Question: Lesson title: Solving Linear Programming: Maximization and Minimization Problems (FOR INFORMATION TECHNOLOGY)\nDIRECTIONS: Choose one (1) of the problems below 6ither the maximization or minimization problem.\nSolve the problem thoroughly, showing all pertinent solutions, computations, and graphs.\n\nI. Maximization Problem\n\nAn IT department develops two types of software projects: Mobile Applications ($x$) and Web Applications ($y$). The department wants to maximize its total project revenue.\nEach mobile application earns 6000, while each web application earns 8000.\n\nResource Mobile App ($x$) Web App ($y$) Available\nProgramming hours 3 hours 4 hours 36 hours\nTesting hours 2 hours 1 hour 18 hours\nProfit 6000 8000\n\nProblem: How many mobile applications and web applications should the IT department develop to maximize its total revenue? Task: Using the graphical method, determine the optimal number of mobile and web applications and the maximum possible revenue.
1. **State the problem:**\nWe want to maximize the total revenue from developing mobile applications ($x$) and web applications ($y$) given resource constraints on programming and testing hours.\n\n2. **Define variables:**\nLet $x$ = number of mobile applications developed\nLet $y$ = number of web applications developed\n\n3. **Objective function:**\nMaximize total revenue $$Z = 6000x + 8000y$$\n\n4. **Constraints:**\n- Programming hours: Each mobile app requires 3 hours, each web app requires 4 hours, total available is 36 hours.\n$$3x + 4y \leq 36$$\n- Testing hours: Each mobile app requires 2 hours, each web app requires 1 hour, total available is 18 hours.\n$$2x + y \leq 18$$\n- Non-negativity constraints:\n$$x \geq 0, \quad y \geq 0$$\n\n5. **Graphical method:**\nPlot the constraints on the $xy$-plane:\n- For programming hours: when $x=0$, $y=9$; when $y=0$, $x=12$.\n- For testing hours: when $x=0$, $y=18$; when $y=0$, $x=9$.\n\n6. **Find feasible region:**\nThe feasible region is bounded by the lines $3x + 4y = 36$, $2x + y = 18$, and the axes $x=0$, $y=0$.\n\n7. **Find corner points of feasible region:**\n- Intersection with axes:\n - $(0,0)$\n - $(0,9)$ from $3x + 4y = 36$\n - $(12,0)$ from $3x + 4y = 36$\n - $(0,18)$ from $2x + y = 18$ (outside programming constraint, so not feasible)\n - $(9,0)$ from $2x + y = 18$\n- Intersection of constraints $3x + 4y = 36$ and $2x + y = 18$:\nSolve system:\n$$3x + 4y = 36$$\n$$2x + y = 18$$\nMultiply second equation by 4:\n$$8x + 4y = 72$$\nSubtract first equation from this:\n$$8x + 4y - (3x + 4y) = 72 - 36$$\n$$5x = 36$$\n$$x = \frac{36}{5} = 7.2$$\nSubstitute back to find $y$:\n$$2(7.2) + y = 18 \Rightarrow 14.4 + y = 18 \Rightarrow y = 3.6$$\n\n8. **Evaluate objective function at corner points:**\n- At $(0,0)$: $Z = 6000(0) + 8000(0) = 0$\n- At $(0,9)$: $Z = 6000(0) + 8000(9) = 72000$\n- At $(12,0)$: $Z = 6000(12) + 8000(0) = 72000$\n- At $(7.2,3.6)$: $Z = 6000(7.2) + 8000(3.6) = 43200 + 28800 = 72000$\n- At $(9,0)$: $Z = 6000(9) + 8000(0) = 54000$ (less than others)\n\n9. **Conclusion:**\nThe maximum revenue is 72000, achieved at multiple points: $(0,9)$, $(12,0)$, and $(7.2,3.6)$.\nSince $x$ and $y$ represent number of projects, fractional projects may not be practical, so the department can choose either 12 mobile apps and 0 web apps, or 0 mobile apps and 9 web apps for maximum revenue.\n\n10. **Answer:**\n- Optimal number of mobile applications ($x$): 12 or 0\n- Optimal number of web applications ($y$): 0 or 9\n- Maximum revenue: 72000