Maximaliseer output bij gegeven kosten
", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "Stap 1: Formuleer het maximalisatieprobleem.
\\begin{eqnarray}
\\max_{A,K} \\phi(A,K) &=& \\var{c} \\cdot A^{\\var{alpha}}\\cdot K^{\\var{beta}}\\\\
\\mbox{als} && w_A \\cdot A + w_K \\cdot K = \\var{kostengeg} \\\\
\\end{eqnarray}
Stap 2: Bepaal de Lagrangefunctie.
\\begin{eqnarray}
L(A, K, \\lambda) &=& \\phi(A,K)+\\lambda (\\var{kostengeg} - w_A \\cdot A -w_K \\cdot K) \\\\
&=& \\var{c} \\cdot A^{\\var{alpha}}\\cdot K^{\\var{beta}} +\\lambda (\\var{kostengeg} - \\var{wA} \\cdot A - \\var{wK} \\cdot K)
\\end{eqnarray}
Stap 3: Schrijf de eerste orde voorwaarden neer om de kritische punten van deze Lagrangefunctie te berekenen.
\\begin{eqnarray}
\\mbox{(1) }\\quad \\frac{\\partial {{L}}}{\\partial A} & = & \\var{c} \\cdot \\var{alpha} \\cdot A^{\\var{alpha-1}}\\cdot K^{\\var{beta}} - \\lambda \\cdot \\var{wA}= 0\\\\
\\mbox{(2) }\\quad \\frac{\\partial {{L}}}{\\partial K} & = & \\var{c} \\cdot \\var{beta} \\cdot A^{\\var{alpha}}\\cdot K^{\\var{beta-1}} - \\lambda \\cdot \\var{wK}= 0 \\\\
\\mbox{(3) }\\quad \\frac{\\partial {{L}}}{\\partial \\lambda} & = & \\var{kostengeg} - \\var{wA} \\cdot A -\\var{wK} \\cdot K = 0
\\end{eqnarray}
Stap 4: Los dit stelsel vergelijkingen op naar $A$, $K$, en $\\lambda$.
Het oplossen van $\\lambda$ uit de vergelijkingen (1) en (2) levert na gelijkstelling:
\\begin{eqnarray}
\\frac{\\var{c} \\cdot \\var{alpha} \\cdot A^{\\var{alpha-1}}\\cdot K^{\\var{beta}}}{\\var{wA}} & = & \\frac{\\var{c} \\cdot \\var{beta} \\cdot A^{\\var{alpha}}\\cdot K^{\\var{beta-1}}}{\\var{wK}}\\\\
K & = & \\frac{\\var{wA*beta/gcd(wA*beta,wK*alpha)}}{\\var{wK*alpha/gcd(wA*beta,wK*alpha)}} \\cdot A
\\end{eqnarray}
Substitutie hiervan in (3) geeft de optimale waarde van $A$:
\\[ A^* = \\var{Aopl} ,\\]
waardoor
\\[ K^* = \\var{Kopl} .\\]
De maximale output bedraagt dan \\[ \\var{output} .\\]
\n{toonfiguuromgekeerd()}
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