Chapter 7.3, Problem 48E

To determine

To calculate: the number of each type of ad the department runs each month.

Answer to Problem 48E

The number of each type of ad the department runs each month is 30 monthly ads on television, 10 monthly ads on radio and 20 monthly ads in newspaper.

Explanation of Solution

Given information:

A health insurance company advertises on television, on radio, and in the local newspaper. The advertising budget is $42,000 per month. A television ad costs $1000, a radio ad costs $200, and a newspaper ad costs $500. The department wants to run 60 ads per month and have as many television ads as radio and newspaper ads combined.

Calculation:

Let us consider x for television ads, y for radio ads, z for newspaper ads. Then the total budget per month would be,

  $1000x+200y+500z=42000\cdots \left(i\right)$

The department wants to run 60 ads per month.

  $x+y+z=60\cdots \left(ii\right)$

Department wants to have as many television ads as radio and newspaper ads combined.

  $x=y+z$

Put the value of x in $\left(i\right)$ and $\left(ii\right)$

we get

  $\begin{array}{l}1000\left(y+z\right)+200y+500z=42000\\ 1000y+1000z+200y+500z=42000\\ 1200y+1500z=42000\\ 4y+5z=140\cdots \left(iii\right)\end{array}$

Now

  $\begin{array}{l}y+z+y+z=60\\ y+z+y+z=60\\ 2y+2z=60\\ y+z=30\\ y=30-z\cdots \left(iv\right)\end{array}$

Now put this value in $\left(iii\right)$

  $\begin{array}{l}4\left(30-z\right)+5z=140\\ 120-4z+5z=140\\ 120+z=140\\ z=140-120\\ z=20\end{array}$

Put $z=20$ in $\left(iv\right)$

We get

  $\begin{array}{l}y=30-20\\ y=10\end{array}$

Now put $z=20$ and $y=10$ in $x=y+z$then

  $\begin{array}{l}x=10+20\\ x=30\end{array}$

Conclusion:

Hence the number of each type of ad the department runs each month is 30 monthly ads on television, 10 monthly ads on radio and 20 monthly ads in newspaper.