
Production Scheduling In a factory, machine 1 produces 8-inch (in.) pliers at the rate of 60 units per hour (h) and 6-in. pliers at the rate of 70 units/h. Machine 2 produces 8-in. pliers at the rate of 40 units/h and 6-in. pliers at the rate of 20 units/h. It costs to operate machine 1, and machine 2 costs 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-h day. Which combination of machines will cost the least money to operate?

To solve: The given linear programming problem.
Answer to Problem 24AYU
Solution:
The minimum expense is when Machine 1 is operated for 30 minutes and machine 2 is operated for 5 hours and 15 minutes.
Explanation of Solution
Given:
Machine | 8 inch pliers | 6 inch pliers | Cost of operation |
1 | 60 units/hour | 70 units/hour | |
2 | 40 units/hour | 20 units/hour |
In 10 hours of production schedule requires atleast, 240 units of 8 inch pliers and 140 units of 6 inch pliers.
Calculation:
Begin by assigning symbols for the two variables.
Number of hours Machine 1 operates.
Number of hours Machine 2 operates.
a. If is the total cost of operating the two machines, then
The goal is to minimize subject to certain constraints on and . Because and represents time, the only meaningful values of and are non-negative.
Therefore, .
From the given data we get
Therefore, the linear programming problem may be stated as
Minimize,
Subject to
The graph of the constraints is illustrated in the figure below.
Corner points are | Value of objective function |
300 | |
210 | |
200 | |
500 | |
800 |
The minimum expense is when Machine 1 is operated for 30 minutes and machine 2 is operated for 5 hours and 15 minutes.
Chapter 11 Solutions
Precalculus
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