A company installs different POS computer systems. POS system A requires two hours to configure and one hour for assemble. POS B requires three hours to configure and one hour to assemble. POS C requires two hours to configure and two hours to assemble. The company has up to 100 labor hours for configure and 800 labor hours for assemble each week. If the profit of POS systems A, B, and C is $700, $ 800, and $1000 respectively how many of each system should be installed each week to maximize profit? What is the maximum profit each week?
This question is strange...why do the POS systems take longer to configure than they do to assemble, yet there are more labor hours for assembly than configuration. The minimum configuration time is two hours and the minimum assembly time is one hour. Yet for two hours config and two hours of assembly (one hour over minimum assembly) you get ~$300 more in profit. Thus, POS C will be the most profitable because the configuration hours determine maximum profit.
50 (100 configuration hours/2 hours per unit) POS C systems can be installed for a profit of $50,000 (assembly hours used = 100 which is less than 800)
33 POS B systems can be installed for a profit of $26,400 (assembly hours used = 99 which is less than 800)
50 POS A systems can be installed for a profit of $35,000 (assembly hours used = 100 which is less than 800)
To determine the optimal number of each POS system to install per week and the maximum profit, let's use linear programming, a mathematical technique for optimizing tasks within certain constraints.
Let's define the variables:
Let x = the number of POS system A to install per week.
Let y = the number of POS system B to install per week.
Let z = the number of POS system C to install per week.
Now let's set up the constraints based on the given labor hour limitations:
The configure time constraint: 2x + 3y + 2z ≤ 100 (total configure hours should not exceed 100).
The assemble time constraint: x + y + 2z ≤ 800 (total assemble hours should not exceed 800).
Next, let's define the objective function to maximize the profit:
Maximize P = 700x + 800y + 1000z (the total profit based on the number of POS systems installed).
Now we have a linear programming problem with constraints and an objective function. We can use optimization techniques or software to solve it and find the optimal solution.
Without optimization software, we can use a technique called the corner point method to solve the problem manually. This method involves evaluating the objective function at each corner point of the feasible region defined by the constraints and selecting the point that yields the maximum profit.
To find the corner points, we need to solve each constraint separately to determine the intersection points:
For the configure time constraint:
2x + 3y + 2z = 100
Solving for z, we get z = 50 - (2/3)x - (3/2)y.
For the assemble time constraint:
x + y + 2z = 800
Solving for z, we get z = 400 - (1/2)x - (1/2)y.
By setting x, y, and z to appropriate values, we can calculate the corresponding profits using the objective function P = 700x + 800y + 1000z.
Now we need to evaluate the objective function at each corner point:
1. When x = 0, y = 0, and z = 50, P = 700(0) + 800(0) + 1000(50) = $50,000.
2. When x = 400, y = 0, and z = 0, P = 700(400) + 800(0) + 1000(0) = $280,000.
3. When x = 0, y = 400, and z = 0, P = 700(0) + 800(400) + 1000(0) = $320,000.
4. When x = 0, y = 0, and z = 200, P = 700(0) + 800(0) + 1000(200) = $200,000.
5. When x = 0, y = 200, and z = 300, P = 700(0) + 800(200) + 1000(300) = $500,000.
Using these results, we can see that the maximum profit is achieved when x = 0, y = 200, and z = 300, with a profit of $500,000.
Therefore, the optimal solution is to install 0 POS system A, 200 POS system B, and 300 POS system C each week, resulting in a maximum profit of $500,000 per week.