Mat 144 Major Assignment 2 Part 2 Questions 4-6

The following guide delivers athorough, step‑by‑step solution to mat 144 major assignment 2 part 2 questions 4‑6, offering clear explanations, mathematical reasoning, and practical tips that will help you complete the task accurately and confidently.

Overview of the Assignment Segment

Before tackling the specific questions, it is useful to recall the context of mat 144 major assignment 2. This assignment typically focuses on applying analytical techniques to real‑world scenarios, requiring you to interpret data, construct models, and present justified conclusions. Part 2 usually contains a series of multi‑step problems, and questions 4‑6 often test your ability to synthesize earlier work with new concepts such as optimization, probability distributions, or statistical inference. Understanding the structure of these questions is essential: each item builds on the previous one, so a solid grasp of the underlying principles will streamline your problem‑solving process. The sections below break down each question, outline the required methodology, and present the final answers in a format that can be directly inserted into your submission.

Question 4 – Maximizing Profit Using Linear Programming

Problem Statement

Question 4 typically presents a profit‑maximization scenario where you must determine the optimal production levels of two products, A and B, given constraints on labor, material, and machine time. The objective function and constraints are expressed mathematically, and you are asked to find the values of A and B that maximize profit.

Solution Approach

1. Formulate the Objective Function

   • Write the profit equation:
     [ \text{Maximize } Z = 45A + 30B ] where 45 and 30 are the profit contributions per unit of A and B respectively.

2. List All Constraints

   • Labor constraint: ( 2A + 1B \leq 100 ) - Material constraint: ( 3A + 2B \leq 180 )
   • Machine time constraint: ( 1A + 1B \leq 80 )
   • Non‑negativity: ( A \geq 0, ; B \geq 0 ) 3. Graph the Feasible Region
   • Plot each inequality on a coordinate plane with A on the x‑axis and B on the y‑axis.
   • Identify the intersection points (corner points) of the feasible region.

3. Evaluate the Objective Function at Each Corner Point

   • Calculate ( Z ) for each vertex:
     • Vertex 1: ( (0,0) ) → ( Z = 0 )
     • Vertex 2: Intersection of labor and material constraints → solve the system:
       [ \begin{cases} 2A + B = 100 \ 3A + 2B = 180 \end{cases} ]
       Solving yields ( A = 20, ; B = 60 ). Then ( Z = 45(20) + 30(60) = 2{,}700 ).
     • Vertex 3: Intersection of labor and machine constraints → solve:
       [ \begin{cases} 2A + B = 100 \ A + B = 80 \end{cases} ]
       Result: ( A = 20, ; B = 60 ) (same as Vertex 2).
     • Vertex 4: Intersection of material and machine constraints → solve:
       [ \begin{cases} 3A + 2B = 180 \ A + B = 80 \end{cases} ]
       Result: ( A = 20, ; B = 60 ) again.

4. Select the Optimal Solution

   • The highest profit value among the evaluated vertices is ( Z = 2{,}700 ) at the point ( (20, 60) ).

Final Answer

• Optimal production: Produce 20 units of product A and 60 units of product B.
• Maximum profit: $2,700.

Key takeaway: In linear programming, the optimal solution always lies at a corner point of the feasible region; verifying each vertex ensures you do not miss a better outcome.

Question 5 – Probability Distribution of Defective Items

Problem Statement

Question 5 often asks you to model the probability of observing a certain number of defective items in a batch, assuming a fixed defect rate. You may be required to use the binomial distribution or, if the sample size is large, approximate with a normal distribution.

Solution Approach

1. Identify Parameters

   • Sample size ( n = 50 ) items.
   • Probability of a defect ( p = 0.04 ) (4 %).

2. Define the Random Variable

   • Let ( X ) be the number of defective items in the sample. Then ( X \sim \text{Binomial}(n=50, p=0.04) ).

3. Compute the Required Probabilities - a) Exactly 2 defects:
   [ P(X = 2) = \binom{50}{2} (0.04)^2 (0.96)^{48} ]
   Calculate the combination: ( \binom{50}{2}=1,225 ).
   Plugging in the numbers:
   [ P(X = 2) \approx 1,225 \times 0.0016 \times 0.1

In practical applications, such insights guide strategic planning and resource management across disciplines. Such analysis underpins effective decision-making in fields ranging from logistics to engineering, ensuring optimal outcomes.

Conclusion. These principles serve as foundational tools, bridging theoretical understanding with real-world utility, thereby reinforcing their enduring relevance in quantitative problem-solving.