Consumption一 budgets ome Examples Different preferences result in different optimal choices. Here are some examples. ex preferences. The tangency point is not the optimum The second is the case of perfect complements. The consumer purchases an equal amount of both goods. The third is the case of perfect substitutes. The consumer spends all income on the cheaper of the two goods The dual The problem the consumer faces can be writ max(r1, I2) s.t. P11+p2I2<m The solution is a demand function telling the consumer how much of each good to purchase given prices and income, for example I1=I1(P1, P2, m)and 12= I2(P1, P2, m). The problem could be solved from an alternative perspective. Suppose the consumer wanted to know what the mum expenditure was to achieve a particular utility level t, then: PiE1+P2I2 s.t. u(r1I2)>T This problem is called the Dual Both are solved by the Lagrangian technique. The mathematics lectures will cower this method. The diagram below illustrates the fact that both problems generate the same solution zConsumption — Budgets 11 Some Examples • Different preferences result in different optimal choices. Here are some examples. . .. ... ... ... .... ...... ....... ....... ........ ........ ......... .......... ............ ............... ....... . ... ... .... ...... ........ ......... .......... ............. ..................... ... ................................................................................................................................................................................................................................................................................ .................................................................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................................................. . . . ........................................................................................................................................ ............................................................................................................. . . . . ........ ........ ..... . ........ ........ ..... ..................................................... . . . . . .. . . . . . . . . . . . . . . . . . .. .. .. .. .......... x2 x1 x2 x1 x2 x 0 x1 ∗ 2 = 0 x ∗ 1 x ∗ 2 x ∗ 1 x ∗ 2 = 0 x ∗ 1 Non-convex preferences Perfect complements Perfect substitutes . . . . . ............. ............. ............. ............. ......... • The first diagram illustrates an example of non-convex preferences. The tangency point is not the optimum. • The second is the case of perfect complements. The consumer purchases an equal amount of both goods. • The third is the case of perfect substitutes. The consumer spends all income on the cheaper of the two goods. Consumption — Budgets 12 The Dual • The problem the consumer faces can be written mathematically as: max u(x1, x2) s.t. p1x1 + p2x2 ≤ m • The solution is a demand function telling the consumer how much of each good to purchase given prices and income, for example x1 = x1(p1, p2, m) and x2 = x2(p1, p2, m). • The problem could be solved from an alternative perspective. Suppose the consumer wanted to know what the minimum expenditure was to achieve a particular utility level u, then: min p1x1 + p2x2 s.t. u(x1, x2) ≥ u • This problem is called the Dual. Both are solved by the Lagrangian technique. The mathematics lectures will cover this method. The diagram below illustrates the fact that both problems generate the same solution x ∗ . ............. ............. ............. ............. ......... . . . . . ............. ............. ............. ............. ......... . . . . . ................................................................................................................................................................................................................................................................................ .................................................................................................................................................................................................................................................................................. . . . 0 0 x2 u = u x1 x2 x1 x ∗ 2 x ∗ 1 x ∗ 2 x ∗ 1 . ........ ........ ..... . ..................... . ................................................................................................ . ................................................................................................ . ............................................................................ .