A company manufactures x units of Product A and y units of Product B, on two machines, I and II. It has been determined that the company will realize a profit of $3/unit of Product A and a profit of $6/unit of Product B. To manufacture a unit of Product A requires 6 min on Machine I and 5 min on Machine II. To manufacture a unit of Product B requires 9 min on Machine I and 4 min on Machine II. There are 5 hr of machine time available on Machine I and 3 hr of machine time available on Machine II in each work shift. How many units of each product should be produced in each shift to maximize the company's profit?

Question

A company manufactures x units of Product A and y units of Product B, on two machines, I and II. It has been determined that the company will realize a profit of $3/unit of Product A and a profit of $6/unit of Product B. To manufacture a unit of Product A requires 6 min on Machine I and 5 min on Machine II. To manufacture a unit of Product B requires 9 min on Machine I and 4 min on Machine II. There are 5 hr of machine time available on Machine I and 3 hr of machine time available on Machine II in each work shift. How many units of each product should be produced in each shift to maximize the company's profit?

(x,y)=?

What is the maximum return?

Answer 1

To solve this problem, we can use linear programming. Let's define the decision variables:

x = number of units of Product A produced
y = number of units of Product B produced

Now, let's set up the objective function and constraints:

Objective function:
Maximize Profit = 3x + 6y (profit per unit of Product A multiplied by the number of units of Product A, plus profit per unit of Product B multiplied by the number of units of Product B)

Constraints:
6x + 9y ≤ 300 (total machine time on Machine I cannot exceed 5 hours or 300 minutes)
5x + 4y ≤ 180 (total machine time on Machine II cannot exceed 3 hours or 180 minutes)
x ≥ 0 (non-negativity constraint for Product A)
y ≥ 0 (non-negativity constraint for Product B)

Now, we can solve this linear programming problem using any solver or graphing software. The solution will provide us with the values of x and y that maximize the profit.

The maximum return is the value of the objective function when x and y are substituted with the optimal values.

Answer 2

To solve this problem, we can use the concept of linear programming, specifically the concept of maximizing a linear objective function subject to linear constraints.

Let's define the decision variables:
x = units of Product A to be produced
y = units of Product B to be produced

Now, let's set up the objective function, which is to maximize the company's profit:
Profit = 3x + 6y

Next, let's set up the constraints:
1. Machine I constraint: Each unit of Product A requires 6 minutes on Machine I, and each unit of Product B requires 9 minutes on Machine I. So, the total time on Machine I needed for the production of x units of Product A and y units of Product B should not exceed the available machine time on Machine I, which is 5 hours or 300 minutes: 6x + 9y <= 300.
2. Machine II constraint: Each unit of Product A requires 5 minutes on Machine II, and each unit of Product B requires 4 minutes on Machine II. So, the total time on Machine II needed for the production of x units of Product A and y units of Product B should not exceed the available machine time on Machine II, which is 3 hours or 180 minutes: 5x + 4y <= 180.
3. Non-negativity constraint: The number of units produced cannot be negative: x >= 0, y >= 0.

Now, let's solve this linear programming problem to find the values of x and y that maximize the profit.

There are several methods to solve this, such as graphical method and simplex method. Here, we will use the simplex method to solve this linear programming problem.

The simplex method involves converting the inequalities into equations and solving the resulting system of equations using matrix operations. However, since this is a straightforward problem, we can simply solve it step by step.

1. Write the equations for the constraints:
6x + 9y = 300 (equation 1)
5x + 4y = 180 (equation 2)

2. Let's rewrite equation 1 to solve for x:
x = (300 - 9y) / 6

Now, substitute this value of x into equation 2:
5((300 - 9y) / 6) + 4y = 180

Simplify this equation:
(1500 - 45y + 24y) / 6 + 4y = 180
(1500 - 21y) / 6 + 4y = 180
(1500 - 21y + 24y) / 6 = 180
(1500 + 3y) / 6 = 180

Multiply both sides of the equation by 6 to eliminate the fraction:
1500 + 3y = 1080

Solve for y:
3y = 1080 - 1500
3y = -420
y = -420 / 3
y = -140

Since y cannot be negative, there is no feasible solution for y.

This means that there is no feasible production plan that will maximize the company's profit.

Therefore, the problem might have been set up incorrectly or there might be other constraints that are not mentioned. Without any feasible solution, we cannot determine the maximum return.

Answer 3

maximize p=3x+6y subject to

6x+9y <= 300
5x+4y <= 180

Now just draw the lines and find the vertices. There are many helpful online sites to verify your work. One is at

http://www.zweigmedia.com/RealWorld/LPGrapher/lpg.html