Chapter 3.2, Problem 45E

To determine

To find: the number of students of each group to maximize daily income of the center.

Answer to Problem 45E

Preschool children = 30

School age children = 20

Explanation of Solution

Given information:

Charges for preschool for all day = $18 per day

Charges for preschool after school = $6 per day

Maximum children allowed in the center =50

Number of child care worker =10

Each worker at preschool can look after maximum 3 children.

Each worker at school-age can look after maximum 5 children.

Calculation:

Let x and y be the children of preschool and school-age; and the maximum income be $f\left(x,y\right)$ .

The total number of preschool and school-age children is as follows:

  $\begin{array}{c}x=\left(\text{Number of child care worker}\right)\left(\text{Maximum children look after by one worker }\right)\\ =\left(10\right)\left(3\right)\\ =30\end{array}$

  $\begin{array}{c}y=\left(\text{Number of child care worker}\right)\left(\text{Maximum children look after by one worker }\right)\\ =\left(10\right)\left(5\right)\\ =50\end{array}$

The inequalities and the function derived from the given data are shown below.

  $\begin{array}{l}x+y\le 50\\ x\le 30\\ y\le 50\\ f\left(x,y\right)=18x+6y\end{array}$

The graph for the given inequalities is as follows:

  

The vertices of the polygon boundary are $\left(0,0\right),\left(0,50\right),\left(30,20\right)$ and $\left(30,0\right)$ . Put these points in the $f\left(x,y\right)$ and check the maximum value as follows:

  $\begin{array}{c}f\left(0,0\right)=18\left(0\right)+6\left(0\right)\\ =0\\ f\left(0,50\right)=18\left(0\right)+6\left(50\right)\\ =300\\ f\left(30,20\right)=18\left(30\right)+6\left(20\right)\\ =660\\ f\left(30,0\right)=18\left(30\right)+6\left(0\right)\\ =540\end{array}$

When $x=30\text{ and }y=20$ , the daily income of the center is maximum i.e.660. Thus, center will have maximum daily income when the number of students of preschool and school age student should be 30 and 20 respectively.