Question

Suppose an economy consists of 11 people: 10 farmers and one miner. All individuals have identical preferences, u(f,g) = ln(f) + 2ln(g), where f is a kilo of food, and g is a gram of gold. Each farmer is endowed with a 50 kilos of food and no gold. The miner is endowed with no food and one gram of gold. The farmers and the miner can trade food and gold at prices pf and pg. In the following, normalize the price of food to one, pf = 1.

a) For given prices (pf = 1 and pg), solve for a farmer’s optimal demand for food and gold given that the farmer has an endowment that has value 50.

b) For given prices (pf = 1 and pg), solve for the miner’s optimal demand for food and gold given that the miner has an endowment that has value 1 × pg .

c) Determine the equilibrium price of gold, pg*. How does the value of the miner’s endowment compare to that of the farmers altogether.

d) Suppose instead of 10 farmers, there are 100 farmers. In equilibrium, how valuable is the miner’s 1 gram of gold in terms of food, then?

Answer 1

a)

Utility function: Log f + 2 Log g

Individual farmer's budget constraint:

Pf*f + Pg*g = wf*Pf + wg*Pg

where wf is the endowment of food and wg is the endowment of gold.

So, f + Pg*g = 50

Hence the optimization problem is:

Max: Log f + 2 Log g subject to f + Pg*g = 50

Setting up Lagrange we get,

L = Log f + 2 Log g + (50 - f - Pg*g)

Getting the first order conditions,

dL/df = 1/f - = 0 ............1

dL/dg = 2/g - Pg = 0 ..............2

dL/d = 50 - f - Pg*g = 0 ..............3

Dividing equation 1 and 2 we get,

g/2f = 1/Pg

g = 2f/Pg

Put the above in equation 3 we get,

50 - f - Pg*g = 0

50 - f - Pg*2f/Pg = 0

3f = 50

f = 50/3

Now, g = 2f/Pg

g = 100/3Pg ........... 4

b)

Similary for gold miner,

Utility function: Log f + 2 Log g

Miner's budget constraint:

Pf*f + Pg*g = wf*Pf + wg*Pg

where wf is the endowment of food and wg is the endowment of gold.

So, f + Pg*g = Pg

Hence the optimization problem is:

Max: Log f + 2 Log g subject to f + Pg*g = Pg

Setting up Lagrange we get,

L = Log f + 2 Log g + (Pg - f - Pg*g)

Getting the first order conditions,

dL/df = 1/f - = 0 ............1

dL/dg = 2/g - Pg = 0 ..............2

dL/d = Pg - f - Pg*g = 0 ..............3

Dividing equation 1 and 2 we get,

g/2f = 1/Pg

g = 2f/Pg

Put the above in equation 3 we get,

Pg - f - Pg*g = 0

Pg - f - Pg*2f/Pg = 0

3f = Pg

f = Pg/3

Now, g = 2f/Pg

g = 2/3

c)

At equilibrium the total demand of gold will be equal to the total supply of gold,

Individual farmer's demand for gold is : g = 100/3Pg (from equation 4)

Hence demand of 10 farmers = g = 1000/3Pg

Demand for gold by gold miner = g = 2/3

Hence total demand = 1000/3Pg + 2/3

Total supply = 1

Now,

1000/3Pg + 2/3 = 1

(1000 + 2Pg) = 3Pg

Pg = 1000

d)

If there were 100 farmers, total demand for gold by farmers = g = 10000/3Pg

Hence total demand = 10000/3Pg + 2/3

Total supply = 1

Now,

10000/3Pg + 2/3 = 1

(10000 + 2Pg) = 3Pg

Pg = 10000

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