Computers and Posters
By: Oliver Mulcahy
October 1st 2013
Simplicity through good design, or the ultimate sophisitcation.
If I could dream up a job right here and now, I would work for a company that does nothing but dream up the insane and perfect it. My favorite hobbies are computing and designing. Therefore, I chose to make this imaginary company produce computers and posters. Random? Yes. Fun? Even more so. Posters leave a striking impact on people and I love studying computers more than anything. Enjoy!
Producing Posters and Computers
Computers are costly. Let's say it cost a good $1000 to produce a single computer. Because of the rare materials inside the computer, the government only allows me to produce 900 a month. Posters are a little easier. A well produced poster will cost the company $90. However, we only use the finest ink and paper. The government again restricts the production of our posters to 2000 a month.
Because the goal is profit, I will sell the computers for $1,500. As for the posters, people already expect something great for cheap, and people do not think $90 is cheap. Therefore, I will have limited profit and sell posters for $100.
In total, my profits will be $500 per computer sold and $10 per poster sold.
C = # of computers per month
P = # of posters per month
C ≤ 900 (Computers must be less than 900)
P ≤ 2000 (Posters must be less than 2000)
1000c + 100p ≤ 10,000 (We have a budget of $10,000)
500c + 10p
This is my graph overview. The green is the equation P ≤ 2000. The amount of posters must be below this. In the pink is the equation C ≤ 900. The amount of computers must be less than 900. And in the blue is the equation 1000c + 100p ≤ 10,000. The vertices in this graph are not very good. However, even if it's very slight, the blue does make accurate vertices. The next graph will show this zoomed much closer.
The vertices are much more visible here. 2 sets of vertices appear: 10 computers and 0 posters (10,0) or 0 computers and 100 posters (0,100). These are my two possible solutions. Usually, there is a third set of vertices but because the blue line is so incredibly thin compared to my other restricitons this is not possible.
Profit equation: 500c + 10p
Set 1: 500(10) + 10(0) = $5000
Set 2: 500(0) + 10(100) = $1000
Because I want the most profit, producing 10 computers and no posters is the most profitable system that respects all restrictions. If I were to do this again, I would decrease the money limit, and change the values of the posters or computers to make a much more realistic goal.
- Oliver Mulcahy