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Let b= b12= b21. Then the cost function is c(w1, w2, y)=W111+w2. c2=1b11 +w2b22+2byw12ly Answer 1.6 (a) The production set is defined by ≥V+ which means that if the firm wants to produce(/, g2) it needs at least a amount of labor. Since the labor is an input, it should be negative in the definition of implicit production function. This means that we can choose 33=- and define G(,,)=V听++ with Y={(m,m,)>0,m>0,功≤0 The production process is then defined by G(y1, y2, y3)<0 for y E Y. We first have G √+ 0,G=1>0, yi t y2 thus Assumption 2. 1 is satisfied. The 2nd order conditions are yiy1y19192 Gy2 Gy2yn gg2y2 + y+2 0 Gu Gy2 gys 0 Gy Gya 1 yI Cy1y1 Uy1y2 Cy193 y1 Uy191 Cy1y2 9191y192 0. GGGG GoaL G Gya Guay gaya grays Therefore, Assumptions 2.2 is satisfi (b)The problem is 丌=maxp1h+P2y-a t.2+v2=x2 6Let b ≡ b12 = b21. Then the cost function is c(w1, w2, y) = w1x1 + w2x2 = [w1b11 + w2b22 + 2b √w1w2]y. Answer 1.6. (a) The production set is defined by x ≥ t y2 1 + y2 2 which means that if the firm wants to produce (y1, y2) it needs at least x amount of labor. Since the labor x is an input, it should be negative in the definition of implicit production function. This means that we can choose y3 = −x and define G(y1, y2, y3) ≡ t y2 1 + y2 2+y3 with Y = {(y1, y2, y3) | y1 > 0, y2 > 0, y3 ≤ 0}. The production process is then defined by G(y1, y2, y3) ≤ 0 for y ∈ Y. We first have Gy1 = s y1 y2 1 + y2 2 > 0, Gy2 = s y2 y2 1 + y2 2 > 0, Gy3 = 1 > 0, thus Assumption 2.1 is satisfied. The 2nd order conditions are            0 Gy1 Gy2 Gy1 Gy1y1 Gy1y2 Gy2 Gy2y1 Gy2y2            =             0 √ y1 y2 1+y2 2 √ y2 y2 1+y2 2 √ y1 y2 1+y2 2 y2 2 (y2 1+y2 2) 3/2 − y1y2 (y2 1+y2 2) 3/2 √ y2 y2 1+y2 2 − y1y2 (y2 1+y2 2) 3/2 y2 1 (y2 1+y2 2) 3/2             = − 1 sy2 1 + y2 2 < 0 and                0 Gy1 Gy2 Gy3 Gy1 Gy1y1 Gy1y2 Gy1y3 Gy2 Gy2y1 Gy2y2 Gy2y3 Gy3 Gy3y1 Gy3y2 Gy3y3                =                0 Gy1 Gy2 1 Gy1 Gy1y1 Gy1y2 0 Gy2 Gy2y1 Gy2y2 0 1 0 00                = −        Gy1y1 Gy1y2 Gy2y1 Gy2y2        = 0. Therefore, Assumptions 2.2 is satisfied. (b) The problem is π = max p1y1 + p2y2 − wx¯ s.t. y2 1 + y2 2 = ¯x2 6
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