PROBLEM I. A nightclub manager realizes that demand for drinks is more elastic a

Question

PROBLEM I. A nightclub manager realizes that demand for drinks is more elastic among students and tries to determine the optimal pricing schedule. Specifically, he estimates that the demand functions are given by g.-30-6P1 for students and = 24-4p for non-students. Assume that drinks cost the nightclub S2 each. Q1. If the market cannot be segmented, what is the uniform monopoly price? (a) $3.10 (b) $3.30 (c) S3.50 (d) S3.70 (e) S4.20 Q2. If the nightclub can charge according to whether or not the customer is a student but is limited to linear pricing, what price (per drink) should be set for students? (a) $3.10 (b) S3.25 (c) $3.40 (d) S3.50 (e) $3.80 Q3. Under the same conditions of Q2, what price (per drink) should be set for non-students? (a) s3.80 (b) $4.00 (c) S4.30 (d) $4.60 (e) $5.20

Explanation / Answer

(Q1) (d)

Without segmentation, p1 = p2 = p

q1 = 30 - 6p

q2 = 24 - 4p

Market demand: q = q1 + q2 = 30 - 6p + 24 - 4p

q = 54 - 10p

10p = 54 - q

p = 5.4 - 0.1q

Profit is maximized when Marginal revenue (MR) equals Marginal cost (MC).

Total revenue (TR) = p x q = 5.4q - 0.1q2

MR = dTR/dq = 5.4 - 0.2q

Equating with MC,

5.4 - 0.2q = 2

0.2q = 3.4

q = 17

p = 5. - (0.1 x 17) = 5.4 - 1.7 = $3.7

(Q2) (d)

With segmentation, profit is maximized when MR1 = MC and MR2 = MC

For students,

q1 = 30 - 6p1

6p1 = 30 - q1

p1 = (30 - q1) / 6

TR1 = p1 x q1 = (30q1 - q12) / 6

MR1 = dTR1/dq1 = (30 - 2q1) / 6

Equating with MC,

(30 - 2q1) / 6 = 2

30 - 2q1 = 12

2q1 = 18

q1 = 9

p1 = (30 - 9) / 6 = 21 / 6 = $3.5

(Q3) (b)

For non-students, q2 = 24 - 4p2

4p2 = 24 - q2

p2 = 6 - 0.25q2

TR2 = p2 x q2 = 6q2 - 0.25q22

MR2 = dTR2/dq2 = 6 - 0.5q2

Equating with MC,

6 - 0.5q2 = 2

0.5q2 = 4

q2 = 8

p2 = 6 - (0.25 x 8) = 6 - 2 = $4

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