# Lecture 8: Isoquant and Cost Analysis

Firms minimize costs and ﻿$RTS$﻿ are important. Firms seek ﻿$LSIC$﻿ - lowest sufficient ISO-cost (cheapest). The HAIC of firms ﻿$\rightarrow$﻿ enough to produce ﻿$y^*$﻿. ﻿$A: \$﻿ ﻿$\mu RTS > \frac{\mu P_1}{\mu P_2}$﻿ so better at ﻿$c$﻿

﻿$B: \ \mu RTS < \frac{\mu P_1}{\mu P_2}$﻿ so better at ﻿$c$﻿

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

### ﻿$LR$﻿ & ﻿$SR$﻿ Technique

What happens if ﻿$q=2 \ \uparrow$﻿ to ﻿$q=4$﻿? ﻿$SR(4)>C(4) \rightarrow$﻿ Short run costs

﻿$\uparrow$﻿ than long run costs

With Numbers:

A: suff ﻿$K^{1/4}L^{1/4}=2=KL=16$﻿

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

﻿$4L^2 = 2^4 \rightarrow 2L = 2^2 \Rightarrow L=2 \ \ K = 8 \ \ c = 4$﻿ B: ﻿$SR$﻿ suff ﻿$KL = 4^4$﻿

﻿$\bar{K} = 8 \ \ 8L=4^4$﻿

﻿$\underset {SR}{L} = 32$﻿ ﻿$K=8$﻿ ﻿$c=34$﻿ C: ﻿$LR$﻿ suff ﻿$\left.\begin{matrix} LK=4^4\\ K=4L \end{matrix}\right\}$﻿ ﻿$4L^2 = 4^4$﻿

﻿$L=8$﻿ ﻿$h=32$﻿ ﻿$c=16$﻿

First concept of cost curve analysis: Isocost comes from Isoquants

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

﻿$SAC$﻿ ﻿$TAN$﻿ ﻿$LAC$﻿﻿$,$﻿ ﻿$S \mu C = L$﻿﻿$\mu C$﻿ at ﻿$y^*$﻿

﻿$K=8$﻿ ﻿$SR$﻿ ﻿$q=2$﻿ - Derive ﻿$LR$﻿ costs

T: ﻿$K=4L$﻿ S: ﻿$KL = q^4 \rightarrow 4L^2=q^4$﻿ ﻿$L = \frac{1}{2} q^2$﻿ ﻿$K = 2q^2$﻿

﻿$c=wL + rK$﻿

﻿$c = q^2$﻿ ﻿$Ac = q$﻿ ﻿$\mu C = 2q$﻿

Derive ﻿$SR$﻿ Cost: ﻿$\bar{K}=8$﻿ ﻿$S: 8L = q^4 \rightarrow L = \frac{1}{8} q^4$﻿

﻿$SvC= \frac{1}{8} ​ q ^ 4$﻿ ﻿$SfC = 2$﻿ ﻿$SC = \frac{1}{8} q ^4 +2$﻿ ﻿$S \mu C = \frac {1}{2} q^3$﻿

﻿$Ac$﻿'s tangent & ﻿$\mu c$﻿'s equal ### Perfect Subs:Long & Short Runs

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

﻿$TSR < 2$﻿ so no labour

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

﻿$c= wL + rK = 50$﻿

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

SC: ﻿$4(20)+2(25) = 130$﻿

For ﻿$q>100$﻿

SR ﻿$k=25$﻿ ﻿$q= 4(25) +5L=100+5L$﻿

﻿$L = \frac {1}{5} (q-100)$﻿

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

LR ﻿$q=4k$﻿ ﻿$k = \frac{q}{4}$﻿ ﻿$c= 2(\frac{q}{4}) - 0.5q$﻿

At ﻿$q=100$﻿ ﻿$k$﻿ is optimal at ﻿$SC = C=50$﻿ ﻿$q>100$﻿ ﻿$SC>C$﻿

### Changes in Input Prices

If ﻿$w \uparrow$﻿

SE: o costL ﻿$\uparrow$﻿ ﻿$L \downarrow$﻿ ﻿$K \uparrow$﻿

OE: o cost ﻿$q \downarrow$﻿ ﻿$L \downarrow$﻿ ﻿$K \downarrow$﻿ ﻿$\Rightarrow$﻿ ﻿$\downarrow$﻿ if OE>SE

TE: ﻿$L \downarrow$﻿ ﻿$K$﻿? ﻿$\uparrow$﻿ if SE>OE If ﻿$r \uparrow$﻿

SE: o costL ﻿$\uparrow$﻿ ﻿$L \uparrow$﻿ ﻿$L \downarrow$﻿ ﻿$K \downarrow$﻿

OE: o cost ﻿$q \downarrow$﻿ ﻿$L \downarrow$﻿ ﻿$K \downarrow$﻿ ﻿$\Rightarrow$﻿ ﻿$\downarrow$﻿ if OE>SE

TE: ﻿$L \downarrow$﻿ ﻿$K$﻿? ﻿$\uparrow$﻿ if SE>OE

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

### SE & OE

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

﻿$q=L^{1/2} + k^{1/2}=2L^{1/2}$﻿ ﻿$L=K= \frac{q^2}{4}$﻿

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

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

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

﻿$12= K^{1/2} +L^{1/2}= (4L)^{1/2} + L^{1/2}﻿$﻿

﻿$=2L^{1/2} =16$﻿

﻿$K=64$﻿ B﻿$(16, \ 64)$﻿ ﻿$C_B = 16(2)+64=96$﻿ C: ﻿$K=4L$﻿ ﻿$q= (4L)^{1/2} + L^{1/2} = 3L ^ {1/2}$﻿

﻿$L = \frac{q^2}{q}$﻿ ﻿$k = \frac{4 q^2}{q}$﻿ ﻿$c = \frac{2 q^2}{q}+ \frac{4 q^2}{q}= \frac{6}{9} q^2$﻿

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

﻿$A \rightarrow B$﻿: SE ﻿$B \rightarrow C$﻿: OE Here SE﻿$=$﻿ OE

### Multi Firm Plant: RTS Matter

If ﻿$\mu_{CM} >\mu_{CB}$﻿ ﻿$q_B \uparrow$﻿ ﻿$q_M \downarrow$﻿ why?

﻿$\Rightarrow$﻿ Lower cost without output sacrifice

If ﻿$﻿\mu_{CM} <\mu_{CB}$﻿ ﻿$q_B \downarrow$﻿ ﻿$q_M \uparrow$﻿

Only when ﻿$\mu C_m = \mu C_B$﻿ ﻿$q_B$﻿ & ﻿$q_m$﻿ are optimal

﻿$q_s = 2k^{1/2}L^{1/2}$﻿ ﻿$\left.\begin{matrix} S: \ 2^4KL=q^4\\ T: \ K=L \end{matrix}\right\}$﻿﻿$\begin{matrix} 2^4L^2=q^4 & L=\frac{1}{4}q^2 & c=8q^2 \\ 2^2L=q^2 & k=\frac{1}{2}q^2 & \mu_C =16q \end{matrix}$﻿

﻿$q_m = 4k^{1/4}L^{1/4}$﻿ ﻿$\left.\begin{matrix} S: \ 4^4KL=q^4\\ T: \ L=K \end{matrix}\right\}$﻿﻿$\begin{matrix} 4^4L^2=q^4 & L=\frac{1}{16}q^2 & c= \frac{1}{16}q^2 \\ 4^2L=q^2 & c=16L+16K=2q^2 & m_C =\mu q \end{matrix}$﻿

Want ﻿$q=200$﻿ Want ﻿$\mu_{Cm} = \mu_{CS}$﻿

﻿$q_m + q_s =200$﻿ ﻿$16q_s = 4q_m \rightarrow 4q_s = q_m$﻿

﻿$4q_s +q_s =200$﻿ ﻿$q_s = 40$﻿ ﻿$q=160$﻿

### Other Examples: RTS Affects AC & MC

If CRTS AC ﻿$\rightarrow$﻿ horizontal if ﻿$2 \times q$﻿, need ﻿$2 \times k$﻿ ﻿$8 \times 2 \times L \ \therefore C \times 2$﻿

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

If DRTS and ﻿$2 \times q$﻿, need ﻿$3 \times k$﻿ & ﻿$3 \times L$﻿ ﻿$\therefore 3 \times C$﻿

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

### IRTS AC & ﻿$\mu C$﻿ Are Downward Slopping

#### Case 1: If both DRTS #### Case 2: CRTS1, DRTS2 Concepts ﻿$\mu C$﻿'s equal #### Case 3L IRTS? 