Supplement%20to%20Handout%205 - Supplementary Notes...

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Unformatted text preview: Supplementary Notes lt-sziAf'fW A general solution for the maximization problem of two inputs—one output CeD production. Suppose the production function is given by q: LQKfi where a > 0,13 > O and a+fi <1 Step 1: Set up the maximization problem Ti' =pL‘1Kfi — wL —— T'K Step 2: Find FOC g:— * paL““1K’9—w:0 8 i = pfiLflKfid—wto Step 3: Find the relation between L and K posLQ‘lKfi w pfiLaKfi*1 _ Then, It follows that Step 4: Plug it into one of equations in the FOC J6 paLafi‘fifil = w T G Then, Denote L“, K * and q* as the profit maximizing amount of labor, capital and corn, respectively. Let 6:1u(a+fi)>0. Hence, I; 2 p1/9a(1‘5l/6fi5/waill/0T—3/9 Plug E“ into (1) to get K“. K* = pl/Baa/‘Bfifl—a)/6wia/9r(a—1)/6 and (1* = (ma WW Notice that these are functions of p, w and 7'. You can check how L*, K* and q* change with respect to p, w and r by taking the partial derivative of each function with respect to each parameter. 815* 1 2 _ (um—1 (1—m/a fi/fi (rs—1W 43/9 6p 6p a B m r >0 3-“ _ (fig—1) (1/0) (hm/a 6/6 «mum—1 we 5w _ 6 p a 6 w T <0 an ar : _gPU/m—1a(143)Ngfl/wa—IVSTEfi/m—1 < 0 Similarly, the partial derivatives of K* with respect to p, w and 'r can be derived. Then, for (1*, you can use the chain rule to see the Sign. a * a; 315* a , m 6‘; = am) 1 6p (K*)5+B(L*) (W)fl 166p 6 * 815* K 79: ~ a(L*)“‘15w(K*)’3+fi(L*)“(K*)fi'163w 851* _ >5: owl 615* at ,6 * C! * {3’1 a? a(L) 6? (K) +£(L) (K) 81" ...
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Supplement%20to%20Handout%205 - Supplementary Notes...

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