# Linear equations, management decisions, production scheduling, and financing

1. Solve the system of linear equations, using the Gauss-Jordan elimination method.

a) 2x + 3y = 2

x + 3y = -2

x - y = 3

b) 3x - 2y = 5

-x + 3y = -4

2x - 4y = 6

c) x - 2y = 2

7x - 14y = 14

3x - 6y = 6

2. Management Decisions. The Management of Hartman Rent-A-Car has allocated $1,008,000 to purchase 60 new automobiles to add to their existing fleet of rental cars. The company will choose from compact, mid-sized, and full-sized cars costing $12,000, $19,200, and $26,400 each, respectively. Find formulas giving the options available to the company. Give two specific options.

3. Manufacturing-Production Scheduling. Kane Manufacturing has a division that produces two models of fireplace grates, model A and model B. To produce each model A grate requires 3lbs of cast iron and 6 min of labor. To produce each model B grate requires 4lbs of cast iron and 3 min of labor. The profit for each model A grate is $2.00, and the profit for each model B grate is $1.50. If 1000lbs of cast iron and 20 labor-hours are available for the production of fireplace grates per day, how many grates of each model should the division produce in order to maximize Kane's profit? What is the optimal profit?

4. Finance-Allocation of Funds. Madison Finance has a total of $20 million earmarked for homeowner and auto loans. On the average, homeowner loans have a 10% annual rate of return, whereas auto loans yield a 12% annual rate of return. Management has also stipulated that the total amount of homeowner loans should be greater than or equal to 4 times the total amount of automobile loans. Determine the total amount of loans of each type that Madison should extend to each category in order to maximize its returns What are the optimal returns?

#### Solution Preview

Please see the attached PDF documents as well as the attached MS Word document for the full solutions to the questions asked.

Thank you for using BrainMass.

31. assume X A and Y B are produced.

Constraints are:

3x+4y<=1000

6x+3y<=20

We want to maximize profit: 2x+1.5y

The dashed area represents the constraints.

When 2x+1.5y passes y intercept, it has maximal values.

In this case, when x=0, y=20/3, 2x+1.5y has maximal values.

However, x and y are integers. Therefore, when y=6, x=0, we could have maximal ...

#### Solution Summary

The expert examines linear equations, management decisions, production scheduling and financing.