Alandscaper is offering a 25% discount on her $150 lawn care package to new customers. she has a budget of $200 for brochures and mailers and 3 hours to distribute the materials. she estimates that 5% of the people who get a brochure and 12% of the people who get a mailer will sign up for the service. brochures cost $0.20 each, and mailers cost $0.99 each including postage. it takes 1 hour to pass out 100 brochures, but it only takes 40 minutes to stamp 100 mailers

I set this problem up with a system of linear inequalities. I made brochures equal x, and mailers equal y. My first inequality was .6x+.4y≤180 because each brochure takes 6 tenths of a minute to pass out, and each mailer takes 4 tenths of a minute to stamp, and the combined time must be less than 180 minutes. My next inequality was .2x+.99y≤200 because the brochures were 2 tenths of a dollar, and the mailers were 99 hundredths of a dollar, and the combined money spent must be less than 200. x≥0 and y≥0 are understood as you cannot have a negative amount of brochures/mailers. I then graphed the inequalities to get vertices of (0, 202) and (300, 0). Then I solved the first two inequalities as a system of equations to get (191.17, 163.4). I rounded down to get (191, 163). I did not use an objective function for this problem, because I needed to round how many people bought a lawn care package. I first solved for (0, 202). 202*.12 equals 24.24, but I rounded down to 24 because you cant have .24 of a person. Then I multiplied 24 by 112.5 (which is the lawn care package with a 25% discount) to get 2700. The cost of 202 mailers (202*.99) is $199.98, so I subtracted that from 2700 to get $2500.02 of revenue. I worked the 2 other vertices the same way, and I ended up getting that 191 brochures and 163 mailers gave the most revenue, at $3175.43.

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Maximizing the difference between total revenue and total cost;