A company produces two products, A and B. Product A sells for $50 and takes 2 hours to make. Product B sells for $80 and takes 3 hours to produce. If the company wants to maximize revenue and has 120 production hours available, how many of each product should they produce?

["Title: How to Maximize Revenue: Optimizing Production of Two Products Under Time Constraints", "In today’s competitive business environment, companies must make strategic decisions to maximize revenue efficiently—especially when resources are limited. One classic challenge is determining the optimal quantity of two products to produce when each has different revenue potential and time requirements.", "In this scenario, a company produces two distinct products:\n- Product A sells for $50 and takes 2 hours to manufacture.\n- Product B sells for $80 and takes 3 hours to produce.", "The company has a total of 120 production hours available. The goal is to determine how many units of Product A and Product B to produce to maximize total revenue.", "---", "### Understanding the Problem", "Let:\n- $ x $ = number of units of Product A\n- $ y $ = number of units of Product B", "Revenue function to maximize:\n$$\n\ ext{Revenue} = 50x + 80y\n$$\nSubject to the time constraint:\n$$\n2x + 3y \leq 120\n$$\nAnd $ x \geq 0 $, $ y \geq 0 $ (non-negative production levels).", "---", "### Analyzing Cost per Hour and Revenue Efficiency", "To make informed decisions, evaluate the revenue generated per hour for each product:\n- Product A: $ \frac{50}{2} = 25 $ dollars per hour\n- Product B: $ \frac{80}{3} \approx 26.67 $ dollars per hour", "Although Product B generates slightly more revenue per hour, the optimal solution depends on combinations within the time limit—not just individual efficiency.", "---", "### Solving the Optimization Problem", "We use linear programming logic by evaluating corner points of the feasible region defined by $ 2x + 3y = 120 $.", "Case 1: Produce only Product A\nSet $ y = 0 $, then $ 2x \leq 120 $ → $ x = 60 $\nRevenue = $ 50 \ imes 60 = 3000 $", "Case 2: Produce only Product B\nSet $ x = 0 $, then $ 3y \leq 120 $ → $ y = 40 $\nRevenue = $ 80 \ imes 40 = 3200 $", "Case 3: Mixed production — solve for maximum revenue", "Express $ x $ in terms of $ y $:\n$ 2x = 120 - 3y $ → $ x = 60 - 1.5y $\nSubstitute into revenue:\n$$\n\ ext{Revenue} = 50(60 - 1.5y) + 80y = 3000 - 75y + 80y = 3000 + 5y\n$$", "Revenue increases with $ y $, so the maximum occurs when $ y $ is as large as possible—within feasible integer values.", "To find optimal integer solution, test values of $ y $ from 0 to 40:", "- At $ y = 40 $, $ x = 0 $ → Revenue = $ 3200 $\n- At $ y = 39 $, $ x = 1.5 $ → Not valid (must be whole units)\n- At $ y = 38 $, $ x = 3 $ → Revenue = $ 50×3 + 80×38 = 150 + 3040 = 3190 $\n- Continue decreasing $ y $ reduces revenue", "Thus, maximum occurs when no Product A is made, and all 120 hours are used for Product B.", "---", "### Final Recommendation", "To maximize revenue within 120 production hours:\n- Produce 0 units of Product A\n- Produce 40 units of Product B\n- Total revenue: $3,200", "This solution leverages Product B’s higher per-hour margin effectively while fully utilizing available capacity.", "---", "### Key Takeaways", "- When resources are limited, strategic trade-offs between time, cost, and revenue are essential.\n- Product B offers better revenue per hour despite higher labor time, making it more profitable overall under the same constraints.\n- Businesses should model production under constraints to identify optimal output mixes.", "Maximizing revenue doesn’t always mean producing everything—sometimes specialization yields the best results.", "---", "By applying mathematical optimization, this company can reliably complete its production plan to achieve maximum earnings efficiently."]