# A political candidate plans to spend no more than \$9000 on news paper and radio advertising with no more than twice being spent on newspaper ads at \$50 each than radio ads at \$150 each. it is assumed each newspaper ad is read by 8000 people and each radio ad is heard by 6000 people. how many way of each should be used to maximize the number of people who hear or see the message?

1
by sweetysiri92
radio ads cost \$15 or \$150. check again.

2014-08-22T14:36:08+05:30

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there seem to be some typing mistake in question. so i do different cases:

case I :  radio ads cost \$15 each

Follow procedure explained for case II.
Constraints are: 10 x + 3 y <= 1800      and  5 x <= 3y
Maximization function:  Reach = 8000 x + 6000 y

We get a triangle with O(0,0) , A(120,200) and B(180,0).
Maximum Reach  is at A.
Maximum reach : 2.16 million
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Case II  : radio ads cost \$150 each.
Let us x newspaper ads are given and y radio ads are given.
The Spend on newspaper ads = x * \$50
The spend on radio ads = y * \$150

1st constraint :
x * \$50 + y * \$150 <= \$9000
x + 3 y <= 180        ---- constraint 1

2nd constraint
The spend on Newspaper ads <= 2 * the Spend on radio ads

x * \$50  <= 2 * y * \$150
x <= 6 y        or    x  - 6y <= 0            -------------  constraint 2

If we plot graphs of the two straight lines for the two constraints
they meet at A (120,20).
We also find x value for y = 0 from constraint 1.  x = 180
We have a triangle with O(0,0) , B(180,0)

The maximization FUNCTION is  Number of people the ads reach.
Reach = 8000 x + 6000 y

Find value of Reach at O, A and B. Choose the one with MAXIMUM reach.
Here it is B(180,0)    Reach is 1.44 million
So    X = 180  Y = 0
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case III :

Constraint :   total money spent is \$9000.  Money spent on newspaper ads is at most Half compared money spent on radio ads.

x * 50 <= 1/2 y * 150            => 2 x < 3y
and x*50 + y*150 <= 9000    => x + 3 y <= 180

solving these constraints ,  we get  x = 60  and y = 40 for maximum reach