# Maths CBA 2 Examples: A Guide for Leaving Certificate Students

Updated: Mar 27

Are you a Leaving Certificate student looking for examples of how to approach your Maths CBA 2? Look no further! In this post, we will provide two examples of how to tackle this important assessment.

## 1. What is the Maths CBA 2?

First, let's start with a brief overview of what the Maths CBA 2 is and why it's important. The Maths CBA 2, also known as the Classroom-Based Assessment 2, is a part of the Leaving Certificate Maths examination. It consists of two tasks, each worth 25% of the total CBA 2 marks. The CBA 2 is designed to assess students' ability to apply their mathematical knowledge and skills to real-world problems.

## 2. Example 1: Maximizing the Area of a Rectangular Enclosure

"A farmer has 100 meters of fencing and wants to use it to create a rectangular enclosure for his cows. He wants the enclosure to have the largest possible area. Determine the dimensions of the enclosure that will maximize the area."

Step 1: Determine the formula for the area of a rectangle.

To solve this problem, we need to use the formula for the area of a rectangle, which is A = lw, where l is the length and w is the width of the rectangle.

Step 2: Use the given information to set up an equation for the perimeter of the rectangle.

Since the farmer has 100 meters of fencing, and he wants to use it all for the enclosure, we can set the perimeter of the rectangle equal to 100, or l + w + l + w = 100.

Step 3: Solve the equation to find the dimensions that will maximize the area.

We can rearrange the equation from Step 2 to get l + w = 50, and then use the area formula from Step 1 to find the dimensions that will maximize the area. Plugging in the value of l + w into the formula, we get A = (50)(w) = 50w. To maximize the area, we need to maximize the value of w. Since w is a side length of the rectangle, it must be a positive number. Therefore, the maximum value of w is 50, and the dimensions of the enclosure that will maximize the area are 50 meters by 50 meters.

## 3. Example 2: Maximizing a Company's Profit with Linear Programming

"A company produces plastic containers in two sizes: small and large. The small containers cost $5 each to produce and the large containers cost $10 each to produce. The company sells the small containers for $10 each and the large containers for $15 each. The company must make at least 100 containers each day and has $1000 to spend on materials each day. Determine the number of small and large containers the company should produce each day to maximize its profit."

Step 1: Use linear programming to set up the objective function and constraints.

To solve this problem, we need to use some basic concepts from linear programming. Linear programming is a mathematical method used to find the maximum or minimum value of a linear objective function, subject to a set of linear constraints.

In this case, the objective function is the company's profit, which is equal to the revenue from selling the containers minus the cost of producing them. The revenue from selling the small containers is $10 per container, and the revenue from selling the large containers is $15 per container. The cost of producing the small containers is $5 per container, and the cost of producing the large containers is $10 per container. Therefore, the company's profit can be represented by the following linear function:

Profit = (10x1 + 15x2) - (5x1 + 10x2)

where x1 is the number of small containers and x2 is the number of large containers.

The linear constraints in this problem are the minimum number of containers the company must make each day and the maximum amount of money the company can spend on materials each day. These constraints can be represented by the following equations:

x1 + x2 >= 100

5x1 + 10x2 <= 1000

To find the values of x1 and x2 that maximize the company's profit, we can use a graphical method known as the feasible region method. This involves graphing the linear constraints on a coordinate plane and identifying the feasible

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