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

newspaper ads = 60 radio ads = 40

Maximum reach is 0.720 million people

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