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Desi Pizza, Ltd, provides delivery and carryout service to the city of Surrey an

ID: 1216464 • Letter: D

Question

Desi Pizza, Ltd, provides delivery and carryout service to the city of Surrey and surrounding areas. An analysis of the daily demand for its super-supremo-wombo-combo paneer-dripping pizzas reveals the following demand equation:

Qx = 2,000 - 100Px – 2.5PS + 0.01CSP + 500S

where Qx is the quantity measured by the number of pizzas per day, Px is the price ($), PS is a price index for soda pop, CSP is the college student population and S, a binary or dummy variable, equals 1 on Friday, Saturday and Sunday and zero otherwise.

Currently Px = $10, PS = 100, and CSP = 25,000

(a)    Calculate the quantity demanded of pizzas on Tuesdays and Fridays.

(b)    Calculate the price elasticity of demand for Desi pizzas on theses two days.

(c)     If Desi Pizza, Ltd is primarily interested in maximizing revenue, should they charge the same price everyday of the week? If not, what prices should they charge?

(d)    If the cost per pizza is $5.00 and Desi Pizza, Ltd., behaves as a profit-maximizing monopolist, should they charge the same price everyday of the week? If not, what prices should they charge?

Explanation / Answer

Qx = 2,000 - 100Px – 2.5PS + 0.01CSP + 500S = 2,000 - (100 x 10) - (2.5 x 100) + (0.01 x 25,000) + 500S

= 2,000 - 1,000 - 250 + 250 + 500S

Qx = 1,000 + 500S

(a)

(i) On Tuesday, S = 0

Qx = 1,000

(ii) On Friday, S = 1

Qx = 1,000 + (500 x 1) = 1,000 + 500 = 1,500

(b)

(i) On Tuesday, Price elasticity = (dQx / dPx) x (Px / Qx) = - 100 x (10 / 1,000) = - 1

(ii) On Friday, Price elasticity = (dQx / dPx) x (Px / Qx) = - 100 x (10 / 1,500) = - 0.67

(c) Qx = 2,000 - 100Px – 2.5PS + 0.01CSP + 500S

Qx = 2,000 - 100Px - (2.5 x 100) + (0.01 x 25,000) + 500S

= 2,000 - 100Px - 250 + 250 + 500S

Qx = 2,000 - 100Px + 500S

Since demand function is different on different days, uniform price should not be charged.

(i) On Friday, Saturday & Sunday: S = 1

Qx = 2,000 - 100Px + 500 = 2,500 - 100Px

100Px = 2,500 - Qx

Px = 25 - 0.01Qx

Total revenue, TR = Px. Qx = 25Qx - 0.01Qx2

TR is maximized when dTR / dQx = 0

25 - 0.02Qx = 0

0.02Qx = 25

Qx = 1,250

Px = 25 - (0.01 x 1,250) = 25 - 12.5 = 12.5

(ii) On Monday, Tuesday, Wednesday & Thursday: S = 0

Qx = 2,000 - 100Px + 0 = 2,000 - 100Px

100Px = 2,000 - Qx

Px = 20 - 0.01Qx

Total revenue, TR = Px. Qx = 20Qx - 0.01Qx2

TR is maximized when dTR / dQx = 0

20 - 0.02Qx = 0

0.02Qx = 20

Qx = 1,000

Px = 20 - (0.01 x 1,000) = 20 - 10 = 10

(d) Monopolist maximizes profit by equating MR with MC (= 5) where MR = dTR / dQx

(i) On Friday, Saturday & Sunday,

dTR / dQx = 25 - 0.02Qx

Equating with MC,

25 - 0.02Qx = 5

0.02Qx = 20

Qx = 1,000

Px = 25 - (0.01 x 1,000) = 25 - 10 = 15

(ii) On Monday, Tuesday, Wednesday & Thursday:

dTR / dQx = 20 - 0.02Qx

Equating with MC,

20 - 0.02Qx = 5

0.02Qx = 15

Qx = 750

Px = 20 - (0.01 x 750) = 20 - 7.5 = 12.5