1. Consider a quantity of water held in storage that is to be used in two periods (periods 1 and 2). The willingness to pay for water, costs of extraction from storage, and other parameters are as follows:

Inverse demand: p(q) = 50 - 6q

Cost of water extraction: c(q) = 0.9q

Initial amount of water in storage: Si = 6

Interest rate: r = 0.3

(a) Derive the optimal quantities of water to extract in periods 1 and 2: q1* and q2*. Also, calculate the value of the Lagrange multiplier at the optimum: λ*.

(b) Draw a graph to visualize the optimum. Be sure to label q1*, q2* and λ* on the graph. (Hint: your graph should have two vertical axes.)

(c) Repeat parts (a) and (b) for each of the following changes to the problem. The changes below are not cumulative (for each change, all other parameters revert to their values from original problem). For each case below, explain how the value of the Lagrange multiplier will change and why.

i. The interest rate rises to r = 0.50.

ii. A drop in the price of energy lowers extraction costs to c(q) = 0.1q.

iii. An increase to water demand in period 1 only to p(q) = 70 - 6q. (Water demand in period 2 remains the same as before.)

iv. The initial stock of water becomes Si = 10. (Hint: Lagrange multipliers cannot be negative at an optimum.)