6.3 - MATH
1081
 Monday,
March
14
 
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Unformatted text preview: MATH
1081
 Monday,
March
14
 
 Chapter
6
–
Section
3
 
 FURTHER
BUSINESS
APPLICATIONS
 
 Homework
#9
(due
3/28):
 
 Section
6.3
#10,
18,
22
 
 
 
 
 
 
 
 
 
 Section
9.1
#28,
46,
48
 
 Clicker
Check­in:

Choose
any
letter
to
check
in
now.
 
 Suppose
that
you
are
a
manufacturer.

Each
time
you
run
a
 batch
through
production,
there
are
expenses
of
set
up
in
the
 factory.

So,
to
minimize
those
costs,
you
will
want
a
few
 production
runs
as
possible.
 
 On
the
other
hand,
a
large
production
run
introduces
the
 expense
of
storing
the
product
until
the
consumer’s
purchase
 it.

So,
to
minimize
storage
costs,
you
want
as
small
a
batch
of
 product
at
a
time
as
possible.
 
 Where
is
the
point
that
minimizes
total
cost
overall?
 
 This
is
called
the
economic
lot
size. To
make
this
more
concrete,
suppose
a
company
manufactures
 12,000
units
of
a
product
each
year
(for
which
the
demand
is
 uniform
throughout
the
year)
and
that
the
product
can
be
 manufactured
in
several
small
batches
(of
equal
size)
 throughout
the
year.
 
 The
company
could,
for
example,
schedule
one
production
run
 at
the
start
of
the
year
making
all
12,000
units;
or
they
could
 schedule
4
production
runs
(once
every
3
months)
making
 3000
in
each
batch;
or
they
could
schedule
12
production
runs
 (one
at
the
start
of
each
month)
making
just
1000
in
each
 batch.
 
 
 
 
 
 Assuming
uniform
demand
 through
the
year,
the
 inventory
in
the
warehouse
 would
look
like
one
of
the
 following:
 
 So,
the
average
in
storage
is
 half
the
batch
size. 
 Let’s
say
the
number
in
each
batch
is
 q .

This
is
the
variable
in
 our
problem,
as
we
can
decide
how
many
per
batch
(and
 therefore
how
many
batches
in
the
year)
we
make.
 
 Then
in
our
current
scenario,

 q we
have
 
units
in
storage
throughout
the
year
and,
 2 12000 we
will
have
a
total
of
 
production
runs.
 q 
 Suppose
it
costs
$2
to
store
a
unit
of
the
product
for
a
year,
 $500
to
set
up
the
factory
for
each
production
run,
and
$5
per
 unit
to
produce.
 
 This
means
that
the
total
cost
for
the
year
will
be
given
by
the
 function
 
 ⎛ 12000 ⎞ ⎛ q⎞ 
 
 
 
 C (q ) = 2 ⎜ ⎟ + 500 ⎜ ⎟ + 5 (12000 ) .
 ⎝ 2⎠ ⎝ q ⎠ product storage set up 
 This
is
the
function
that
we
must
minimize.

(What
makes
 sense
for
the
domain
of
this
function?)

What
is
the
value
of
 q 
 that
does
the
trick?
 If
we
go
back
and
look
at
this
problem
in
general,
we
can
 derive
the
general
formula
(and
avoid
the
calculus
every
time).
 
 Namely,
if
we
let
 
 q = number of units in each batch 
 
 k = cost of storing one unit for a year 
 f = fixed setup costs for a production run 
 
 
 g = cost of production of each unit 
 
 M = total annual demand for the product 
 
 we
get
the
following
cost
function
 ⎛M⎞ ⎛ q⎞ 
 
 C (q ) = k ⎜ ⎟ + f ⎜ ⎟ + g ( M ) 
 ⎝ 2⎠ ⎝ q⎠ product storage set up Keeping
in
mind
that
the
only
actual
variable
in
that
problem
is
 q ,
we
find
that
there
is
one
critical
number
found
at
 
 2 fM 
 
 
 
 
 
 q= .
 k 
 Why
isn’t
the
cost
per
unit,
 g ,
part
of
this
formula?
 
 
 As
we
did
in
our
specific
example,
this
can
be
shown
to
be
a
 minimum,
and
since
it
is
the
only
critical
number,
that
is
where
 the
absolute
min
occurs.
 
 A
very
similar
problem
is
one
regarding
the
economic
order
 quantity.

In
this
case,
we
are
ordering
and
storing
product
to
 sell.

Each
time
we
order,
there
are
fees
for
ordering
(in
 addition
to
the
cost
per
unit
ordered)
and
there
are
fees
to
 store
any
product
currently
on
inventory.
 
 How
many/how
often
should
we
order
to
minimize
costs,
 2 fM assuming
that
we
know
the
annual
demand?


q = 
 k 

 q = number of units in each batch k = cost of storing one unit for a year 
 f = fixed setup costs to place an order M = total annual demand for the product In
this
section,
we
also
consider
another
application
that
is
 actually
not
one
of
optimization.

This
application,
called
 elasticity
of
demand,
is
a
measure
of
how
much
a
change
in
the
 price
of
the
product
affects
the
demand.
 
 
 If
the
price
goes
up,
generally
the
demand
will
drop.

However,
 depending
on
what
the
item
is,
the
sensitivity
of
demand
 varies.
 
 To
quantify
this
notion
of
elasticity,
we
can
compare
the
 percent
change
in
price
to
the
percent
change
in
demand.
 
 If
a
“small”
percent
change
(increase)
in
price
results
in
a
 “large”
percent
change
(drop)
in
demand,
then
raising
the
price
 will
reduce
revenue.

This
is
probably
something
we
want
to
 avoid.
 
 
 
 If
 E < 1,
then
the
relative
change
in
demand
is
less
than
the
 relative
change
in
price.

So,
total
revenue
increases
as
price
 increases.
 
 If
 E > 1,
then
the
relative
change
in
demand
is
greater
than
the
 relative
change
in
price
and
total
revenue
decreases
as
price
 increases.
 
 What
happens
when
 E = 1?
 
 
 This
is
called
unit
elasticity
and
indicates
that
the
relative
 change
in
demand
is
equal
to
the
relative
change
in
price
at
 that
point.


This
is
the
point
at
which
total
revenue
is
 maximized.

(See
page
393
for
further
detail.)
 
 Example:

Suppose
for
a
certain
commodity,
 q = 48, 000 − 10 p 2 .


 Find
 E 
and
determine
at
which
price
(if
any)
 revenue
is
maximized.
 
 Clicker
Check­out:

Choose
any
letter
to
check
out
now.
 
 
 Next
time
–
Section
9.1
 
 ...
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This note was uploaded on 05/01/2011 for the course MATH 1081 taught by Professor Johanson during the Spring '08 term at Colorado.

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