 (185.5) = 4832.315 and elasticity = -24.084 * 185.5 / 4832.315 = -0.925

(d) At price $201.6, Q = 9299.983 - 24.084 (201.6) = 4444.555 and elasticity = -24.084 * 201.6 / 4444.555 = -1.092

Since the demand is elastic at price $201.6, revenue can be increased by decreasing the price since quantity will increase more proportionately as compared to fall in price.

However, at price $185.5, elasticity = -0.925, i.e., demand is inelastic. Hence, to increase revenue, price needs to be increased because the change in quantity (decrease) will be less proportionate than the change (increase) in price in this case.

(e) Since P = 386.140 - 0.042 Q,

TR = PQ = (P = 386.140 - 0.042 Q) Q = 386.140Q - 0.042 Q2

MR = d(TR)/dQ = 386.140 - 0.042 (2) Q = 386.140 - 0.082 Q

Using the above functions, the P and MR can be drawn as follows:

(f) At maximum revenue, MR = 0 =>386.140 = 0.082 Q => Q = 386.140 / 0.082 = 4650 (up to 0 decimal place)

and P = 386.140 - 0.042 (4650) = 193.07

Hence, the revenue maximizing price is less than $210 and more than $185.5. Therefore, it is consistent with answer in (d) that price needs to be reduced from $210 and increased from $185.5 so as to maximize profit. Because profit is maximized at price = $193.07 where MR = 0 and elasticity = 1.