Defining Units

Algebra 1 Tutorial

Defining Units

Intro

Labor

Imagine doing some labor. All things being equal, you would expect to
do twice as much work in 2 hours as you would in only 1 hour. Using
much the same reasoning, you would also expect, if you had another
person helping you, to perform twice as much labor in a given 1 hour
period as the amount of work you could do by yourself in the same 1
hour period.

By using the above reasoning, humankind has devised a definition for
labor which allows us to use our understanding of counting and
grouping to make predictions (or at least estimates) concerning the
amount of time and number of people required to perform a given
quantity of work.

definition: person hour
person\cdothour;

The amount of labor performed by 1 person in a period of 1 hour is,
by definition, 1 person\cdothour of labor.

From the definition of person\cdothour and the definition for counting
multiplication, it is obvious that, if you have n people performing
labor for t hours, n t person\cdothours of labor will be
performed.

Suppose it has been found that 20 carpenters can build a shopping
plaza in 56 days. From this information, estimate how long it will
take 70 carpenters to build the same shopping plaza.

From past experience, we know that it takes

(1)   \begin{eqnarray*}     20 \times 56 &=& 2 \times (50 + 6) \times 10 \nonumber\\     20 \times 56 &=& (100 + 12) \times 10 \nonumber\\     20 \times 56 &=& 1120 \nonumber   \end{eqnarray*}

person\cdotdays of labor to build the given structure.
Therefore, we might estimate that it takes 70 carpenters

(2)   \begin{eqnarray*}     \frac{1120}{70} &=& \frac{112 \times 10}{7 \times 10} \nonumber\\     \frac{1120}{70} &=& \frac{11^42}{7} \nonumber\\     \frac{1120}{70} &=& 16 \nonumber   \end{eqnarray*}

days to complete this job.

If it is desired that the shopping plaza in the above illustration
be completed in 10 days, then we would estimate that it will require

(3)   \begin{eqnarray*}     \frac{1120}{10} &=& 112 \nonumber   \end{eqnarray*}

carpenters on the project.

Suppose that at a factory that manufactures widgets, it requires 2.7
person\cdothours of labor to manufacture 3000 widgets. How many
people are required if it is desired that 3 \times 10^7 widgets are
produced in a total production time of 30 hours?

The amount of labor required per widget is

(4)   \begin{eqnarray*}     \frac{2.7}{3 \times 10^3} &=& 0.9 \times 10^{-3} \nonumber\\     \frac{2.7}{3 \times 10^3} &=& 9 \times 10^{-4} \nonumber   \end{eqnarray*}

9 \times 10^{-4} person\cdothours. Since we wish to manufacture
3 \times 10^{7} widgets, we require

(5)   \begin{eqnarray*}     9 \times 10^{-4} \times 3 \times 10^7 &=& 27 \times 10^3 \nonumber\\     9 \times 10^{-4} \times 3 \times 10^7 &=& 2.7 \times 10^4 \nonumber   \end{eqnarray*}

person\cdothours of labor. Since we also wish
to complete this work in a period of 30 hours, we will require

(6)   \begin{eqnarray*}     \frac{2.7 \times 10^4}{30} &=& 0.09 \times 10^4 \nonumber\\     \frac{2.7 \times 10^4}{30} &=& 900 \nonumber   \end{eqnarray*}

workers at this factory.

Re-purposing the Concept of Labor

We can make definitions similar in spirit to the definition of person\cdothour, in situations where the underlying reasoning is
the same. In this section, we will provide an illustration where it
would be advantageous to have the following definition;

Definition: ox\cdotweek

ox\cdotweek of grass;

The amount of grass that 1 ox eats in a 1 week period of time is
said to be 1 ox\cdotweek of grass.

If 12 oxen eat the grass growing on \frac{10}{3}
acres\footnote{The acre is an archaic unit of rectilinear area that
still remains in common use in English speaking countries for the
purpose of measuring land area.} in 4 weeks and 21 oxen eat the
grass growing on 10 acres in 9 weeks, how many oxen will eat up 24
acres in 18 weeks, assuming that the grass is equal on every acre,
grows uniformly, and, that the each oxen eats the same amount of
grass each week?

Let the amount of initial grass per acre be \rho, and, let r be
the amount of grass that grows on one acre in one week. If we also
let the area of the pasture be A, then the amount of grass that
grows in a period of time t may be given by r t A, and the
amount of initial grass may be given by \rho A, therefore the total
amount of grass Y may be given by;

(7)   \begin{eqnarray*}     Y = \rho A + r t A \nonumber\\     Y = (\rho + r t) A   \end{eqnarray*}

From the definition of ox\cdotweek, Equation ?? and the problem statement, we have;

(8)   \begin{eqnarray*}     (\rho + 4 r) \frac{10}{3} &=& 12 \times 4 \nonumber\\     \rho + 4 r &=& 12 \times 4 \times \frac{3}{10} \nonumber\\     \rho + 4 r &=& 48 \times 0.3 \nonumber\\     \rho + 4 r &=& 12.0 + 2.4 \nonumber\\     \rho + 4 r &=& 14.4   \end{eqnarray*}

and

(9)   \begin{eqnarray*}     (\rho + 9 r) 10 &=& 21 \times 9 \nonumber\\     \rho + 9 r &=& 21 \times 9 \times \frac{1}{10} \nonumber\\     \rho + 9 r &=& 189 \times \frac{1}{10} \nonumber\\     \rho + 9 r &=& 18.9   \end{eqnarray*}

From Equations 8 and 9, we have;

(10)   \begin{eqnarray*}     \rho + 9 r - (\rho + 4 r) &=& 18.9 - 14.4 \nonumber\\     \rho + 9 r - \rho - 4 r &=& 4.5 \nonumber\\     9 r - 4 r &=& 4.5 \nonumber\\     (9 - 4) r &=& 4.5 \nonumber\\     5 r &=& 4.5 \nonumber\\     r &=& 0.9   \end{eqnarray*}

From Equations 8 and 10, we have;

(11)   \begin{eqnarray*}     \rho + 4 \times 0.9 &=& 14.4 \nonumber\\     \rho + 3.6 &=& 14.4 \nonumber\\     \rho &=& 14.4 - 3.6 \nonumber\\     \rho &=& 10.^{-1}8   \end{eqnarray*}

From Equations 7, 10, and
11 the amount of grass on 24 acres after 18 weeks
will be;

(12)   \begin{eqnarray*}   (10.8 + 0.9 \times 18) \times 24 &=& (10.8 + 16.^{7}2) \times 24 \nonumber\\   (10.8 + 0.9 \times 18) \times 24 &=& 27.^10 \times 24 \nonumber \end{eqnarray*}

Therefore, a total of 27 \times 24 ox\cdotweeks of grass will be
found on 24 acres after 18 weeks. The number of oxen that will eat
this much grass in 18 weeks will then be given by;

(13)   \begin{eqnarray*}   \frac{27 \times 24}{18} &=& 27 \times \frac{24}{18} \nonumber\\   \frac{27 \times 24}{18} &=& 27 \times \frac{4 \times 6}{3 \times 6} \nonumber\\   \frac{27 \times 24}{18} &=& 27 \times \frac{4}{3} \nonumber\\   \frac{27 \times 24}{18} &=& 9 \times 4 \nonumber\\   \frac{27 \times 24}{18} &=& 36 \nonumber \end{eqnarray*}

Therefore, 36 oxen will eat the grass growing on 24 acres in 18 weeks.

Sample Problem

Please see the intro section.

Solution

Please see the intro section.



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