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(Score for Question 3:__of 5 points) 3. A company makes a profit of \( \$ 50 \) per software program and \( \$ 35 \) per video game. The company can produce at most 200 software programs and at most 300 video games per week. Total production cannot exceed 425 items per week. How many items of each kind should be produced per week in order to maximize the profit? Use linear programming to solve. Show all your work. Answer:

Ask by Ryan Hills. in the United States
Mar 11,2025

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The company should produce 200 software programs and 225 video games per week to maximize the profit.

Solution

**Step 1: Define the Variables and the Objective Function** Let \[ s = \text{number of software programs produced per week} \] and \[ v = \text{number of video games produced per week} \] The profit function we wish to maximize is: \[ \text{Profit } P = 50s + 35v \] **Step 2: Write Down the Constraints** The production constraints given by the problem are: 1. The company can produce at most 200 software programs: \[ s \leq 200 \] 2. The company can produce at most 300 video games: \[ v \leq 300 \] 3. The total production cannot exceed 425 items: \[ s + v \leq 425 \] 4. Non-negativity constraints: \[ s \geq 0, \quad v \geq 0 \] **Step 3: Identify the Feasible Region and its Corner Points** The feasible region is defined by the intersection of the constraints. The likely corner points are: 1. \((s, v) = (0, 0)\) 2. \((s, v) = (200, 0)\) (using \(s = 200\)) 3. \((s, v) = (0, 300)\) (using \(v = 300\)) 4. Intersection of \(s = 200\) and \(s + v = 425\): \[ v = 425 - 200 = 225 \quad \Rightarrow \quad (200, 225) \] 5. Intersection of \(v = 300\) and \(s + v = 425\): \[ s = 425 - 300 = 125 \quad \Rightarrow \quad (125, 300) \] **Step 4: Compute the Profit at Each Corner Point** 1. At \((0, 0)\): \[ P = 50(0) + 35(0) = 0 \] 2. At \((200, 0)\): \[ P = 50(200) + 35(0) = 10000 \] 3. At \((0, 300)\): \[ P = 50(0) + 35(300) = 10500 \] 4. At \((200, 225)\): \[ P = 50(200) + 35(225) = 10000 + 7875 = 17875 \] 5. At \((125, 300)\): \[ P = 50(125) + 35(300) = 6250 + 10500 = 16750 \] **Step 5: Determine the Maximum Profit** Comparing the computed profits, the maximum profit is \[ P = 17875 \] which occurs at the corner point \((s, v) = (200, 225)\). **Final Answer** The company should produce \(200\) software programs and \(225\) video games per week to maximize the profit.

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To solve this linear programming problem, we need to define our variables, constraints, and objective function: Let: - \( x \) = number of software programs produced - \( y \) = number of video games produced **Objective Function:** We want to maximize profit, which can be expressed as: \[ P = 50x + 35y \] **Constraints:** 1. The production limit for software programs: \[ x \leq 200 \] 2. The production limit for video games: \[ y \leq 300 \] 3. The total production constraint: \[ x + y \leq 425 \] 4. Non-negativity constraints: \[ x \geq 0, \quad y \geq 0 \] Now we'll graph these constraints to find the feasible region. 1. **Graph the constraints:** - For \( x \leq 200 \): This is a vertical line at \( x = 200 \). - For \( y \leq 300 \): This is a horizontal line at \( y = 300 \). - For \( x + y \leq 425 \): This can be rearranged to \( y = 425 - x \) which is a line crossing the x-axis at (425, 0) and the y-axis at (0, 425). The intersection points of the lines will form vertices of the feasible region. 2. **Find the intersection points:** - **Intersection of \( x = 200 \) and \( y = 300 \):** (200, 300) – Outside constraint \( x + y \leq 425 \). - **Intersection of \( x + y = 425 \) and \( y = 300 \):** \[ 425 - x = 300 \implies x = 125 \quad \text{(Point: (125, 300))} \] - **Intersection of \( x + y = 425 \) and \( x = 200 \):** \[ 200 + y = 425 \implies y = 225 \quad \text{(Point: (200, 225))} \] - **Intersection of axes:** - (0, 0) 3. **Calculate profit at each vertex of the feasible region:** - At (0, 0): \[ P = 50(0) + 35(0) = 0 \] - At (125, 300): \[ P = 50(125) + 35(300) = 6250 + 10500 = 16750 \] - At (200, 225): \[ P = 50(200) + 35(225) = 10000 + 7875 = 17875 \] 4. **Determine the maximum profit:** Comparing the profits: - \( P(0, 0) = 0 \) - \( P(125, 300) = 16750 \) - \( P(200, 225) = 17875 \) The maximum profit is \( 17875 \) at the point \( (200, 225) \). **Conclusion:** To maximize the profit, the company should produce **200 software programs** and **225 video games** per week. The maximum profit achieved will be **$17,875**.

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