Lecture 8: Isoquant and Cost Analysis

Firms minimize costs and RTSRTS are important. Firms seek LSICLSIC - lowest sufficient ISO-cost (cheapest). The HAIC of firms \rightarrow  enough to produce yy^*.

A: A: \  μRTS>μP1μP2\mu RTS > \frac{\mu P_1}{\mu P_2} so better at cc

B: μRTS<μP1μP2B: \ \mu RTS < \frac{\mu P_1}{\mu P_2} so better at cc

C:C:  Optimal μRTS=μP1μP2\mu RTS = \frac{\mu P_1}{\mu P_2}

LRLR & SRSR Technique

What happens if q=2 q=2 \ \uparrow to q=4q=4?

SR(4)>C(4)SR(4)>C(4) \rightarrow Short run costs

\uparrow than long run costs

With Numbers:

A: suff K1/4L1/4=2=KL=16K^{1/4}L^{1/4}=2=KL=16

TAN: TRS=wr=10.25=4TRS = \frac{w}{r}=\frac{1}{0.25}=4 \rightarrow K=4LK=4L

4L2=242L=22L=2  K=8  c=44L^2 = 2^4 \rightarrow 2L = 2^2 \Rightarrow L=2 \ \ K = 8 \ \ c = 4

B: SRSR  suff KL=44KL = 4^4

Kˉ=8  8L=44\bar{K} = 8 \ \ 8L=4^4

LSR=32\underset {SR}{L} = 32  K=8K=8 c=34c=34

C: LRLR suff LK=44K=4L}\left.\begin{matrix} LK=4^4\\ K=4L \end{matrix}\right\} 4L2=444L^2 = 4^4

L=8L=8 h=32h=32 c=16c=16

First concept of cost curve analysis: Isocost comes from Isoquants

At y, SC=C y^*, \ SC = C  because fixed cost is optimal 

SACSAC TANTAN LACLAC,, SμC=LS \mu C = L μC\mu C at yy^*

K=8K=8 SRSR q=2q=2 - Derive LRLR costs

T: K=4LK=4L S: KL=q44L2=q4KL = q^4 \rightarrow 4L^2=q^4 L=12q2L = \frac{1}{2} q^2 K=2q2K = 2q^2

c=wL+rKc=wL + rK

c=q2c = q^2  Ac=qAc = q μC=2q\mu C = 2q

Derive SRSR Cost: Kˉ=8\bar{K}=8 S:8L=q4L=18q4S: 8L = q^4 \rightarrow L = \frac{1}{8} q^4

SvC=18q4SvC= \frac{1}{8} ​ q ^ 4 SfC=2SfC = 2 SC=18q4+2SC = \frac{1}{8} q ^4 +2 SμC=12q3S \mu C = \frac {1}{2} q^3

AcAc's tangent & μc\mu c's equal

Perfect Subs:Long & Short Runs

TSR:μPvμPk=54TSR: \frac{\mu P_v}{\mu P_k} = \frac{5}{4}  o-cost: wr=2\frac{w}{r} = 2

TSR<2TSR < 2 so no labour

A: q=100=4k+5Lq=100 = 4k +5L  k=25k=25 A(0,25)A(0, 25)

c=wL+rK=50c= wL + rK = 50

B: q=200q=200  Kˉ=25\bar{K} = 25  L=20L=20 B(20,25)B(20,25)

SC: 4(20)+2(25)=1304(20)+2(25) = 130

For q>100q>100

SR k=25k=25 q=4(25)+5L=100+5Lq= 4(25) +5L=100+5L

L=15(q100)L = \frac {1}{5} (q-100)

SC =4(15(q100)+2(25))=0.8q30= 4 (\frac{1}{5}(q-100)+2(25))=0.8q -30

LR q=4kq=4k k=q4k = \frac{q}{4} c=2(q4)0.5qc= 2(\frac{q}{4}) - 0.5q

At q=100q=100 kk is optimal at SC=C=50SC = C=50 q>100q>100 SC>CSC>C

Changes in Input Prices

If ww \uparrow

SE: o costL \uparrow  LL \downarrow KK \uparrow 

OE: o cost qq \downarrow LL \downarrow KK \downarrow \Rightarrow \downarrow if OE>SE

TE: LL \downarrow KK? \uparrow if SE>OE

If rr \uparrow

SE: o costL \uparrow  LL \uparrow LL \downarrow KK \downarrow

OE: o cost qq \downarrow LL \downarrow KK \downarrow \Rightarrow \downarrow if OE>SE

TE: LL \downarrow KK? \uparrow if SE>OE

Size of SE depends on substitutability: σ%Δ(K/L)%Δ(TRS)\sigma \equiv \frac { \% \Delta (K/L)} { \% \Delta (TRS)}


A: TRS=TRS= o-cost μPLμPK=wr\frac{\mu P_L}{\mu P_K}=\frac{w}{r} (KL)1/2=1(\frac{K}{L})^{1/2}=1 K=LK=L

q=L1/2+k1/2=2L1/2q=L^{1/2} + k^{1/2}=2L^{1/2} L=K=q24L=K= \frac{q^2}{4}

c=wL+rK=q22c=wL +rK = \frac{q^2}{2} μC=q\mu C = q p=μCp = \mu C q=12q=12

C=1222=72C = \frac{12^2}{2} = 72 L=K=36L=K=36 A(36, 36)A(36, \ 36)

B: w=2w=2 (KL)1/2=2(\frac{K}{L})^{1/2} = 2 K=4LK=4L

12=K1/2+L1/2=(4L)1/2+L1/212= K^{1/2} +L^{1/2}= (4L)^{1/2} + L^{1/2} 

=2L1/2=16=2L^{1/2} =16

K=64K=64 B(16, 64)(16, \ 64) CB=16(2)+64=96C_B = 16(2)+64=96

C: K=4LK=4L  q=(4L)1/2+L1/2=3L1/2q= (4L)^{1/2} + L^{1/2} = 3L ^ {1/2}

L=q2qL = \frac{q^2}{q} k=4q2qk = \frac{4 q^2}{q} c=2q2q+4q2q=69q2c = \frac{2 q^2}{q}+ \frac{4 q^2}{q}= \frac{6}{9} q^2

μC=P\mu C = P q=9q=9 L=9L=9 K=36K=36 c=54c=54

ABA \rightarrow B: SE BCB \rightarrow C : OE Here SE== OE

Multi Firm Plant: RTS Matter

If μCM>μCB\mu_{CM} >\mu_{CB} qBq_B \uparrow qMq_M \downarrow why?

\Rightarrow  Lower cost without output sacrifice

If μCM<μCB \mu_{CM} <\mu_{CB}  qBq_B \downarrow  qMq_M \uparrow

Only when μCm=μCB\mu C_m = \mu C_B  qBq_B  & qmq_m are optimal

qs=2k1/2L1/2q_s = 2k^{1/2}L^{1/2} S: 24KL=q4T: K=L}\left.\begin{matrix} S: \ 2^4KL=q^4\\ T: \ K=L \end{matrix}\right\}24L2=q4L=14q2c=8q222L=q2k=12q2μC=16q\begin{matrix} 2^4L^2=q^4 & L=\frac{1}{4}q^2 & c=8q^2 \\ 2^2L=q^2 & k=\frac{1}{2}q^2 & \mu_C =16q \end{matrix}

qm=4k1/4L1/4q_m = 4k^{1/4}L^{1/4} S: 44KL=q4T: L=K}\left.\begin{matrix} S: \ 4^4KL=q^4\\ T: \ L=K \end{matrix}\right\}44L2=q4L=116q2c=116q242L=q2c=16L+16K=2q2mC=μq\begin{matrix} 4^4L^2=q^4 & L=\frac{1}{16}q^2 & c= \frac{1}{16}q^2 \\ 4^2L=q^2 & c=16L+16K=2q^2 & m_C =\mu q \end{matrix}

Want q=200q=200 Want μCm=μCS\mu_{Cm} = \mu_{CS} 

qm+qs=200q_m + q_s =200 16qs=4qm4qs=qm16q_s = 4q_m \rightarrow 4q_s = q_m

4qs+qs=2004q_s +q_s =200 qs=40q_s = 40 q=160q=160

Other Examples: RTS Affects AC & MC

If CRTS AC \rightarrow horizontal if 2×q2 \times q, need 2×k2 \times k 8×2×L C×28 \times 2 \times L \ \therefore C \times 2

AC =cq= \frac{c}{q} if ×2\times 2 still cq\frac{c}{q}. If AC is horizontal, so is μC\mu C

If DRTS and 2×q2 \times q , need 3×k3 \times k & 3×L3 \times L 3×C\therefore 3 \times C

So AC =Cq= \frac{C}{q}  will be 3c2q,\frac{3c}{2q}, upward slopping means μC\mu C  is above it

IRTS AC & μC\mu C Are Downward Slopping

Case 1: If both DRTS

Case 2: CRTS1, DRTS2 Concepts μC\mu C's equal

Case 3L IRTS?

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