# Essay about Solution Jehle Reny

Thomas Herzfeld September 2010 Contents

1 Mathematical Appendix 1.1 Chapter A1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Chapter A2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Consumer Theory 2.1 Preferences and Utility . . . . . . 2.2 The Consumer’s Problem . . . . . 2.3 Indirect Utility and Expenditure . 2.4 Properties of Consumer Demand 2.5 Equilibrium and Welfare . . . . . 3 Producer Theory 3.1 Production . . . . . 3.2 Cost . . . . . . . . . 3.3 Duality in production 3.4 The competitive ﬁrm 2 2 6 12 12 14 16 18 20 23 23 26 28 30

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1 Mathematical Appendix For any concave function, 2 f (x) ≤ 0, the case the two functions are convex: 2 h(x) ≥ 0. g(x) ≤ 0, it should hold 2 h(x) ≤ 0. In 2 f (x) ≥ 0 and 2 g(x) ≥ 0, it should hold

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A1.48 Let f (x1 , x2 ) = −(x1 − 5)2 − (x2 − 5)2 . Prove that f is quasiconcave. Answer Proof: f is concave iﬀ H(x) is negative semideﬁnite and it is strictly concave if the Hessian is negative deﬁnite. H= −2 0 0 −2

2 2 zT H(x)z = −2z1 − 2z2 < 0, for z = (z1 , z2 ) = 0

Alternatively, we can check the leading principal minors of H: H1 (x) = −2 < 0 and H2 (x) = 4 > 0. The determinants of the Hessian alternate in sign beginning with a negative value. Therefore, the function is even strictly concave. Since f is concave, it is also quasiconcave. A1.49 Answer each