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1. (a) (i) /,", = -2 < 0, /,'; = 0 < 0, and f{\f£ - (//;)2 = 0 > 0, so / is concave.

(ii) f(x) = (x — >') + (—*") is a sum of two concave functions, hence concave.

(b) F(u) = —e~u is strictly increasing and concave (F'(u) = e~u > 0 and F"(U) = -e~u < 0). By Theorem 17.6, part (c), z = -eis concave.

3. /,", = -12, = -2, and f[\f2 - (ffo2 = 24 - (2a + 4)2 = -4a2 -16a + 8. Because /," < 0, the function is never convex. It is concave iff -4a2 - 16a + 8 > 0, that is, iff -2 - V6 < a < -2 + V6.

5. Draw your own figure and use definition [17.14] of Section 17.6 to show that / is convex.

7. (100. 300) is a stationary point for jr. Moreover, = -0.08 < 0, tt", = -0.02 < 0. and - (t^,)2 = (-0.08)(-0.02) - (0.01)2 = 0.015 > 0, so (100. 300) is a (global) maximum point for x.

9. Direct verification of second-order conditions.

( -^(lnx)û"2(ln y)b(a - 1 -lnx) — (lnx^-'any)*"1

ûb h I —(lnxr-^lny)'-1 -(lny)^2(lnxr(fc-1-lny) /

Here /,", < 0 because 0 < a < 1. Also |H| is equal to f[\f£ - (f{'2)2, or to cib

— (lnx)2"-2(Iny)2b'2-[l - (a + b) + (1 - a) lny + (1 - b)\nx + lnx lny ]

This is positive, so / is strictly concave.

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