A hospital performs 3 types of operations. The first type takes 30 minutes, and they charge $1000 for it. The second type takes 1 hour, and they charge $1500 for it. The third operation takes 2 hours, and they charge $2000 for it. The hospital can schedule up to 95 total operations per day. Moreover, they have 5 operating rooms, each of which can be used for 10 hours per day. Determine how many of each type of operation the hospital would perform to maximize its revenue.
Number of 30 min operations =?
Number of 1h operations =?
Number of 2h operations = 0
I can't figure out the rest, help please? Thanks in advance !
To determine the number of each type of operation the hospital should perform to maximize revenue, we can set up a linear programming problem.
Let's define:
x = number of 30-minute operations
y = number of 1-hour operations
z = number of 2-hour operations
First, we need to establish the constraints:
1. The hospital can schedule up to 95 total operations per day:
x + y + z ≤ 95
2. The hospital has 5 operating rooms, each of which can be used for 10 hours per day. Let's assume that the operating rooms are completely utilized for each operation:
0.5x + y + 2z ≤ 5 * 10 = 50
3. Since we want to maximize revenue, we will express the revenue as the objective function:
Revenue = 1000x + 1500y + 2000z
Now, we need to find the values of x, y, and z that maximize the revenue while satisfying the constraints.
One approach to solve this linear programming problem is by using graphical methods, but since we have multiple constraints, it's more efficient to use a software solver or a programming tool like Python.
Here's a Python code snippet that utilizes the PuLP library to solve the problem:
```python
from pulp import *
# Create the problem variable
problem = LpProblem("Maximize Revenue", LpMaximize)
# Define the decision variables
x = LpVariable("x", lowBound=0, cat='Integer')
y = LpVariable("y", lowBound=0, cat='Integer')
z = LpVariable("z", lowBound=0, cat='Integer')
# Define the objective function
problem += 1000 * x + 1500 * y + 2000 * z, "Revenue"
# Add the constraints
problem += x + y + z <= 95
problem += 0.5 * x + y + 2 * z <= 50
# Solve the problem
status = problem.solve()
# Print the optimal solution
print("Number of 30 min operations =", value(x))
print("Number of 1h operations =", value(y))
print("Number of 2h operations =", value(z))
```
Running this code will output the optimal number of each type of operation that maximizes revenue.