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# Midterm 1 Answers Continued - x 2 and less x 1 Technical...

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7. Answer. We are given: U ( x 1 ,x 2 ) = min( 1 3 x 1 ,x 2 ). These are Leontief preferences/perfect complements, so 1 3 x 1 = x 2 . Substitute into the budget constraint. p 1 x 1 + p 2 x 2 = M 3 p 1 x 2 + p 2 x 2 = M x 2 (3 p 1 + p 2 ) = M x * 2 = M (3 p 1 + p 2 ) We know that 1 3 x 1 = x 2 , so: x * 1 = 3 x * 2 = 3 M (3 p 1 + p 2 ) Alternatively, one could have substituted x 2 for x 1 in the beginning step, and arrive at the same answers. Common Errors Errors in basic algebra. (1-2 pts deducted, depending on severity of error.) Took Lagrangian of the min function. (No credit.) Rearranged budget constraint. (No credit.) Demand function was 1 3 x * 1 = x * 2 . (1 point credit, for realizing Leontief preferences/perfect complements.) 8. (Removed from exam.) Answer. We are given: MU 1 = 2 x 1 MU 2 = 3 x 2 p 1 = 4 p 2 = 1 1

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We want to know if the bundle (6 , 2) is optimal, and if not, ﬁnd which direction to modify the bundle. Compare the MU p ratios to determine which good gives the most marginal utility per dollar. MU 1 p 1 = 2 * 6 4 = 3 MU 2 p 2 = 3 * 2 1 = 6 MU 2 p 2 > MU 1 p 1 x 2 generates more marginal utility per dollar spent on it, so we conclude that we must purchase more
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Unformatted text preview: x 2 and less x 1 . Technical aside. Many students tried to use the ﬁrst order conditions MRS = p 1 p 2 to ﬁnd the optimal bundle. What they failed to realize was that the preferences were not convex. To ﬁnd the original optimal bundle, we integrate over the marginal utilities of both goods, separately. U ( x 1 ,x 2 ) = Z MU 1 dx 1 = Z 2 x 1 dx 1 = x 2 1 + c 1 U ( x 1 ,x 2 ) = Z MU 2 dx 2 = Z 3 x 2 dx 2 = 3 2 x 2 2 + c 2 So, we conclude that: U ( x 1 ,x 2 ) = x 2 1 + 3 2 x 2 2 + c Where c , c 1 , c 2 are arbitrary constants. The indiﬀerence curves of this utility function are elliptical, centered around the origin. It is clear then that preferences are not convex, so the ﬁrst order conditions MRS = p 1 p 2 cannot be used. (Nonconvex preferences imply that points of tangencies are local minima, not maxima.) 2...
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Midterm 1 Answers Continued - x 2 and less x 1 Technical...

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