math problem solve [closed]

Question

You are the advertising chief for a video game company. You have a budget of \$200,000.00 to spend each week on television advertising time. Advertising time is \$5,000 per minute for prime time and \$3,000 per minute for off time. Prime time is considered to be from 8pm to 11pm everyday.

Make a schedule for your weekly budget. Include the days, times, and number of minutes each ad will run. If you don't want your budget cut, spend as close as you can to all of your money.

Answer 1

If the point is just to spend 200k in increments of 3k and 5k, you could note that 3+5=8 and 8 divides evenly into 200. There are many other solutions. If there is another point, it should be provided.

Answer 2

If you want to advertise for $x$ minutes during prime-time throughout the week and $y$ minutes during off-prime-times, then we have that $$5x+3y = 200$$ Let us call the impact per minute during prime-time as $a$ and the impact per minute during off-prime-time as $b$. Then our goal is to maximize $$ax+by \,\,\,\,\,\,\,\,\,\,\,\,\, (\star)$$ subject to the constraint $5x+3y = 200$ where $x,y \in \mathbb{Z}^+$. The solution to $5x+3y = 200$ for $x,y \in \mathbb{Z}^+$ is given by $$(x,y) = (3k+1,65-5k)$$where $k \in \{0,1,2,\ldots,13\}$.Hence, our goal is to maximize $$a(3k+1) + b(65-5k) = k(3a-5b) + a+65b$$ for $k \in \{0,1,2,\ldots,13\}$. Hence, if we have that the impact during the prime-time $a$ is greater than or equal to five-thirds the impact during the off-prime-times $b$ i.e. $a \geq \dfrac53 b$, then choose $k=13$ i.e. invest all your money for prime-time ads i.e. $(x,y) = (40,0)$, else invest most of the money during off-prime-time ads and the whatever little remains invest in prime-time ads i.e. $(x,y) = (1,65)$. This is what you get with a simple linear objective, $(\star)$, to maximize impact. You can do better if you have a better model for the impact during prime-times and off-prime-times.