# Determine the most profitable mixture of groups you should contact and the most money you can raise a monthYou are working to raise money for the homeless by sending information letters and making...

Determine the most profitable mixture of groups you should contact and the most money you can raise a month

You are working to raise money for the homeless by sending information letters and making follow up calls to local labor organizations and church groups. You discover that each church group requires 2 hours of writing and 1 hour of follow up, while, for each labor union, you need 2 hours of letter writing and 3 hours of follow up. You can raise \$100 from each church group and \$200 from each union local, and you have a maximum of 16 hours of letter-writing time and a maximum of 12 hours follow up time available per month.

embizze | Certified Educator

Let c be the number of churches contacted; l the number of labor organizations contacted; and R the total money raised.

**This is a linear programming problem **

The equation we want to optimize (in this case maximize) is:

`R=100c+200l`  (\$100 from each church, \$200 from each labor org.)

The constraints are:

`c>=0,l>=0` (The so-called natural constraints)

`2c+2l<=16` ** The number of hours writing is 2 hours per contact for both churches and labor org; the total hours must be less than or equal to 16**

`c+3l<=12` **The hours of follow-up are 1 hour per church, and 3 hours per labor org; the total hours available are less than or equal to 12**

We graph the constraints -- I used c as the independent variable (horizontal axis) and l the dependent variable (vertical axis) though it does not matter.

By theory, the optimal points can only occur at the "corners"; the options are (c,l)=(0,0),(0,4),(8,0),(6,2) -- The point (6,2) can be found from the graph or by algebraically solving the simultaneous system `2c+2l=16;c+3l=12`

** The program will not allow me to shade the feasible region. The feasible region is "beneath" both lines, above the horizontal axis, and to the right of the vertical axis. **

Then we substitute the points into the objective function:

`R_("(0,0)")=0+0=0`

`R_("(8,0)")=800+0=800`

`R_("(0,4)")=0+800=800`

`R_("(6,2)")=600+400=1000`

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The optimal solution is to contact 6 churches and 2 labor organizations with total recepts of \$1000

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