Problem 18 - Algebra and Trigonometry

Question

Use the two steps for solving a linear programming problem, given in the box on page \(888,\) to solve the problems. A large institution is preparing lunch menus containing foods A and \(\mathrm{B}\). The specifications for the two foods are given in the following table: $$\begin{array}{cccc} {} & {} & {\text { Units of }} & {\text { Units of }} \\ {} & {\text { Units of Fat }} & {\text { Carbohydrates }} & {\text { Protein }} \\ {\text { Food }} & {\text { per Ounce }} & {\text { per Ounce }} & {\text { per Ounce }} \\ {\mathrm{A}} & {1} & {2} & {1} \\ {\mathrm{B}} & {1} & {1} & {1} \end{array}$$ Each lunch must provide at least 6 units of fat per serving. no more than 7 units of protein, and at least 10 units of carbohydrates. The institution can purchase food A for 0.12 per ounce and food B for 0.08 per ounce. How many ounces of each food should a serving contain to meet the dietary requirements at the least cost?

Step-by-Step Solution

Answer

The precise values of x (ounces of food A) and y (ounces of food B) that minimize the cost of each serving while still meeting the dietary requirements depend on the intersection points of the constraints and could be calculated using linear programming techniques or graphing utilities. These values will provide the cheapest combination of foods A and B that meet all dietary requirements.

Step 1: Formulate the Constraints of the Problem

To solve this linear programming problem, we need to first set up the constraints that the optimum solution needs to satisfy. These constraints arise from the nutritional requirements specified.\n\nLet \(x\) be the ounces of food A and \(y\) be the ounces of food B in the serving.\n\nBased on the table given, we can set up the following constraints:\n\n1) Fat Units: \(x + y \geq 6\) (each lunch must provide at least 6 units of fat)\n2) Protein Units: \(x + y \leq 7\) (no more than 7 units of protein)\n3) Carbohydrate Units: \(2x + y \geq 10\) (at least 10 units of carbohydrates)

Step 2: Setting up the Objective Function

The institution wants to minimize the cost of each serving while still meeting the dietary requirements, hence, we need to find the minimum value (minima) of the cost. This minimum cost corresponds to our objective function in this case. The cost of each ounce of food A is $0.12 and of food B is $0.08. This leads to our objective function:\n\nCost = \(0.12x + 0.08y\)

Step 3: Solving the Linear Programming Problem

To solve this linear programming problem, we need to locate the feasible region satisfying all the constraints and where the objective function is minimized. This is typically the intersection point (or points) of the constraint boundaries. This can be done using a graphical method or by using linear programming techniques such as the Simplex Method. For this problem, these calculations might be complex and would typically involve using a graphing utility or linear programming software to determine the precise values of x and y that minimize the cost while still meeting all the constraints.

Step 4: Interpreting the Solution

The solution (values of x and y) obtained will represent the number of ounces of each food A and B that the serving should contain to meet the dietary requirements at the minimum possible cost. If there is more than one solution (which may happen in the case of a tie for minimum cost), the institution can choose any of them based on other non-specified preferences (such as taste, availability, etc.).

Key Concepts

Constraints in Linear Programming

In any linear programming problem, constraints are a set of conditions or limitations that the solution must satisfy. They are expressed in the form of linear inequalities or equations. For the lunch menu problem, the constraints come from the nutritional requirements that need to be met for each serving. We are dealing with the following constraints:

• Fat Units Constraint: Each serving needs at least 6 units of fat. This is expressed by the inequality: \(x + y \geq 6\).
• Protein Units Constraint: The total protein should not exceed 7 units per serving, given by: \(x + y \leq 7\).
• Carbohydrate Units Constraint: At least 10 units of carbohydrates should be provided, represented as: \(2x + y \geq 10\).

Incorporating these constraints ensures that the dietary requirements are met adequately. Breaking up the problem into constraints helps in determining the feasible region for solving the linear programming problem.

Understanding the Objective Function

The objective function in a linear programming problem is the function we want to optimize, which means we either aim to maximize or minimize it. In the context of the lunch menu, the goal is to minimize the total cost of the menu while fulfilling the dietary constraints.

• The cost of food A per ounce is \(0.12.
• The cost of food B per ounce is \)0.08.

This gives us the objective function:\[\text{Cost} = 0.12x + 0.08y\]Here, \(x\) represents the ounces of food A and \(y\) denotes the ounces of food B. By minimizing this function, we ensure that the institution spends the least amount of money possible while still meeting all required nutritional standards.

Defining the Feasible Region

The feasible region is the set of all possible solutions that satisfy all the constraints of the problem. This region is typically represented graphically in a plane as the area where all the constraints overlap. For the given problem, the feasible region is determined by the inequalities:

• \(x + y \geq 6\)
• \(x + y \leq 7\)
• \(2x + y \geq 10\)

The intersection of these constraints forms a polygonal area on a graph. Any point within this area is a potential solution that satisfies all the constraints. Finding the optimal solution involves determining which point within this region minimizes the objective function's value.

Minimization Strategy in Linear Programming

Minimization in linear programming is focused on finding the lowest possible value of the objective function while satisfying all constraints. In the case of the lunch menu, we are looking to minimize the cost. Here is how the minimization strategy works:

• The feasible region, as discussed, represents all combinations of quantities for foods A and B that meet nutritional constraints.
• The task is to find the point within the feasible region that results in the lowest value of the cost function \(0.12x + 0.08y\).

Graphical methods or linear programming techniques like the Simplex Method can be employed for this purpose. By evaluating the objective function at each vertex of the feasible region, you can determine which point offers the minimum cost. This method allows you to efficiently choose the best combination of foods A and B for cost savings.