<PRE><B>Question 1166337</B>
.
A company has $14,830 available per month for advertising. Newspaper ads cost $190 each and can&#39;t run 
more than 24 times per month. Radio ads cost $590 each and can&#39;t run more than 32 times per month at this price.
Each newspaper ad reaches 5700 potential customers, and each radio ad reaches 6700 potential customers. 
The company wants to maximize the number of ad exposures to potential customers.
Determine the most profitable / (effective) way to do it.
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&lt;pre&gt;
To find the maximum number of ad exposures, let&#39;s formulate the problem in terms 
of objective function and constraints.  


Let x be the number of newspaper ads and y be the number of radio ads. 


The objective function is 

    P = 5700*x + 6700*y.       (1)

It is the number of possible expositions, and we want to maximize it.


The constraints are: 

    190*x + 590*y ≤ 14830 (the budget),     (2)

    x ≤ 24 (newspaper ad limit), y ≤ 32 (radio ad limit),     (3)

    x ≥ 0, y ≥ 0 (non-negativity).     (4)


Now it is presented as a typical Linear Programming problem. But it can be easily solved MENTALLY
using &quot;the most aggressive&quot; logical strategy/methodology.


From expression (1) for the objective function, we see that the contribution of each single newspaper ad
(in terms of the number of potential expositions, 5700) is comparable with (or distinct insignificantly from) 
the contribution of each single radio ad (6700).


But each newspaper ad is much cheaper ($190) than each radio ad ($590). So, it is clear that 
the most profitable strategy is  to make as many newspaper ads as possible (x=24), and then
to spend the rest of the budget for the radio ads. 


Thus the most effective solution is to make 24 newspaper ads, spending 24*190 = 4560 dollars for it.

The rest of the budget is then  $14830 - $4560 = $10270.

This amount can be / (should be) spent for radio ads.

It provides the number of radio ads  y = {{{10270/590}}} = 17.40678,
and we should round this decimal number to the closest lesser integer number, which is y = 17.


So, the answer to the problem&#39;s question is THIS:

    24 newspaper ads and 17 radio ads provide the greatest possible number of expositions (~ potential customers),
    which is then  5700*24 + 6700*17 = 250700.
&lt;/pre&gt;

Solved.


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&lt;H3&gt;Post-solution note&lt;/H3&gt;

In this concrete problem, the presented method has one important advantage comparing with the traditional form
geometric solution of Linear Programming problems.


Working in the frame of traditional Linear Programming geometric method, you will get the solution with non-integer decimals, 
so, you will be forced to use other arguments to complete the traditional solution.


Working in the way, presented here in the solution above, you will get the answer in integer numbers without any complications.



In other words, this problem is for INTEGER Linear Programming - not for regular Linear Programming.
But integer Linear Programming problems require their special solution methodology 
(which is not studied in the school Math) and requires special solvers.
Or, as it is done in my solution above - a special logical treatment.



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