Kane Manufacturing has a division that produces two models of hibachis, model A and model B. To produce each model A hibachi requires 6 lb of cast iron and 12 min of labor. To produce each model B hibachi requires 7 lb of cast iron and 6 min of labor. The profit for each model A hibachi is $8, and the profit for each model B hibachi is $7.50. If 2200 lb of cast iron and 40 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?

Question

Kane Manufacturing has a division that produces two models of hibachis, model A and model B. To produce each model A hibachi requires 6 lb of cast iron and 12 min of labor. To produce each model B hibachi requires 7 lb of cast iron and 6 min of labor. The profit for each model A hibachi is $8, and the profit for each model B hibachi is $7.50. If 2200 lb of cast iron and 40 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?

Answer 1

6a + 7b <= 2200 pound iron

12a + 6b <=2400 minutes labor

graph those (a on x and b on y)
line 1 (iron)
(367,0) and (0,314) solution on or below)
line 2 (labor)
(200,0) and (0,400) (solution on or below)
intersection at (75,250)
so test 3 points
(0,314)
(75,250)
(200,0)
with profit = 8x + 7.5 y
point 1
p = 8(0) + 7.5*314 = $2355
point 2
p = 8(75)+7.5(250) = $ 2475 winner
point 3
p = 8(200) +7.5(0) = $1600
point 1 p =

Answer 2

To solve this problem, we can use the concept of linear programming, which involves finding the maximum or minimum value of a linear objective function, subject to linear equality or inequality constraints.

Let's set up the variables:
Let x be the number of model A hibachis produced each week.
Let y be the number of model B hibachis produced each week.

Objective Function:
We want to maximize the total profit, given by:
Profit = (Profit per Model A hibachi) * (Number of Model A hibachis) + (Profit per Model B hibachi) * (Number of Model B hibachis)
Profit = 8x + 7.5y

Constraints:
1. Cast Iron constraint: We have 2200 lb of cast iron available for the production of hibachis each week.
(Weight of cast iron per Model A hibachi) * (Number of Model A hibachis) + (Weight of cast iron per Model B hibachi) * (Number of Model B hibachis) <= 2200
6x + 7y <= 2200

2. Labor constraint: We have 40 hours of labor available for the production of hibachis each week.
(Time taken for Model A hibachi) * (Number of Model A hibachis) + (Time taken for Model B hibachi) * (Number of Model B hibachis) <= 40
12x + 6y <= 40

3. Non-negative constraint:
Since we cannot produce a negative number of hibachis, x and y should be greater than or equal to 0.
x >= 0, y >= 0

Now, we can solve this linear programming problem using graphical method or the simplex method. However, we'll solve this using the graphical method.

Step 1: Graph the feasible region:
Plot the graphs of the above two inequalities (constraint equations) on a graph to form a feasible region.

Step 2: Find the corner points:
Identify the coordinates of the corner points formed by the intersecting lines/inequalities.

Step 3: Evaluate the objective function:
Calculate the profit at each corner point by substituting the values into the objective function. Determine the maximum value.

Let's solve this step-by-step:

1. Plot the inequalities on a graph and identify the feasible region.

Answer 3

To determine the number of hibachis to be produced for each model, we can use linear programming to maximize the profit.

Let's define the decision variables:
- Let x be the number of model A hibachis produced per week.
- Let y be the number of model B hibachis produced per week.

Now, let's establish the constraints based on the available resources:
- The cast iron constraint: Each model A hibachi requires 6 lb of cast iron, and each model B hibachi requires 7 lb. So the total cast iron constraint is: 6x + 7y ≤ 2200 lb.
- The labor constraint: Each model A hibachi requires 12 min of labor, and each model B hibachi requires 6 min. So the total labor constraint is: 12x + 6y ≤ 40 labor-hours.

We also have to consider the non-negativity constraints:
- x ≥ 0
- y ≥ 0

Next, let's determine the objective function to maximize the profit:
- The profit for each model A hibachi is $8, so the profit contribution for model A is 8x.
- The profit for each model B hibachi is $7.50, so the profit contribution for model B is 7.5y.
- The total profit is given by the objective function: Z = 8x + 7.5y.

Now, let's solve the linear programming problem using the constraints and objective function.

The solved values will provide the number of hibachis to be produced for each model in order to maximize the profit.

To find the largest profit the company can realize, plug in the values of x and y obtained from the linear programming solution into the objective function Z = 8x + 7.5y.

To determine if there is any raw material left over, compare the amount of cast iron used in production (6x + 7y) with the available cast iron (2200 lb). If the amount used is less than the available, then there will be raw material left over.