The solution are: P1. p2. 所+n Pi +p2 丌=(12+n2- Therefore the supplies are 0-() +西2 Answer 1.7 (a)5. The consumer's problem is v(p, I)=max -- st.p11+p2x2=1 Let L=---++A(I-P1-1-p2 C2). The FOC's a2 imply T1=v22 x2. Substituting this into the budget constraint will immediately 2+√P1P2 By symmetry, we also have I n1=p1+√P2 (b)Substituting the consumer's demands into the utility function will give us v(p, I)=-P1+vp1p2_ P2+VP1P2 p1+p+2√1P 匝+VP2)2 (c) Let u=v(p, e),i.e (V+√P)2 which immediately gives us the expenditure function e(,a)=(+V)2The solution are: y1 = p1x¯ sp2 1 + p2 2 , y2 = p2x¯ sp2 1 + p2 2 . π = ¯x t p2 1 + p2 2 − w  . Therefore the supplies are: (y1, y2) = ⎧ ⎪⎪⎨ ⎪⎪⎩  √ p1x¯ p2 1+p2 2 , √ p2x¯ p2 1+p2 2  if w ≤ sp2 1 + p2 2 (0, 0) if w > sp2 1 + p2 2. Answer 1.7. (a) [5]. The consumer’s problem is ⎧ ⎪⎨ ⎪⎩ v(p, I) = max − 1 x1 − 1 x2 s.t. p1x1 + p2x2 = I Let L ≡ − 1 x1 − 1 x2 + λ(I − p1x1 − p2x2). The FOC’s 1 x2 1 = λp1, 1 x2 2 = λp2 imply x1 = tp2 p1 x2. Substituting this into the budget constraint will immediately give us x∗ 2 = I p2 + √p1p2 . By symmetry, we also have x∗ 1 = I p1 + √p1p2 . (b) Substituting the consumer’s demands into the utility function will give us v(p, I) = −p1 + √p1p2 I − p2 + √p1p2 I = −p1 + p2 + 2√p1p2 I = −( √p1 + √p2)2 I . (c) Let u = v(p, e), i.e. u = −( √p1 + √p2)2 e which immediately gives us the expenditure function: e(p, u) = −( √p1 + √p2)2 u . 7