A certain factory, the daily output is Q(K,L)=60K^{1/2}L^{1/3} units, where K denotes the capital...

Question:

A certain factory, the daily output is {eq}Q(K,L)=60K^{1/2}L^{1/3} {/eq} units, where {eq}K {/eq} denotes the capital investment measured in units of {eq}$1,000 {/eq} and {eq}L {/eq} the size of the labor force measured in worker-hours. The current capital investment is {eq}\$10,000 {/eq} and {eq}1,000 {/eq} worker hours of labor are used each day.

a.Show that {eq}Q(3K,3L)=3Q(K,L). {/eq}

b.Estimate the change in output that will result if the capital increases by {eq}\$1,000 {/eq} and labor increases by {eq}2 {/eq} worker-hours.

c. Determine if the function has a relative minimum or a relative maximum.

Radical Function Modeling Daily Output of Factory


A radical function (function with square roots and cube roots) is used to model the daily output of a factory as a function of capital investment and labor hours. Using linear differentials we estimate the change in the daily output when there is a change in the capital investment and labor hours. The concepts used here include partial derivatives from Calculus.


Answer and Explanation:


a) Given {eq}Q(K,L)=60K^{1/2}L^{1/3}, {/eq} then,

{eq}Q(3K,3L)=60 (3K)^{1/2} (3L)^{1/3} = 60 (3^{1/2})( 3^{1/3}) K^{1/2} L^{1/3}=60 (3^{5/6}) K^{1/2}L^{1/3} \ne 3Q(K,L). {/eq}

Hence, NO, the statement {eq}Q(3K,3L)= 3Q(K,L) {/eq} is FALSE.


b) Let {eq}\Delta Q, \; \Delta K, \; \Delta L {/eq} denote the changes in Q, K and L, respectively. Then we have, using linear differentials, that

{eq}\displaystyle \Delta Q = 60 \left( \frac {K^{1/2} L^{-2/3} \Delta L}{3}+ \frac {K^{-1/2} L^{1/3} \Delta K}{2} \right) \qquad (1) {/eq}

From the question,

{eq}K=10, \; L=1000, \; \Delta K=1, \; \Delta L=2 \qquad (2) {/eq}

Using (2) in (1) gives us

{eq}\displaystyle \Delta Q = 60 \left( \frac {10^{1/2} 1000^{-2/3} (2)}{3}+ \frac {10^{-1/2} 1000^{1/3} (1)}{2} \right) = 96.13. {/eq}

The change in output is 96.13 units.


c. Since the partial derivatives are

{eq}\displaystyle Q_K=\frac {30L^{1/3}}{K^{1/2}} \; and \; Q_L=\frac {20K^{1/2}}{L^{2/3}} {/eq} which when set equal to zero simultaneously provide us no critical points owing to division by zero, the function has no relative minimum or maximum.


Learn more about this topic:

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What is a Radical Function? - Definition, Equations & Graphs

from Math 105: Precalculus Algebra

Chapter 9 / Lesson 6
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