Problem 37 - Algebra 2

Question

BUSINESS A factory is making skirts and dresses from the same fabric. Each skirt requires 1 hour of cutting and 1 hour of sewing. Each dress requires 2 hours of cutting and 3 hours of sewing. The cutting department can cut up to 120 hours each week and the sewing department can sew up to 150 hours each week. If profits are \(\$ 12\) for each skirt and \(\$ 18\) for each dress, how many of each should the factory make for maximum profit?

Step-by-Step Solution

Answer

Produce 120 skirts and 0 dresses for maximum profit.

Step 1: Define the Variables

Let's define two variables: Let \( x \) be the number of skirts produced and \( y \) be the number of dresses produced. The goal is to maximize the profit expressed by the objective function, which is \( P = 12x + 18y \).

Step 2: Establish the Constraints

The problem provides constraints based on the manufacturing process: 1. Cutting hours constraint: \( x + 2y \leq 120 \) (each skirt needs 1 hour and each dress needs 2 hours).2. Sewing hours constraint: \( x + 3y \leq 150 \) (each skirt needs 1 hour and each dress needs 3 hours).3. Non-negativity constraints: \( x \geq 0 \), \( y \geq 0 \).

Step 3: Solve the System of Inequalities Graphically (Feasible Region)

Plot the inequalities on a graph:- For \( x + 2y \leq 120 \), draw the line \( x + 2y = 120 \) and shade the region below it.- For \( x + 3y \leq 150 \), draw the line \( x + 3y = 150 \) and shade the region below it.- The feasible region is bounded by the lines and axes, and it consists of the overlapping shaded area.

Step 4: Identify the Corner Points

Determine the vertices of the feasible region:- The intersection of \( x + 2y = 120 \) and \( x + 3y = 150 \) is found by solving the system: \[ x + 2y = 120 \] \[ x + 3y = 150 \]Subtract the first equation from the second: \[ y = 30 \]Substitute \( y = 30 \) into \( x + 2y = 120 \): \[ x + 2(30) = 120 \implies x = 60 \]So, the intersection point is \((60, 30)\).- The intersection with the axes: \((x = 0, y = 60)\) for \( x + 2y = 120 \) and \((x = 150, y = 0)\) for \( x + 3y = 150 \).- Also, calculate where these lines intercept the axes.

Step 5: Calculate Profit at Each Corner Point

Evaluate the profit function \( P = 12x + 18y \) at each vertex of the feasible region:- At (0, 0), \( P = 12(0) + 18(0) = 0 \)- At (120, 0), \( P = 12(120) + 18(0) = 1440 \)- At (0, 60), \( P = 12(0) + 18(60) = 1080 \)- At (60, 30), \( P = 12(60) + 18(30) = 1260 \)

Step 6: Determine the Maximum Profit

Compare the profit amounts calculated for each vertex:- Maximum profit occurs at \((120, 0)\) with profit \(P = 1440\).

Key Concepts

Objective Function

In any linear programming problem, the objective function is the main goal we are trying to achieve. It is a mathematical expression that represents the profit, cost, or any measurable outcome we want to optimize. In business problems, like the one about skirts and dresses, the objective function usually involves profit maximization.

Here, the objective function is given by the formula:

• \( P = 12x + 18y \), where \( P \) is the profit.
• \( x \) is the number of skirts produced.
• \( y \) is the number of dresses produced.

The coefficients of 12 and 18 represent the profit from each skirt and dress, respectively. Our aim is to find values of \( x \) and \( y \) that maximize \( P \). This is the essence of defining an objective function.

Constraints

Constraints are conditions or restrictions expressed as inequalities that dictate the feasible solutions in a linear programming problem. These are derived from the available resources or limitations in the process.

In our skirt and dress example, the constraints are based on the fabric cutting and sewing hours available each week. They are:

• Cutting hours: \( x + 2y \leq 120 \) (since each skirt needs 1 hour and each dress needs 2 hours).
• Sewing hours: \( x + 3y \leq 150 \) (since each skirt needs 1 hour and each dress needs 3 hours).
• Non-negativity: \( x \geq 0 \), \( y \geq 0 \) (since it doesn't make sense to produce a negative number of items).

Constraints define the boundary of our feasible region where we look for the best possible solution.

Feasible Region

The feasible region is the area on a graph that represents all possible combinations of solutions that satisfy the problem’s constraints. It is a vital concept in linear programming as it visually demonstrates the set of possible solutions.

For the skirts and dresses problem, we plot the constraints as linear equations, then shade the area that meets both conditions:

• The line \( x + 2y = 120 \) represents the cutting constraint.
• The line \( x + 3y = 150 \) represents the sewing constraint.

The feasible region is where these shaded regions overlap, including the axes as boundaries. The key is that any solution to maximize the objective function must lie within this feasible region. Each corner or vertex of this region is a potential candidate for being the optimal solution.

Profit Maximization

Profit maximization is the primary goal in most business-oriented linear programming problems. It involves determining the combination of products or solutions that ensure the highest possible profit.

In solving the skirts and dresses problem, after establishing the objective function and feasible region, we evaluate the profit at each vertex of the feasible region:

• At \((0, 0)\), \(P = 0\).
• At \((120, 0)\), \(P = 1440\).
• At \((0, 60)\), \(P = 1080\).
• At \((60, 30)\), \(P = 1260\).

We calculate the profit by substituting the values of \( x \) and \( y \) into the objective function. The optimal solution is where the profit is highest, which in this case happens at \((120, 0)\), leading to a maximum profit of \(1440\).