Difference between revisions of "022 Sample Final A, Problem 8"

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::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
\frac{\partial f}{\partial l}(k,l) & = & {\displaystyle 200k^{0.6}\left(0.4l^{\,0.4-1}\right)}\\
+
\displaystyle{\frac{\partial f}{\partial l}}(k,l) & = & {\displaystyle 200k^{0.6}\left(0.4l^{\,0.4-1}\right)}\\
 
  \\
 
  \\
 
& = & 200k^{0.6}\left(\frac{2}{5}l^{-0.6}\right)\\
 
& = & 200k^{0.6}\left(\frac{2}{5}l^{-0.6}\right)\\
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::<math>\begin{array}{rcl}
 
::<math>\begin{array}{rcl}
\frac{\partial f}{\partial k}(k,l) & = & {\displaystyle 200\left(0.6k^{0.6-1}\right)l^{0.4}}\\
+
\displaystyle{\frac{\partial f}{\partial k}}(k,l) & = & {\displaystyle 200\left(0.6k^{0.6-1}\right)l^{0.4}}\\
 
\\
 
\\
 
  & = & 200\left(\frac{3}{5}k^{-0.4}\right) l^{\,0.4}\\
 
  & = & 200\left(\frac{3}{5}k^{-0.4}\right) l^{\,0.4}\\

Latest revision as of 17:09, 6 June 2015

Find ther marginal productivity of labor and marginal productivity of capital for the following Cobb-Douglas production function:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(k, l) = 200k^{\,0.6}l^{\,0.4}.}

(Note: You must simplify so your solution does not contain negative exponents.)

Foundations:  
The word 'marginal' should make you immediately think of a derivative. In this case, the marginal is just the partial derivative with respect to a particular variable.
The teacher has also added the additional restriction that you should not leave your answer with negative exponents.

 Solution:

Marginal productivity of labor:  
We take the partial derivative with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle l} :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial f}{\partial l}}(k,l) & = & {\displaystyle 200k^{0.6}\left(0.4l^{\,0.4-1}\right)}\\ \\ & = & 200k^{0.6}\left(\frac{2}{5}l^{-0.6}\right)\\ \\ & = & {\displaystyle \frac{80k^{0.6}}{l^{\,0.6}}.} \end{array}}


Marginal productivity of capital:    
Now, we take the partial derivative with respect to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} :

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{\partial f}{\partial k}}(k,l) & = & {\displaystyle 200\left(0.6k^{0.6-1}\right)l^{0.4}}\\ \\ & = & 200\left(\frac{3}{5}k^{-0.4}\right) l^{\,0.4}\\ \\ & = & {\displaystyle \frac{120l^{\,0.4}}{k^{0.4}}.} \end{array}}


Final Answer:  
Marginal productivity of labor:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial f}{\partial l}(k,l)\,=\,\displaystyle{\frac{80k^{0.6}}{l^{\,0.6}}}.}
Marginal productivity of capital:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial f}{\partial k}(k,l)\,=\,\displaystyle{\frac{120l^{\,0.4}}{k^{0.4}}}.}

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