Question

Suppose you are the sole owner of a small oil reservoir, and you choose to extract all crude oil from this reservoir over a period of 3 years. The information for the depletion and sale from your field are given as follows: - t=0,1,2(3 periods ) - Q=15 units of crude oil beneath your land. - p=$12 per unit, each period (constant price). - r=10% (interest rate) - MC(qt)=2qt where qt is the amount of oil you extract in period t. How much will you optimally extract in each period? (30pts) (i.e. solve for q0,q1,q2 like we did in class)

Answer 1

Answer: The optimal amount of oil to extract in each period is 5.45 units in period 0, 4.95 units in period 1, and 4.50 units in period 2.

Step-by-step explanation:

To find the optimal extraction we have to maximize

Σ[(p - MC)q/(1+r)^t] for t = 0,1,2

with the condition Σq <= Q

p = $12 per unit, each period

MC(qt) = 2qt (production costs)

r = 10% (interest rate)

and Q = 15 units of crude oil beneath your land

Using the Lagrange multiplier method, we achieve the following Lagrangian function:

L = Σ[(p - MC)q/(1+r)^t] - λ(Σq - Q)

Taking the derivative of the Lagrangian with respect to qt and equating it equal to zero, we get:

dL/dqt = (p - 2qt)/(1+r)^t - λ = 0

Solving for qt, we get:

qt = (p/(2(1+r)^t)) (since λ = (p - 2qt)/(1+r)^t)

Plugging in the given values, we get:

quantity at t=0 = (12/(2(1+0.1)^0)) = 5.45 units

quantity at t=1 = (12/(2(1+0.1)^1)) = 4.95 units

quantity at t=2 = (12/(2(1+0.1)^2)) = 4.50 units

To check if these values are accurate, we need to make sure that the total amount of oil extracted over the three-year period does not exceed Q:

q0 + q1 + q2 = 5.45 + 4.95 + 4.50 = 14.90 units

Since this is less than Q = 15 units, the values are accurate.

Therefore, the optimal amount of oil to extract in each period is 5.45 units in period 0, 4.95 units in period 1, and 4.50 units in period 2.

Answer 2

Answer~

Step-by-step explanation:

To find the optimal

Maximize Σ[(p - MC)q/(1+r)^t] where t = 0, 1, 2

subject to Σq <= Q

where,

p = $12 per unit, each period

MC(qt) = 2qt (production costs)

r = 10% (interest rate)

Q = 15 units of crude oil beneath your land

Using the Lagrange multiplier method, we can set up the following Lagrangian function:

L = Σ[(p - MC)q/(1+r)^t] - λ(Σq - Q)

Taking the derivative of the Lagrangian with respect to qt and setting it equal to zero, we get:

dL/dqt = (p - 2qt)/(1+r)^t - λ = 0

Solving for qt, we get:

qt = (p/(2(1+r)^t)) (since λ = (p - 2qt)/(1+r)^t)

Plugging in the given values, we get:

q0 = (12/(2(1+0.1)^0)) = 5.45 units

q1 = (12/(2(1+0.1)^1)) = 4.95 units

q2 = (12/(2(1+0.1)^2)) = 4.50 units

To check if these values are feasible, we need to make sure that the total amount of oil extracted over the three-year period does not exceed Q:

q0 + q1 + q2 = 5.45 + 4.95 + 4.50 = 14.90 units

Since this is less than Q = 15 units, the solution is feasible.

Therefore, the optimal amount of oil to extract in each period is 5.45 units in period 0, 4.95 units in period 1, and 4.50 units in period 2.