Transcript Unit 7.

Unit 7.
 Analyses
of LR Production and
Costs as Functions of Output
Palladium is a Car Maker’s Best
Friend?

Palladium is a precious metal used as an input in the
production of automobile catalytic converters, which
are necessary to help automakers meet
governmental, mandated environmental standards for
removing pollutants from automobile exhaust
systems. Between 1992 and 2000, palladium prices
increased from about $80 to over $750 per ounce.
One response at Ford was a managerial decision to
guard against future palladium price increases by
stockpiling the metal. Some analysts estimate that
Ford ultimately stockpiled over 2 million ounces of
palladium and, in some cases, at prices exceeding
$1,000 per ounce. Was this a good managerial
move?
This Little Piggy Wants to Eat

Assume Kent Feeds is producing swine
feed that has a minimal protein content
(%) requirement. Two alternative
sources of protein can be used and are
regarded as perfect substitutes. What
does this mean and what are the
implications for what inputs Kent Feeds
is likely to use to produce their feed?
How Big of a Plant (i.e. K) Do We
Want?

Assume a LR production process utilizing
capital (K) and labor (L) can be represented
by a production function Q = 10K1/2L1/2. If the
per unit cost of capital is $40 and the per unit
cost of L is $100, what is the cost-minimizing
combination of K and L to use to produce 40
units of output? 100 units of output? If the
firm uses 5 units of K and 3.2 units of L to
produce 40 units of output, how much above
minimum are total production costs?
Q to Produce at Each Location?

Funky Foods has two production facilities.
One in Dairyland was built 10 years ago and
the other in Boondocks was built just last
year. The newer plant is more mechanized
meaning it has higher fixed costs, but lower
variable costs (including labor). What would
be your recommendation to management of
Funky Foods regarding 1) total product to
produce and 2) the quantities to produce at
each plant?
LR  Max 

1. Produce Q where MR = MC

2. Minimize cost of producing Q

optimal input combination
Isoquant

The combinations of inputs (K, L) that
yield the producer the same level of
output.

The shape of an isoquant reflects the
ease with which a producer can
substitute among inputs while
maintaining the same level of output.
Typical Isoquant
SR Production in LR Diagram


MRTS and MP

MRTS
= marginal rate of technical substitution
= the rate at which a firm must substitute one input for
another in order to keep production at a given level
= - slope of isoquant
= K
L
= the rate at which capital can be exchanged for 1 more
(or less) unit of labor
 MPK = the marginal product of K =
Q
K

MPL = the marginal product of L =
 Q = MPK K + MPL L
 Q = 0 along a given isosquant
 MPK K + MPL L = 0

 K MPL

L
MPK
Q
L
= ‘inverse’ MP ratio
Indifference Curve & Isoquant
Slopes
Indiff Curve
Isosquant
- slope = MRS
= rate at which consumer is willing
to exch Y for 1X in order to
hold U constant
= inverse MU ratio
= MUX/MUY
For given indiff curve, dU = 0
Derived from diff types of U fns:
1)
Cobb Douglas  U = XY
2)
Perfect substitutes 
U=X+Y
3)
Perfect complements  U =
min [X,Y]
- slope = MRTS
= rate at which producer is able to
exch K for 1L in order to hold
Q constant
= inverse MP ratio
= MPL/MPK
For given isoquant, dQ = 0
Derived from diff types of
production fns:
1)
Cobb Douglas  Q = LK
2)
Perfect substitutes 
Q=L+K
3)
Perfect complements  Q =
min [X,Y]
Cobb-Douglas Isoquants

Inputs are not
perfectly
substitutable
 Diminishing
marginal rate of
technical
substitution
 Most production
processes have
isoquants of this
shape
Linear Isoquants

Capital and labor are perfect substitutes
Leontief Isoquants
Capital and labor
are perfect
complements
 Capital and labor
are used in fixedproportions

Deriving Isoquant Equation


Plug desired Q of output into production function and
solve for K as a function of L.
Example #1 – Cobb Douglas isoquants
–
–
–
–
–

Desired Q = 100
Production fn: Q = 10K1/2L1/2
=> 100 = 10K1/2L1/2
=> K = 100/L (or K = 100L-1)
=> slope = -100 / L2
Exam #2 – Linear isoquants
–
–
–
–
–
Desired Q = 100
Production fn: Q = 4K + L
=> 100 = 4K + L
K = 25 - .25L
=> slope = -.25
Budget Line

= maximum combinations of 2 goods
that can be bought given one’s
income

= combinations of 2 goods whose cost
equals one’s income
Isocost Line

= maximum combinations of 2 inputs
that can be purchased given a
production ‘budget’ (cost level)

= combinations of 2 inputs that are
equal in cost
Isocost Line Equation
TC1 = rK + wL
  rK = TC1 – wL
  K = TC1  w L

r
r
Note: slope = ‘inverse’ input price ratio
K
=
L
= rate at which capital can be exchanged for
1 unit of labor, while holding costs constant.
Increasing Isocost

Changing Input Prices

Different Ways (Costs) of Producing q1

Cost Minimization (graph)

LR Cost Min (math)

 - slope of isoquant = - slope of isocost line

MPL w

MPK r

MPL
w
(r )  (r )
MPK
r

MPL (r ) 1
1
(
)  w(
)
MPK MPL
MPL

r
w

MPK MPL

MCK  MCL
Reducing LR Cost (e.g.)
w
r

MPK MPL
 MC K  MC L
 K, L
SR vs LR Production



Assume a production process:
Q
Q
K
L
R
W

=
=
=
=
=
=
10K1/2L1/2
units of output
units of capital
units of labor
rental rate for K = $40
wage rate for L = $10
Given q =

1/2
1/2
10K L
Q
K
L
TC=40K+10L
40*
2*
8*
160*
100*
5*
20*
400*
40
5
3.2
232
100
2
50
580
* LR optimum for given q
Given q = 10K1/2L1/2, w=10, r=40
 Minimum LR Cost Condition
 inverse MP ratio = inverse input P
ratio
 (MP of L)/(MP of K) = w/r
 (5K1/2L-1/2)/(5K-1/2L1/2) = 10/40
 K/L = ¼
 L = 4K

Optimal K for q = 40?






(Given L* = 4K*)
q = 40 = 10K1/2L1/2
 40 = 10 K1/2(4K)1/2
 40 = 20K
 K* = 2
 L* = 8
 min SR TC = 40K* + 10L*
= 40(2) + 10(8)
= 80 + 80 = $160
 SR
TC for q = 40? (If K = 5)
q = 40 = 10K1/2L1/2
 40 = 10 (5)1/2(L)1/2
 L = 16/5 = 3.2
 SR TC = 40K + 10L
= 40(5) + 10(3.2)
= 200 + 32 = $232
Optimal K for q = 100?
(Given L* = 4K*)






Q =
100 = 10K1/2L1/2
 100 = 10 K1/2(4K)1/2
 100 = 20K
 K* = 5
 L* = 20
 min SR TC = 40K* + 10L*
= 40(5) + 10(20)
= 200 + 200 = $400
SR TC for q = 100? (If K = 2)
Q = 100 = 10K1/2L1/2
  100 = 10 (2)1/2(L)1/2
  L = 100/2 = 50
  SR TC = 40K + 10L
= 40(2) + 10(50)
= 80 + 500 = $580

Two Different costs of q = 100

LRTC Equation Derivation
[i.e. LRTC=f(q)]

 LRTC = rk* + wL*
= r(k* as fn of q) + w(L* as fn of q)

To find K* as fn q
from equal-slopes condition L*=f(k), sub f(k) for L
into production fn and solve for k* as fn q

To find L* as fn q
from equal-slopes condition L*=f(k), sub k* as fn
of q for f(k) deriving L* as fn q
LRTC Calculation Example


Assume
q = 10K1/2L1/2, r = 40, w = 10
L* = 4K (equal-slopes condition)
K* as fn q
q
= 10K1/2(4K)1/2
= 10K1/22K1/2
q
= 20K  K *  .05q
20
LR TC = rk* + wL*
L* as fn q
= 40(.05q)+10(.2q)
 L*
= 4K*
= 2q + 2q
= 4(.05 q)
= 4q
L*
= .2q
Graph of SRTC and LRTC

Optimal* Size Plant (i.e. K)
(*=> to min TC of q1)
Step
Example
1. Solve SR_ prod. fn.
for L = f(q, K)
q  10 K
1/ 2
L1 / 2
 L1 / 2  q /(10 K
1/ 2
)
 L  q 2 / 100 K
2. Set up TC as fn of K
(given q, r, w)
q  40, r  40, w  10
 SR TC  r K  wL
 40 K  10 (40 2 / 100 K )
 40 K  160 K
_
3. Min TC w.r.t.
K
1
TC
160
 40  2  0
K
K
2
K 4
K 2
Expansion Path  LRTC



Technological Progress

Returns to Scale
 a LR production concept that looks at how the
output of a business changes when ALL inputs are
changed by the same proportion (i.e. the ‘scale’ of
the business changes)
 Let q1 = f(L,K) = initial output
q2 = f(mL, mK) = new output
m = new input level as proportion of old input
level
Types of Returns to Scale:
1) Increasing  q2 > mq1
2) Constant  q2 = mq1
3) Decreasing  q2 < mq1
 output ↑ < input ↑


Assume a firm is considering using two different plants (A and
B) with the corresponding short run TC curves given in the
diagram below.
$
TCA
TCB
Q1

Q of output
Explain:
1. Which plant should the firm build if neither plant has been
built yet?
2. How do long-run plant construction decisions made today
determine future short-run plant production costs?
3. How should the firm allocate its production to the above
plants if both plants are up and operating?
Multiplant Production Strategy

Assume:
P =
output price = 70 - .5qT
qT =
total output (= q1+q2)
q1 =
output from plant #1
q2 =
output from plant #2
 MR = 70 – (q1+q2)
TC1 = 100+1.5(q1)2  MC1 = 3q1
TC2 = 300+.5(q2)2  MC2 = q2

  PqT  TC1  TC2
  max 
d d ( PqT ) d (TC1 )


0
dq1
dq1
dq1
 MR  MC1
d d ( PqT ) dTC2


0
dq2
dq2
dq2
 MR  MC2
Multiplant  Max

(#1) MR = MC1
 (#2) MR = MC2
(#1) 70 – (q1 + q2) = 3q1
 (#2) 70 – (q1 + q2) = q2

from (#1), q2 = 70 – 4q1

Sub into (#2),
 70 – (q1 + 70 – 4q1) = 70 – 4q1
 7q1 = 70
 q1 = 10, q2 = 30
  = TR – TC1 – TC2
= (50)(40)
- [100 + 1.5(10)2]
- [300 + .5(30)2]
= 2000 – 250 – 750 = $1000
 If q1 = q2 = 20?


=
-
TR
TC1
TC2
= (50)(40)
- [100 + 1.5(20)2]
- [300 + .5(20)2]
= 2000 – 700 – 500 = $800
Multi Plant Profit Max
(alternative solution procedure)

1.
Solve for MCT as fn of qT
knowing cost min  MC1=MC2=MCT
 MC1=3q1
 q1 = 1/3 - MC1 = 1/3 MCT
MC2 = q2
q2 = MC2
= MCT
 q1+q2 = qT = 4/3 MCT
 MCT = ¾ qT

2.
Solve for profit-max qT
 MR=MCT
 70-qT = ¾ qT
 7/4 qT = 70
 q*T = 40
 MC*T = ¾ (40) = 30
Multi Plant Profit Max
(alternative solution procedure)

3.
Solve for q*1 where MC1 = MC*T
 3q1 = 30
 q*1 = 10

4.
Solve for q*2 where MC2 = MC*T
 q*2 = 30