Chapter 3.4, Problem 26PPS

To determine

To calculate: The number of day-shift workers and night-shift workers to minimize the cost.

Answer to Problem 26PPS

The number of day-shift workers are 8 and night-shift workers are 6. The minimal cost is $776

Explanation of Solution

Given information:

Working schedule of employees at a local movie theater is provided below,

  $\begin{array}{cccc}\text{Time}& \begin{array}{l}\text{noon to}\\ \text{4 P}\text{.M}\text{.}\end{array}& \begin{array}{l}\text{4 P}\text{.M}\text{.}\text{}\text{ to}\\ \text{8 P}\text{.M}\text{.}\end{array}& \begin{array}{l}\text{8 P}\text{.M}\text{.}\text{}\text{ to}\\ \text{midnight}\end{array}\\ \begin{array}{l}\text{Number of}\\ \text{Employees Needed}\end{array}& \text{at least 5}& \text{at least 14}& 6\\ \text{Rates per hour}& \$5.50& \$7.50& \$7.50\end{array}$

Calculation:

Consider the provided information, working schedule of employees at a local movie theater is provided below,

  $\begin{array}{cccc}\text{Time}& \begin{array}{l}\text{noon to}\\ \text{4 P}\text{.M}\text{.}\end{array}& \begin{array}{l}\text{4 P}\text{.M}\text{.}\text{}\text{ to}\\ \text{8 P}\text{.M}\text{.}\end{array}& \begin{array}{l}\text{8 P}\text{.M}\text{.}\text{}\text{ to}\\ \text{midnight}\end{array}\\ \begin{array}{l}\text{Number of}\\ \text{Employees Needed}\end{array}& \text{at least 5}& \text{at least 14}& 6\\ \text{Rates per hour}& \$5.50& \$7.50& \$7.50\end{array}$

Let x denote the number of day-shift workers and y denote the number of night-shift workers.

Since, there is a requirement of at least 5 day-shift workers and 6 night shift workers also 14 workers who worked both in day and night shift.

  $\begin{array}{c}x\ge 5\\ y\ge 6\\ x+y\ge 14\end{array}$

Plot the inequalities and shade the common region $x\ge 5,y\ge 6\text{ and }x+y\ge 14$

  

The function that is the cost function that is to be minimized is,

  $\begin{array}{c}f\left(x,y\right)=5.50\left(4\right)x+7.50\left(4\right)\left(x+y\right)+7.50\left(4\right)y\\ =52x+60y\end{array}$.

Substitute the vertices of the feasible region to find the point at which maximum revenue is there.

Substitute $\left(5,9\right)$ in the function $f\left(x,y\right)=52x+60y$.

  $f\left(5,9\right)=800$

Substitute $\left(8,6\right)$ in the function $f\left(x,y\right)=52x+60y$.

  $f\left(8,6\right)=776$

Since, minimum cost is $776 which is when 8 are day-shift workers and 6 night-shift workers.