Question

The advertising director for Karisma, a chain of four retail stores is considering 3 media possibilities: 1- Ads in the Ahram newspaper, 2- ads in a local UGO trade magazine that is
distributed free to all houses in the city and suburbs, and 3- ads on a
local Alex TV station. The stores are expanding their lines of Do -It-
Yourself tools and the advertising director is interested in a total new
customer exposure level of at least 50% within the city and 60% in the
suburbs. Each TV ad has a new-customer exposure level of 5% in the
city and 3% in the northwest suburbs. The newspaper ads have
corresponding exposure levels per ad of 3.5% and 3%, while the trade
magazine has corresponding exposure levels per ad of 0.5% and 1%.
The relevant costs are $1000 per newspaper ad, $300 per trade
magazine ad, and $2000 per TV ad.all 3 types of media
are used, Karizma would like to ensure that no single medium
consumes more than 45% of the total amount spent.(newspapers ads = 9, magazine ads = 30, TV = 1, Cost$20,000)

Answer 1

Answer on Question #38415 – Math – Linear Algebra

The advertising director for Karisma, a chain of four retail stores is considering 3 media possibilities: 1- Ads in the Ahram newspaper, 2- ads in a local UGO trade magazine that is distributed free to all houses in the city and suburbs, and 3- ads on a local Alex TV station. The stores are expanding their lines of Do -It-Yourself tools and the advertising director is interested in a total new customer exposure level of at least 50% within the city and 60% in the suburbs. Each TV ad has a new-customer exposure level of 5% in the city and 3% in the northwest suburbs. The newspaper ads have corresponding exposure levels per ad of 3.5% and 3%, while the trade magazine has corresponding exposure levels per ad of 0.5% and 1%. The relevant costs are $1000 per newspaper ad, $300 per trade magazine ad, and $2000 per TV ad. All 3 types of media are used, Karizma would like to ensure that no single medium consumes more than 45% of the total amount spent. (newspapers ads = 9, magazine ads = 30, TV = 1, Cost$20,000)

Solution

Let x be the consumption of TV ads; y be the consumption of newspaper ads; z be the consumption of magazine ads.

Compose the next inequalities:

$$5x + 3.5y + 0.5z \geq 50$$

(exposure level of at least 50% within the city)

$$3x + 3y + z \geq 60$$

(exposure level of at least 60% within within the suburbs)

$$x \leq 0.45 \cdot (x + y + z)$$

(for x: no single medium consumes more than 45% of the total amount spent)

$$y \leq 0.45 \cdot (x + y + z)$$

(for y: no single medium consumes more than 45% of the total amount spent)

$$z \leq 0.45 \cdot (x + y + z)$$

(for z: no single medium consumes more than 45% of the total amount spent)

$$x \geq 0, y \geq 0, z \geq 0$$

(all 3 types of media are used).

There exist indefinitely many solutions of this system of inequalities, but x should be greater than 20/7.

Then

$$1000y - \text{cost of newspaper ads}$$ $$300z - \text{cost of trade magazines}$$ $$2000x - \text{cost of TV ads}.$$

But simultaneously $x = 1, y = 9, z = 30$ is not the solution of this system of inequalities because it does not satisfy inequality $z \leq 0.45 \cdot (x + y + z)$. We come to $30 \leq 0.45 \cdot (1 + 9 + 30)$, which is false.