A speculator operates an elevator that has a storage capacity of 5000 bushels of wheat. At the beginning of month 1, the elevator contains 2000 bushels. The estimated buying and selling price of wheat of wheat for the next four (4) months is as follows:

Month	Purchase price for 1000 bushels	Selling price for 1000 bushels
1	40 $	35$
2	50	50
3	70	60
4	70	70

Wheat sold in a given month is removed from the elevator at the beginning of that month. Thus, up to 2000 bushels can be sold in month 1. Wheat purchased in a given month is placed in the silo in the middle of the month, but it cannot be sold until the next month. Knowing the and selling prices of wheat and the storage cost assumption presented below, the speculator below, the speculator wants to know how much wheat to buy and sell in each month to maximize month in order to maximize the total profit accumulated until after the beginning of the fourth month (i.e., after sales that may occur in that last month). To help formulate the storage cost assumptions, let us define:

At = Bushels in the elevator immediately after the amount of wheat sold in period t is removed, t = 1, 2, 3, 4.

Bt = Bushels in the elevator immediately after the quantity purchased in period t is entered, t = 1, 2, 3, 4. t is entered, t = 1, 2, 3.

Assume that the storage cost for period t (t = 1, 2, 3) is (in $):

(0.005*(At+Bt))/2

Define the decision variables and formulate a linear programming model for this problem, then solve it with software and explain the optimal solution to the speculator. Note: this is a deterministic problem (no chance).