SOLUTION: The financial institution has asked your firm to place some spot advertisements. The financial institution wants you to find the best allocation of ads between 2 of the local media

Question 174330: The financial institution has asked your firm to place some spot advertisements. The financial institution wants you to find the best allocation of ads between 2 of the local media outlets.
The company you have chosen to handle the transaction has estimated the average number of potential new customers reached per spot announcement to be 1,500 for radio and 500 for newspaper. The price for each spot announcement is $500 for radio and $200 for newspaper as long as the minimum guaranteed number is exceeded. The customer wants a total of 45,000 potential customers reached for which they will pay $16,000. Find the proper number of spot announcements to fit the customer�s request of potential customers reached and fee charged.
There are two standard plans for radio/newspaper ad combinations that give a reduced price (but same price for each spot) if you use them. These discounts are included in the $16,000 charge. The first spot plan is for the weekend and has one radio spot/weekend and one newspaper spot/weekend. The second spot plan is for during the 5-day work week and has two radio spots/week and four newspaper spots/week. These plans are also available on holidays but not at an extra discounted rate. Find the corresponding number of week and weekend spot plans.

I am DESPERATE at this point. I am an online student and have two classes for each lesson, 1 or 2 discussion boards or 1 DB and an independent project. This is one of my DB's. I have atteneded the classes on this section, read the book and took other lessons on line. I do not grasp linear equations. I have already done the next assignment, algebaric sets and done wonderfully. My teacher has corresponded through e-mail with me about these problems, and I STILL do not understand them. PLEASE HELP!
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
let x be the # of radio ads and y be the number of newspaper ads
:
we know that a radio add reaches 1500 people
we know that a newspaper ad reaches 500 people we need a combination of these two that reaches 45000 people. so lets write and equation to fit that
:
1500(x)+500(y)=45000...eq 1
:
The other stipulation regarding these adds is the cost. we know that the cost of each radio ad is 500 and the cost of each newspaper ad is 200. we have a limit of 16000 dollars to spend altogether so lets write an equation to fit that.
:
500(x)+200(y)=16000....eq 2
:
we now have 2 equations with 2 unknowns....a linear system. Lets solve by elimination
:
1500x+500y=45000...eq 1
500x+200y=16000...eq 2
:
if we multiply eq 2 by -3 we can eliminate the x terms by adding the 2 equations together
:
1500x+500y=45000....eq 1
-1500x-600y=-48000....eq 2(multiplied by -3)
:
now adding the 2 equations together....the x terms are eliminated because 1500x-1500x=0. We are left with
500y-600y=45000-48000---->-100y=-3000
:
newspaper spots
take y's found value and plug it into either of eq 1 or 2
:
500x+200(30)=16000---->500x+6000=16000
:
500x=10000
:
radio spots
:
The second part is a bit more difficult and I hope I have done this correctly
:
We know that the total of radio spots must equal 20 and the number of newspaper spots have to equal 30
on the weekend we have the number of radio and newspaper ads run as 1 each
during the week we have the number of radio and newspaper ads run at 2 radio and 4 newspaper.
:
The number of week and weekend radio ads have to total 20
the number of week and weekend newspaper ads have to total 30
:
lets call the number of week and weekend spots, w and wd, respectively
:
1wd+2w=20....eq 1
1wd+4w=30....eq 2
:
again lets use elimination method
subtract eq1 from eq 2....the wd terms will be eliminated. We are left with
:
4w-2w=10
:
2w=10
:
the number of weekly spots
:
plug w's value into eq 1 or 2 and solve for wd
:
1wd+2(5)=20
:
the number of weekend spots

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