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 Solution (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.