Kane Manufacturing has a division that produces two models of hibachis, model A and model B. To produce each model A hibachi requires 3 lb of cast iron and 6 min of labor. To produce each model B hibachi requires 4 lb of cast iron and 3 min of labor. The profit for each model A hibachi is $7, and the profit for each model B hibachi is $6.50. If 1000 lb of cast iron and 20 labor-hours are available for the production of hibachis each week, how many hibachis of each model should the division produce each week to maximize Kane's profit?
Model A ____ hibachis
Model B ____ hibachis
What is the largest profit the company can realize? $
Is there any raw material left over? (If so, give the amount remaining. If not, enter 0.)
Cast iron ___ lb
labor ___ hr
First of all, "Finite" is not a school subject I've ever heard of and I doubt any tutor here is trained in that subject.
Second of all, you have made no attempt to solve this problem or answer the questions. Tutors will not do your homework for you. We can and will check your work.
To solve this problem, we can use linear programming techniques. Let's define our variables:
Let x = the number of model A hibachis produced per week
Let y = the number of model B hibachis produced per week
Now let's set up the constraints based on the available resources:
1) Cast Iron Constraint: Each model A hibachi requires 3 lb of cast iron, and each model B hibachi requires 4 lb of cast iron. Since we have 1000 lb of cast iron available per week, the constraint is: 3x + 4y ≤ 1000
2) Labor Constraint: Each model A hibachi requires 6 min of labor, and each model B hibachi requires 3 min of labor. Since we have 20 labor-hours available per week (which is equivalent to 1200 minutes), the constraint is: 6x + 3y ≤ 1200
3) Non-negativity Constraint: We cannot produce negative quantities of hibachis, so x ≥ 0 and y ≥ 0
Next, let's define the objective function, which is the profit:
Profit = 7x + 6.50y
To solve this, we can use a linear programming solver or graph the constraints and find the feasible region. Since this is a simplified problem, we can solve it by graphing.
Plotting the constraints on a graph with x and y as the axes, the feasible region is the area of the graph that satisfies all the constraints. The feasible region is bounded by the lines 3x + 4y = 1000 and 6x + 3y = 1200, as well as the x-axis and y-axis.
Now, we need to find the corner points of the feasible region. These corner points can be calculated by solving the system of equations formed by the two constraint lines.
Solving 3x + 4y = 1000 and 6x + 3y = 1200 simultaneously, we find:
x = 200, y = 100 (The intersection point of the two lines - Corner Point 1)
x = 200, y = 0 (The x-axis intercept - Corner Point 2)
x = 0, y = 400 (The y-axis intercept - Corner Point 3)
Now, we substitute these corner points into the objective function Profit = 7x + 6.50y to find the profit at each corner point:
Profit at Corner Point 1 = 7(200) + 6.50(100) = $2,900
Profit at Corner Point 2 = 7(200) + 6.50(0) = $1,400
Profit at Corner Point 3 = 7(0) + 6.50(400) = $2,600
From the profit calculations, we observe that the optimal corner point is Corner Point 1, which gives the maximum profit of $2,900. Therefore, Kane Manufacturing should produce 200 Model A hibachis and 100 Model B hibachis each week to maximize profit.
Additionally, to check if any resources are left over, we can calculate the remaining cast iron and labor using the chosen solution:
Remaining cast iron = 1000 - (3(200) + 4(100)) = 100 lb
Remaining labor = 1200 - (6(200) + 3(100)) = 200 minutes or 3 hours and 20 minutes
Thus, there will be 100 lb of cast iron and 3 hours and 20 minutes of labor remaining.