Q. 24 - Precalculus Enhanced with Graphing Utilities

Question

Production Scheduling In a factory, machine 1 produces 8 -inch (in.) pliers at the rate of 60 units per hour (hr) and 6 -in. pliers at the rate of 70 units/hr. Machine 2 produces 8 -in. pliers at the rate of $40units/hr$ and 6-in. pliers at the rate of $20units/hr$. It costs $\backslash (50/hr$ to operate machine 1 , and machine 2 costs $\backslash )30/hr$ to operate. The production schedule requires that at least 240 units of 8-in. pliers and at least 140 units of 6-in. pliers be produced during each 10-hr day. Which combination of machines will cost the least money to operate?

Step-by-Step Solution

Answer

The minimum value of the cost function is attained when the number of machine 1 is $\frac{1}{2}$ and the numbers of machine 2 are $\frac{21}{4}$. Since the number of machines cannot be fraction, so the next minimum of the cost function is when the number of machine 1 used are 4 and the number of machine 2 used are none. The total cost of the process is $\$2000$.

Step 1. Given information

Let the number of machine 1 and machine 2 used be x and y.
The machine 1 produces 8 inches pliers at the rate of 60 units per hour and 6 inches pliers at a rate of 70 units per hour. The machine 2 produces 8 inches pliers at a rate of 40 units per hour and 6 inches pliers at a rate of 20 units per hour. The factory requires at least 240 units of 8 inches and at least 140 units of 6 inches. We can obtain a set of inequalities using the given information. 

$$\begin{gathered} 60x+40y\ge 240 \\ 70x+20y\ge 140 \end{gathered}$$

Step 2. Finding the values constraints

The total cost of machine 1 to operate for 10 hours is $\$500$ and the total cost of machine 2 to operate for 10 hours is $\$300$. The cost function for the given linear programming problem can be obtained.

$C=500x+300y$

The number of machines cannot be less than zero; therefore we can obtain two more constraints using the conditions.

$$\begin{gathered} x\ge 0 \\ y\ge 0 \end{gathered}$$

Step 3. Sketch the graph of the feasible region using the obtained inequalities in rectangular Cartesian coordinates.

Evaluate the cost pf the function to be minimised at each of the corner of the shaded feasible region

Vertex	Value cost of the function $C=500x+300y$
$(0,7)$	$C=500\left(0\right)+300\left(7\right)=2100$
$\left(\frac{1}{2},\frac{21}{4}\right)$	$C=500\left(\frac{1}{2}\right)+300\left(\frac{21}{4}\right)=1825$
$(4,0)$	$C=500\left(4\right)+300\left(0\right)=2000$

The minimum value of the cost function is attained when the number of machine 1 is $\frac{1}{2}$ and the numbers of machine 2 are $\frac{21}{4}$. Since the number of machines cannot be fraction, so the next minimum of the cost function is when the number of machine 1 used are 4 and the number of machine 2 used are none. The total cost of the process is $\$2000$.