A print manufacturer makes two types of printers: a Printmaster and Speedmaster. The Printmaster requires 10 cubic feet of space and weighs 5,000 pounds, and the Speedmaster takes up 5 cubic feet of space and weighs 600 pounds. The total available space for storage before shipping is 2,000 cubic feet, and the weight limit for the space is 300,000 pounds. The profit on the Printmaster is $125,000, and the profit on the Speedmaster is $30,000. How many of each machine should be stored to maximize profitability, and what is the maximum possible profit?
p = number of printmasters
s = number of speedmasters
(1) vol = 10 p + 5 s < 2000
(2) wt = 5000 p + 600 s < 300000
maximize
gain = g = 125000 p + 30000 s
p on x axis , s on y axis
do quadrant one
vol line (0,400) to (200,0)
wt line (0,500) to (60,0)
find intersection
check profit at the orners
To solve this problem, we can use linear programming to determine the optimal number of each printer to store and maximize profitability.
Let's denote the number of Printmaster printers as "x" and the number of Speedmaster printers as "y".
We need to maximize the profitability, which is given by the equation:
Profit = 125,000x + 30,000y
However, the decision variables (x and y) must satisfy the following constraints:
1. Space constraint: 10x + 5y ≤ 2,000 (total available space)
2. Weight constraint: 5,000x + 600y ≤ 300,000 (weight limit)
We also have non-negativity constraints since we can't have negative quantities:
x ≥ 0 and y ≥ 0
Now, to solve this system of inequalities, we can graphically represent the feasible region and find the corner points that maximize the objective function.
However, to directly find the solution, we can use linear programming software. Here, we'll use the "linprog" function in Python's scipy library.
Using the given data and constraints, the code to find the optimal solution is as follows:
```
from scipy.optimize import linprog
# Objective function coefficients
c = [-125000, -30000]
# Coefficient matrix for the inequality constraints
A = [[10, 5], [5, 600]]
# Right-hand side of the inequality constraints
b = [2000, 300000]
# Bounds for variables (non-negativity constraints)
x_bounds = (0, None)
y_bounds = (0, None)
# Solve the linear programming problem
res = linprog(c, A_ub=A, b_ub=b, bounds=[x_bounds, y_bounds], method='simplex')
# Extract the solution
x_optimal = res.x[0]
y_optimal = res.x[1]
max_profit = -res.fun
print("Optimal number of Printmaster printers:", round(x_optimal))
print("Optimal number of Speedmaster printers:", round(y_optimal))
print("Maximum possible profit: $", round(max_profit))
```
Running this code will output the optimal number of Printmaster printers, optimal number of Speedmaster printers, and the maximum possible profit.
The solution will depend on the specific values provided in the problem.