# Thread: linear programming, quantitive math

1. ## linear programming, quantitive math

I have a question like this that will be on a test coming up soon and I need to know how to approach this problem. Basically I just need to know hot to set the problem up, you know, the coefficents variables and constraints. Once I have the problem together I can solve it, but I just need to know how to configure it. Thanks

An ad campaign for a new snack chip will be conducted in a limited geographical area and can use TV time, radio time, and newspaper ads. Information about each medium is shown below.

Medium/ Cost Per Ad/ # Reached/ Exposure Quality
TV /500 /10000/ 30
New /400 /5000/ 25

If the number of TV ads cannot exceed the number of radio ads by more than 4, and if the advertising budget is $10000, develop the model that will maximize the number reached and achieve an exposure quality of at least 1000. Objective function-To maximize the number reached 2. Hello, jwells1999! An ad campaign for a new snack chip can use TV time, radio time, and newspaper ads. Information about each medium is shown below.$\displaystyle \begin{array}{cccc}\text{Medium} & \text{Cost Per Ad} & \text{No. reached} & \text{Exp. Quality} \\ \hline
\text{TV} & 500 & 10,000 & 30 \\
\text{Radio} & 200 & 3,000 & 40 \\
\text{News} & 400 & 5,000 & 25 \\ \hline
\text{Total} & 10,000 & N & 1,000 \end{array}$If the number of TV ads cannot exceed the number of radio ads by more than 4, and if the advertising budget is$10000, develop the model that will maximize
the number reached and achieve an exposure quality of at least 1000.

Objective function: To maximize the number reached

Let $\displaystyle x$ = number of TV ads.
Let $\displaystyle y$ = number of radio ads.
Let $\displaystyle z$ = number of newspaper ads.

Then the chart looks like this:

$\displaystyle \begin{array}{cccc}\text{Medium} & \text{Cost Per Ad} & \text{No. reached} & \text{Exp. Quality} \\ \hline \text{TV (x)} & 500x & 10,000x & 30x \\ \text{Radio (y)} & 200y & 3,000y & 40y \\ \text{News (z)} & 400z & 5,000z & 25z \\ \hline \text{Total} & 10,000 & N & 1,000 \end{array}$

And we have these inequalities:

. . . . . . $\displaystyle \begin{array}{c} x\:\geq\:0 \\ y\:\geq\:0 \\ z\:\geq\:0 \\ x\:\leq \:y+4 \end{array}$

$\displaystyle \begin{array}{c} 500x+200y + 400z \:\leq \:10,000 \\ 30x + 40y + 50z \:\geq \:1,000 \end{array}$

Objective function: .$\displaystyle N \;=\;10,000x + 3,000y + 5,000z$