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Problem 1

# A company manufactures two products, $$A$$ and $$B$$, on two machines, $$\bar{I}$$ and II. It has been determined that the company will realize a profit of $$\ 3$$ on each unit of product $$A$$ and a profit of $$\ 4$$ on each unit of product $$\mathrm{B}$$. To manufacture a unit of product A requires $6 \mathrm{~min}$$on machine$$\mathrm{I}$$and$$5 \mathrm{~min}$ on machine II. To manufacture a unit of product B requires 9 min on machine $$\mathrm{I}$$ and $4 \mathrm{~min}$$on machine$$\mathrm{II}$$. There are$$5 \mathrm{hr}$ of machine time available on machine $$\mathrm{I}$$ and $$3 \mathrm{hr}$$ 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?

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The company should produce 0 units of product A and 45 units of product B in each shift to maximize their profit, which will be 180 dollars.
See the step by step solution

## Step 1: Define Variables

Let's denote x as the number of units of product A and y as the number of units of product B.

## Step 2: Determine Constraints

We have two machines and the processing time allotted for products A and B are given. To write the inequalities, we will consider the total time consumption of machines I and II for both products, which must not exceed their available time in each shift. Machine I: $$6x + 9y \leq 5(hour)\cdot60(minutes)$$ (time used by products A and B should be under 5 hours) Machine II: $$5x + 4y \leq 3(hour)\cdot60(minutes)$$ (time used by products A and B should be under 3 hours) We also have non-negativity constraints, since the number of products cannot be negative: $$x \geq 0$$ $$y \geq 0$$

## Step 3: Define the Objective Function

The objective is to maximize the profit. The profit function is given by: $$P(x, y) = 3x + 4y$$

## Step 4: Graph the Constraints and Find the Feasible Region

Graph the inequalities: Machine I: $$6x + 9y \leq 300$$ Machine II: $$5x + 4y \leq 180$$ Along with non-negativity constraints: $$x \geq 0$$ $$y \geq 0$$ The feasible region is the area where all constraints are satisfied.

## Step 5: Find the Vertices of the Feasible Region

Now, we need to find the vertices of the feasible region, which are the points where the constraints intersect. It is where the optimal solution will be located. Vertices: 1. Intersection of Machine I and the x-axis (y = 0): $$6x = 300$$, x = 50 2. Intersection of Machine II and the x-axis (y = 0): $$5x = 180$$, x = 36 3. Intersection of Machine I and the y-axis (x = 0): $$9y = 300$$, y = 100/3 4. Intersection of Machine II and the y-axis (x = 0): $$4y = 180$$, y = 45 5. Intersection of Machine I and Machine II: Solve the system of equations $$6x + 9y = 300$$ $$5x + 4y = 180$$ After solving the system of equations, we get x = 20 and y = 20. Our vertices are: 1. (50,0): $$P(50,0) = 3(50) + 4(0) = 150$$ 2. (36,0): $$P(36,0) = 3(36) + 4(0) = 108$$ 3. (0,100/3): $$P(0,100/3) = 3(0) + 4(100/3) = 400/3$$ 4. (0,45): $$P(0,45) = 3(0) + 4(45) = 180$$ 5. (20,20): $$P(20,20) = 3(20) + 4(20) = 140$$

## Step 6: Determine the Optimal Solution

Compare the profit function values at the feasible region's vertices and select the one that maximizes profit. Maximized profit is achieved at the point (0,45), which gives a profit of 180 dollars. Therefore, the company should produce 0 units of product A and 45 units of product B in each shift to maximize their profit.

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