Solve this problem using the simplex method.

A company manufactures checker sets and chess sets. Suppose each day the company has available 1200 boards (which can be used for both games) and 80,000 units of wood for making pieces. Each checker set uses 20 units of wood and each chess set uses 120 units of wood. The distributors the company sells to can take up to 1250 checker sets per day and up to 750 chess sets per day. The company makes a profit of $1.00 on each checker set and $2.00 on each chess set.
(a) How many checker sets and how many chess sets should the company make each day in order to maximize its profits?
?- checker sets
?- chess sets

(b) What is the profit per day using this strategy?
$-? per day

To solve this problem using the simplex method, we need to set up a linear programming model.

Let's define our decision variables:
Let x be the number of checker sets produced per day,
and let y be the number of chess sets produced per day.

Now, let's set up the constraints based on the available resources and the distributors' limits:

1. Constraints for the number of boards:
Each checker set uses 1 board, and each chess set uses 1 board.
So the constraint is: x + y ≤ 1200.

2. Constraints for the units of wood:
Each checker set uses 20 units of wood, and each chess set uses 120 units of wood.
So the constraint is: 20x + 120y ≤ 80,000.

3. Constraints for the distributor's limits:
The distributors can take up to 1250 checker sets and 750 chess sets per day.
So the constraints are: x ≤ 1250, and y ≤ 750.

4. Non-negativity constraints:
The number of sets cannot be negative.
So the constraints are: x ≥ 0, and y ≥ 0.

Next, let's define our objective function, which is to maximize the profit:
The profit per checker set is $1, and the profit per chess set is $2.
So the objective function is: $1x + $2y.

Now, we can set up the initial tableau for the simplex method.

Tableau:

| B | x | y | RHS |
---------------------------------------
Z | 0 | 1 | 2 | 0 |
----------------------------------------
x | 1200 | 1 | 1 | 1200 |
y | 80000 | 20 | 120 | 80000 |
x2 | 1250 | 1 | 0 | 1250 |
y2 | 750 | 0 | 1 | 750 |

Now, we can apply the simplex method to find the optimal solution.

Perform the following steps:

1. Select the most negative coefficient in the Z row. In this case, it is -2 in the y column.

2. Select the pivot element in the y column. In this case, it is 80000.

3. Divide the pivot row (y) by the pivot element (80000), so the pivot element becomes 1.

4. Use row operations to clear the y column below the pivot. Subtract the pivot row multiplied by the appropriate coefficient from each subsequent row:

| B | x | y | RHS |
-------------------------------------------
Z | 2 | 1 | 0 | 15,000 |
-------------------------------------------
x | 800 | 1 | -6.6 | 1000 |
y | 1 | 0 | 0.00105 | 1 |
x2 | 450 | 0 | -0.00125 | 1250 |
y2 | 750 | 0 | 1 | 750 |

5. Repeat steps 1-4 until there are no negative coefficients in the Z row.

In this case, the final tableau is:

| B | x | y | RHS |
--------------------------------------
Z | 15,000 | 1 | 0 | 15,000 |
--------------------------------------
x | 1000 | 1 | 0 | 1000 |
y | 1 | 0 | 0 | 1 |
x2 | 1250 | 0 | 0 | 1250 |
y2 | 750 | 0 | 1 | 750 |

Now, we can interpret the optimal solution from the final tableau:

(a) The company should produce 1000 checker sets and 0 chess sets each day to maximize profits.

(b) The profit per day using this strategy is $15,000.

Therefore, the company should produce 1000 checker sets and no chess sets, resulting in a profit of $15,000 per day.