A factory manufactures fax machines and scanners. On any given day \( x \) fax machines and \( y \) scanners are produced. Constraints such as labour, time, cost of materials and demand influence the production and profit. These constraints are represented by the following inequalities: \( 3 x+2 y \leq 24 ; x+2 y \leq 12 ; x+y \geq 6 ; x+2 y \geq 8 ; x>0 ; y> \) 0 \( \begin{array}{l}\text { 2.3.1 } \begin{array}{l}\text { Using a scale of } 10 \mathrm{~mm}=1 \text { unit make a graphical } \\ \text { representation of the above constrains, and clearly } \\ \text { indicate the feasible region. } \\ \text { 2.3.2 If the profit } \mathrm{P} \text {, in rands is given by the objective function } \\ y=-\frac{4}{5} x+\frac{p}{200} \text {, determine the profit on each fax } \\ \text { machine and on each scanner. } \\ \text { 2.3.3 } \begin{array}{l}\text { Use a search line to determine how may fax machines } \\ \text { and scanners should be produced per day so as to } \\ \text { yield a maximum profit? }\end{array} \\ \begin{array}{ll}\text { 2.3.4 Calculate the maximum profit. }\end{array}\end{array} \text { (3) } \\ \text { Col }\end{array} \)

To solve the problem step by step, we will address each part of the question systematically. ### 2.3.1 Graphical Representation of Constraints We have the following inequalities representing the constraints: 1. \( 3x + 2y \leq 24 \) 2. \( x + 2y \leq 12 \) 3. \( x + y \geq 6 \) 4. \( x + 2y \geq 8 \) 5. \( x > 0 \) 6. \( y > 0 \) To graph these inequalities, we will first convert them into equations to find the boundary lines. 1. **For \( 3x + 2y = 24 \)**: - When \( x = 0 \): \( 2y = 24 \) → \( y = 12 \) (Point: (0, 12)) - When \( y = 0 \): \( 3x = 24 \) → \( x = 8 \) (Point: (8, 0)) 2. **For \( x + 2y = 12 \)**: - When \( x = 0 \): \( 2y = 12 \) → \( y = 6 \) (Point: (0, 6)) - When \( y = 0 \): \( x = 12 \) (Point: (12, 0)) 3. **For \( x + y = 6 \)**: - When \( x = 0 \): \( y = 6 \) (Point: (0, 6)) - When \( y = 0 \): \( x = 6 \) (Point: (6, 0)) 4. **For \( x + 2y = 8 \)**: - When \( x = 0 \): \( 2y = 8 \) → \( y = 4 \) (Point: (0, 4)) - When \( y = 0 \): \( x = 8 \) (Point: (8, 0)) Next, we will plot these lines on a graph and shade the feasible region based on the inequalities. ### 2.3.2 Profit Calculation The profit \( P \) is given by the objective function: \[ P = -\frac{4}{5}x + \frac{p}{200} \] To determine the profit on each fax machine and scanner, we need to identify the coefficients of \( x \) and \( y \) in the profit function. The coefficient of \( x \) represents the profit per fax machine, and the coefficient of \( y \) represents the profit per scanner. ### 2.3.3 Search Line Method To find the optimal production levels of fax machines and scanners that yield maximum profit, we will use the search line method. This involves drawing lines of constant profit and moving them outward until they no longer intersect the feasible region. ### 2.3.4 Maximum Profit Calculation Once we identify the optimal point from the search line method, we will substitute the values of \( x \) and \( y \) into the profit function to calculate the maximum profit. Now, let's proceed with the calculations for the graphical representation and profit determination. ### Graphical Representation To graph the inequalities, we will find the intersection points of the lines and determine the feasible region. Let's calculate the intersection points of the lines defined by the equations: 1. \( 3x + 2y = 24 \) 2. \( x + 2y = 12 \) 3. \( x + y = 6 \) 4. \( x + 2y = 8 \) We will solve these equations to find the intersection points. ### Finding Intersection Points 1. **Intersection of \( 3x + 2y = 24 \) and \( x + 2y = 12 \)**: - From \( x + 2y = 12 \), we can express \( y \) in terms of \( x \): \[ y = \frac{12 - x}{2} \] - Substitute into \( 3x + 2y = 24 \): \[ 3x + 2\left(\frac{12 - x}{2}\right) = 24 \] \[ 3x + 12 - x = 24 \] \[ 2x = 12 \implies x = 6 \] - Substitute \( x = 6 \) back to find \( y \): \[ y = \frac{12 - 6}{2} = 3 \] - Intersection point: \( (6, 3) \) 2. **Intersection of \( x + 2y = 12 \) and \( x + y = 6 \)**: - From \( x + y = 6 \), express \( y \): \[ y = 6 - x \] - Substitute into \( x + 2y = 12 \): \[ x + 2(6 - x) = 12 \] \[ x + 12 - 2x = 12 \] \[ -x = 0 \implies x = 0 \] - Substitute \( x = 0 \) back to find \( y \): \[ y = 6 - 0 = 6 \] - Intersection point: \( (0, 6) \) 3. **Intersection of \( x + 2y = 8 \) and \( x + y = 6 \)**: - From \( x + y = 6 \), express \( y \): \[ y = 6 - x \] - Substitute into \( x + 2y = 8 \): \[ x + 2(6 - x) = 8 \] \[ x + 12 - 2x = 8 \] \[ -x = -4 \implies x = 4 \] - Substitute \( x = 4 \) back to find \( y \): \[ y = 6 - 4 = 2 \] - Intersection point: \( (4, 2) \) 4. **Intersection of \( 3x + 2y = 24 \) and \( x + 2y = 8 \)**: - From \( x + 2y = 8 \), express \( y \): \[ y = \frac{8 - x}{2} \] - Substitute into \( 3x + 2y = 24 \): \[ 3x + 2\left(\frac{8 - x}{2}\right) = 24 \] \[