Answer to Question #197654 in Macroeconomics for Thandeka

Question

Suppose both cooperatives are price-takers; with 𝑃𝐺=12 and 𝑃𝐹 =10 being the market prices of gold and fish respectively. The cost of producing G units of gold is 𝑐𝑔(𝐺,𝑥)=𝐺2+(𝑥−4)2; where 𝑥 is the quantity of mercury effluent discharged into the dam. The cost of producing F units of fish is 𝑐𝐹(𝐹,𝑥)= 𝐹2 + 𝑥𝐹 ; where 𝑐𝐹(𝐹,𝑥) is an increasing function of 𝑥.

a) Determine the profit maximizing level of gold output, profit and level of mercury effluent discharged into the dam. Comment and explain your results.

b) What is the fishing cooperative’s profit maximizing level of fish output and profit? Explain your results.

c) Suppose the two cooperatives were merged to operate as a single firm; determine the socially optimal level of gold and fish output, mercury effluent and profit. Explain, comment on and contrast your results with those obtained in a and b above.

d) Suppose property rights to the dam water are created and assigned to the fishing cooperative. Does this induce efficiency and explain why

Answer 1

PG=12 and PF =10

Cost of producing G units of gold is$= cg(G,x) = G{2} +(x−4)^{2}$

Cost of producing F units of fish is $cF(F,x)= F^{2} + xF$

(a)

For profit-maximizing output, Marginal Revenue (MR) = Marginal Cost (MC)

PG=MR for gold and PF = MR for fish

$cg(G,x) = G^{2 }+(x−4)^{2}$

MC for gold$= \frac{\delta [cg(G,x)]}{\delta G}$

MC for gold = 2G

$2G = 12$

$G= 6 \space units$

Now, Total Revenue of Gold$(TRG) = PG\times G$

$TRG= 12\times 6 = \$72$

$Profit = TR- cg(G,x)$

$Profit = 72 - 36 - (x-4)^{2}$

o find the level of x that maximizes the profit, we differentiate the above function wrt x and equate with zero:

$\frac{\delta Profit } {\delta x} = 2(x-4) = 0$

$2(x-4) = 0$

$x-4 = 0$

$x = 4 units$

$profit = 72 - 36 - (x-4)^{2 }$

$profit = 72 - 36 - (4-4)^{2 }$

$Profit = \$36$

the profit maximizing level of gold output is 6 units, profit is $36 and level of mercury effluent discharged into the dam is 4 units.

(b)

For profit-maximizing output, Marginal Revenue (MR) = Marginal Cost (MC)

PG=MR for gold and PF = MR for fish

$cF(F,x)= F^{2} + xF$

MC for Fish $= \frac{\delta [cF(F,x)]}{\delta F}$

$= 2F + x$

Now, equating MC and pF

$2F + x = 10$

We know x=4

$F= (5 - \frac{4}{2}) units$

$F= 3 \space units$

Total Revenue of Fish$(TRF) = PF\times F$

$TRF= 10\times 3 = 30$

$Profit = TRF - cF(F,x)$

$Profit = 30 - 3^{2} - 4^{3 }$

$Profit = \$9$

fishing cooperative’s profit maximizing level of fish is 3 units output and profit is $9

(c)

Now two cooperatives were merged to operate as a single firm:

Marginal benefit $= 12+10 = 22$

The Marginal cost $= MCG+MCF = 2G + 2F + x$

Total Revenue$= 10F + 12 G$

MB = MC 

$22= 2G+2F+x---------1$

$2F+x=22-2G----------2$

$Profit = 10F + 12 G- G^{2} -(x−4)^{2} - F^{2 }- xF$

$\frac{\delta Profit /}{\delta F} = 10 -2F - x = 0$

From equation $2: 2F+X = 22-2G$

$10-22+2G=0$

$G=\frac{12}{2}$

$G=6 \space units$

since G=6, we from equation 2:

$2F+x=22-12$

$2F+x =10$

$x=10-2F------------3$

First order derivative of profit wrt x:

$\frac{\delta Profit }{\delta x }= -2(x-4)-F = 0$

$\frac{\delta Profit }{\delta x }= -2x+8-F = 0$

From equation 3:

$-20+4F+8 -F =0$

$3F =12$

$F=4 \space units$

Fish quantity has increased from 3 to 4

Putting F in equation 3

$x=10-2\times 4$

$x= 2 \space units$

The emission has reduced from 4 to 2

$Profit = 10F + 12 G- G^{2 }-(x−4)^{2} - F^{2} - xF$

$Profit = 10\times 4 + 12\times 6 -36 - 4 - 16 - 8$

$Profit = \$48$

The profit has increased by $3 (48-45) units when the fishes have increased and emission has reduced.

(d)This would enduce efficiency because mercury wont be discharged to the dam hence the water will be used for the right purpose.