Problem 59 - College Algebra with Modeling and Visualization

Question

Minimizing cost (Refer to Example \(6 .\) ) A breeder is mixing Brand \(A\) and Brand \(B\). Each serving should contain at least 60 grams of protein and 30 grams of fat. Brand A costs 80 cents per unit, and Brand B costs 50 cents per unit. Each unit of Brand A contains 15 grams of protein and 10 grams of fat, whereas each unit of Brand B contains 20 grams of protein and 5 grams of fat. Determine how much of each food should be bought to achieve a minimum cost per serving.

Step-by-Step Solution

Answer

The optimal solution is to buy 2 units of Brand A and 1 unit of Brand B.

Step 1: Define Variables

Let \(x\) be the number of units of Brand A, and \(y\) be the number of units of Brand B.

Step 2: Write Objective Function

The objective is to minimize the cost function \(C = 80x + 50y\), where 80 and 50 are in cents.

Step 3: Write Protein Constraint

Each serving needs at least 60 grams of protein. Brand A provides 15 grams per unit and Brand B provides 20 grams per unit, leading to the constraint \(15x + 20y \geq 60\).

Step 4: Write Fat Constraint

Each serving needs at least 30 grams of fat. Brand A provides 10 grams per unit and Brand B provides 5 grams per unit, leading to the constraint \(10x + 5y \geq 30\).

Step 5: Graph the Constraints

Graph the inequalities \(15x + 20y \geq 60\) and \(10x + 5y \geq 30\) to find the feasible region where both constraints are satisfied. Remember to account for non-negativity \(x \geq 0\), \(y \geq 0\).

Step 6: Solve for Intersection Points

Find the intersection points of the constraint lines by solving the system of equations: \(15x + 20y = 60\) and \(10x + 5y = 30\).

Step 7: Test Vertices of the Feasible Region

Calculate the cost at each vertex of the feasible region. If the intersection points are \((x_1, y_1)\) and \((x_2, y_2)\), check the cost function \(C = 80x + 50y\) at these points and at points such as \((0, y)\) and \((x, 0)\).

Step 8: Determine Minimum Cost

From the calculated costs at each vertex, select the combination of \(x\) and \(y\) that results in the lowest cost. This combination is the optimal solution.

Key Concepts

Objective Function

In linear programming, the objective function is a mathematical expression that we aim to optimize, either by maximizing or minimizing it. In the given exercise, the objective function is designed to minimize the total cost of purchasing Brand A and Brand B. The function is represented as:

• \( C = 80x + 50y \)

Here, \(x\) is the number of units of Brand A, and \(y\) is the number of units of Brand B. The coefficients 80 and 50 are the costs per unit in cents for each brand respectively.
This function reflects the total cost based on the quantities of each brand used.
By setting this as the objective function, we aim to find the optimal amount of each brand that minimizes the total cost while satisfying the nutritional requirements.

Constraints

Constraints are conditions that must be met in a linear programming problem. They define the limits within which the solution must fall, often expressed as inequalities. For the breeder's problem, we have two main constraints related to the nutritional content:

• Protein Constraint: Each serving should include at least 60 grams of protein. This requirement translates into the inequality: \[ 15x + 20y \geq 60 \]
• Fat Constraint: Each serving must also have at least 30 grams of fat, converted into: \[ 10x + 5y \geq 30 \]

These inequalities ensure that any solution involving \(x\) and \(y\) meets the minimum required grams of protein and fat.
In addition, the non-negativity constraints, \(x \geq 0\) and \(y \geq 0\), indicate that the number of units cannot be negative.

Feasible Region

The feasible region in a linear programming problem represents all possible solutions that satisfy the constraints. This region is typically visualized on a graph, plotting the inequalities for the constraints.
In our exercise:

• The inequality \(15x + 20y \geq 60\) forms one boundary.
• The inequality \(10x + 5y \geq 30\) forms another boundary.

The intersection of these constraints along with the non-negativity condition (\(x \geq 0\) and \(y \geq 0\)) forms the feasible region.
This area contains all the pairs \(x, y\) that are possible solutions to the problem. Only solutions within, or on the boundary of, this region can be considered valid for minimizing the cost.

Optimization Problem

An optimization problem involves finding the best solution from a set of feasible solutions. In our case, it means identifying the values of \(x\) and \(y\) that minimize the objective function cost within the feasible region.
To solve the optimization problem:

• First, identify the vertices of the feasible region, which are the intersection points of the constraint lines.
• Calculate the value of the objective function at each vertex.

Each vertex represents a possible solution, but we seek the one that offers the lowest cost.By evaluating the cost at each vertex, we determine the best mix of Brand A and Brand B that achieves the lowest cost while adhering to the nutritional constraints.
This is the optimization process crucial to solving linear programming problems.