COMM 220 Lecture 2: Practice Problem Set3_solution part 1

• School: Concordia University
• Department: Commerce
• Course: COMM 220
• Professor: Krista Riley

Document Summary

Mc(q) = dc/dq = 20 1. 6q + 0. 06q2 (b) atc is minimized when atc=mc. 20 0. 8q + 0. 02q2 = 20 1. 6q + 0. 06q2 q = 0. 8/0. 04 = 20 (c) graph of the atc and mc curves using q = 1, 5, 10, 15, 20, 25, 30, 35 and 40. q. 10 15 20 25 30 35 40: (a) k = 10, q = 50l0. 5100. 5 l0. 5 = q/(50*100. 5) l = (q/(50*100. 5))2 (i) tc(q) = 25l + 100*10 + 10q = 25(q/(50*100. 5))2 + 1000 + 10q. Ac(q) = 1000/q + 10 + 0. 001q (ii) q = 2000 = 50l0. 5100. 5. 1 (b) cost minimization occurs when mrtslk = w/r (i) mrtslk = mpl/mpk = 25l-0. 5k0. 5/25l0. 5k-0. 5 = k/l. Optimal capital-labour ratio: k/l = 25/100 = 1/4 (i. e. , k = l/4 or l = 4k) (ii) plug l=4k into q = 2000 = 50(4k)0. 5k0. 5 k = 20 and l = 4(20) = 80.