a store has a $9000 monthly advertising budget. newspaper ads cost $600 each with a limit of 40 per month…

a store has a $9000 monthly advertising budget. newspaper ads cost $600 each with a limit of 40 per month, radio ads cost $30 each with a limit of 60 per month, and tv ads cost $3000 each with a limit of 10 per month. about 5000 people see each newspaper ad, 3000 hear each radio ad, and 12,000 see each tv ad. how much of each type should be used to maximize the stores exposure? interpret the results.\nif the store wants to maximize exposure, it should use newspaper ads, radio ads, and tv ads.

a store has a $9000 monthly advertising budget. newspaper ads cost $600 each with a limit of 40 per month, radio ads cost $30 each with a limit of 60 per month, and tv ads cost $3000 each with a limit of 10 per month. about 5000 people see each newspaper ad, 3000 hear each radio ad, and 12,000 see each tv ad. how much of each type should be used to maximize the stores exposure? interpret the results.\nif the store wants to maximize exposure, it should use newspaper ads, radio ads, and tv ads.

Answer

Explanation:

Step1: Define variables

Let $x$ be the number of newspaper ads, $y$ be the number of radio ads, and $z$ be the number of TV ads. The objective function (exposure function) $E$ to maximize is $E = 5000x+3000y + 12000z$. The constraints are: Cost constraint: $600x + 30y+3000z\leq9000$ (budget constraint) Quantity constraints: $0\leq x\leq40$, $0\leq y\leq60$, $0\leq z\leq10$.

Step2: Simplify cost - constraint

Divide the cost - constraint $600x + 30y+3000z\leq9000$ by 30 to get $20x + y+100z\leq300$.

Step3: Analyze the coefficients of the objective function

The coefficient of $z$ in the objective function $E$ is the largest ($12000$), so we first try to use as many TV ads as possible within the budget. Since the cost of a TV ad is $3000$ and the budget is $9000$, if we take $z = 3$ (because $3\times3000=9000$), then $x = 0$ and $y = 0$. Let's check if this satisfies all constraints: For the quantity constraints: $z = 3\leq10$, $x = 0\leq40$, $y = 0\leq60$. For the cost constraint: $600\times0+30\times0 + 3000\times3=9000$ which is equal to the budget.

Answer:

0 newspaper ads, 0 radio ads, and 3 TV ads.