Introduction (Grade 11 NSC Matric Mathematics): Revision Notes

Worked Example: Bicycle Manufacturing

Problem: Mr. Hunter manufactures two types of bicycles - Mountees and Roadees. He can produce a maximum of 5 Mountees per day and 3 Roadees per day. Each Mountee needs 1 technician and each Roadee needs 2 technicians to assemble. He has 8 technicians total. The profit on a Mountee is R800 and on a Roadee is R2400. Find the number of each bicycle to manufacture for maximum profit.

Solution:

Step 1: Assign variables

• Let $M$ = number of Mountees produced per day
• Let $R$ = number of Roadees produced per day

Step 2: Organize the information

• Maximum Mountees: $M \leq 5$
• Maximum Roadees: $R \leq 3$
• Technician constraint: $M + 2R \leq 8$
• Profit per Mountee: R800
• Profit per Roadee: R2400

Step 3: Create a table of all possible combinations

Step 4: Apply the technician constraint We eliminate combinations where $M + 2R > 8$:

Step 5: Calculate profits The profit equation is: $P = 800M + 2400R$

Step 6: Find the maximum The maximum profit of R8800 occurs when $M = 2$ and $R = 3$.

Worked Example: Bicycle Problem Using Graphs

Using the same bicycle manufacturing problem:

Step 1: Write constraints as inequalities

• $M \leq 5$ (maximum Mountees)
• $R \leq 3$ (maximum Roadees)
• $M + 2R \leq 8$ (technician constraint)
• $M \geq 0, R \geq 0$ (cannot produce negative bicycles)

Step 2: Write the objective function $P = 800M + 2400R$ (to be maximized)

Step 3: Draw the constraints on a graph

Step 4: Show the technician constraint From $M + 2R \leq 8$, we get $R \leq -\frac{1}{2}M + 4$

Step 5: Identify the feasible region The feasible region is where all constraints are satisfied:

Step 6: Find the optimal point Test the corner points of the feasible region. The maximum profit of R8800 occurs at point A(2,3).

Worked Example: Furniture Store (Minimization)

Problem: A furniture store wants to give away at least 40 prizes (kettles and toasters). They need at least 10 of each type. A kettle costs R120 and a toaster costs R100. Find the combination that minimizes cost.

Step 1: Assign variables

• Let $k$ = number of kettles
• Let $t$ = number of toasters

Step 2: Write constraints

• $k \geq 10$ (minimum kettles)
• $t \geq 10$ (minimum toasters)
• $k + t \geq 40$ (minimum total prizes)

Step 3: Write objective function $C = 120k + 100t$ (to be minimized)

Step 4: Graph the constraints

Step 5: Find the feasible region and optimal point

The minimum cost occurs at point A(10,30), giving a cost of R4200.