1. Linear Programming Project Elizabeth Anderson Aubri Rabideau

2. Scenario You are the assistant of an appliance store. Next month you will order two types of stereo systems. Model A Costs $300. Your profit is $40. Model B cost $400. Your Profit is $60. You expect a profit of at least $4800. You expect to sell at least 100 units. How many of each model should you order to MINIMIZE the cost?

3. Variables X = Model A Stereo Y = Model B Stereo

4. Constraints X + Y ≥ 100 40X + 60Y ≥ 4800 X ≥ 0 Y ≥ 0

5. Objective

6. Objective Function and Reasoning C = 300X+ 400Y C (cost) = $300 per model A Stereo + $400 per model B Stereo

7. GraphGraph

8. C=300(120)+400(0) 36,000 C=300(60)+400(40) 34,000 This was chosen because it had both models being sold instead of one and it has the lowest cost of both models being sold. C=300(0)+400(100) 40,000

9. You are the assistant of an appliance store. Next month you will order two types of stereo systems. Model A Costs $300. Your profit is $40. Model B cost $400. Your Profit is $60. you expect a profit of at least $4800. You expect to sell at least 100 units. How many of each model should you order to MINIMIZE the cost? For the objective function(c=300x+400y) we knew that the cost of the models would need to be included (numbers) then multiplied by either X or Y then add the two numbers together to get the total cost(36,000). When the two profits(1,600 and 3,600) are added together it makes a total of $5200.

10. Summary We found that at 60 of Model A will need to be bought and 40 of Model B will need to be bought. We found this by graphing all the restrictions and finding all of the important points. The point that showed at least 100 products were sold and at least $4800 was made was (60,40). When the point was plugged into the objective function it had the lowest cost of $34,000.