A company produces two types of lamps, $L_1$ and $L_2$. Manufacturing $L_1$ requires 20 minutes of manual work and 20 minutes of mechanical work. Manufacturing $L_2$ requires 30 minutes of manual work and 10 minutes of mechanical work. The available manual work is 100 hours per month, and the available mechanical work is 80 hours per month. The profit per unit is K15 for $L_1$ and K10 for $L_2$. The problem is to determine the quantities of each lamp to maximize the profit.

1. Problem Description

A company produces two types of lamps, $L_1$ and $L_2$. Manufacturing $L_1$ requires 20 minutes of manual work and 20 minutes of mechanical work. Manufacturing $L_2$ requires 30 minutes of manual work and 10 minutes of mechanical work. The available manual work is 100 hours per month, and the available mechanical work is 80 hours per month. The profit per unit is K15 for $L_1$ and K10 for $L_2$. The problem is to determine the quantities of each lamp to maximize the profit.

2. Solution Steps

1. Decision Variables:

Let $x$ be the number of lamps of type $L_1$ produced.

Let $y$ be the number of lamps of type $L_2$ produced.

2. Objective Function:

The objective is to maximize the total profit, which is given by:

$Z = 15x + 10y$

3. Constraints:

Manual work constraint: The total manual work required cannot exceed the available manual work.

$20x + 30y \le 100 \times 60$ (converting hours to minutes)

$20x + 30y \le 6000$

Dividing by 10 gives

$2x + 3y \le 600$

Mechanical work constraint: The total mechanical work required cannot exceed the available mechanical work.

$20x + 10y \le 80 \times 60$ (converting hours to minutes)

$20x + 10y \le 4800$

Dividing by 10 gives

$2x + y \le 480$

Non-negativity constraints:

$x \ge 0$

$y \ge 0$

4. Graphing for Optimum Points:

We need to find the feasible region defined by the constraints.

$2x + 3y \le 600$

$2x + y \le 480$

$x \ge 0$

$y \ge 0$

The intersection points are:

- Intersection of $2x + 3y = 600$ and $2x + y = 480$:

Subtract the second equation from the first:

$2y = 120$

$y = 60$

Substitute $y = 60$ into $2x + y = 480$:

$2x + 60 = 480$

$2x = 420$

$x = 210$

Intersection point: $(210, 60)$

- Intersection of $2x + 3y = 600$ and $x = 0$:

$2(0) + 3y = 600$

$3y = 600$

$y = 200$

Intersection point: $(0, 200)$

- Intersection of $2x + y = 480$ and $x = 0$:

$2(0) + y = 480$

$y = 480$

Intersection point: $(0, 480)$

- Intersection of $2x + 3y = 600$ and $y = 0$:

$2x + 3(0) = 600$

$2x = 600$

$x = 300$

Intersection point: $(300, 0)$

- Intersection of $2x + y = 480$ and $y = 0$:

$2x + (0) = 480$

$2x = 480$

$x = 240$

Intersection point: $(240, 0)$

The corner points of the feasible region are $(0, 0)$, $(240, 0)$, $(210, 60)$, and $(0, 200)$.

5. Solving for maximum values:

Evaluate the objective function $Z = 15x + 10y$ at each corner point:

- $(0, 0)$: $Z = 15(0) + 10(0) = 0$

- $(240, 0)$: $Z = 15(240) + 10(0) = 3600$

- $(210, 60)$: $Z = 15(210) + 10(60) = 3150 + 600 = 3750$

- $(0, 200)$: $Z = 15(0) + 10(200) = 2000$

The maximum profit is 3750, which occurs at $(210, 60)$.

6. Required Answers:

To maximize the profit, the company should manufacture 210 lamps of type $L_1$ and 60 lamps of type $L_2$. The maximum profit is K

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3. Final Answer

The company should manufacture 210 lamps of type $L_1$ and 60 lamps of type $L_2$ to maximize profit. The maximum profit is K

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