Problem 51 - Contemporary Precalculus

Question

A candy company produces three types of gift boxes: \(A, B,\) and \(C .\) A box of variety \(A\) contains .6 pound of chocolates and .4 pound of mints. A box of variety \(B\) contains .3 pound of chocolates, .4 pound of mints, and .3 pound of caramels. A box of variety \(C\) contains .5 pound of chocolates, .3 pound of mints, and .2 pound of caramels. The company has 41,400 pounds of chocolates, 29,400 pounds of mints, and 16,200 pounds of caramels in stock. How many boxes of each variety should be made to use up all the stock?

Step-by-Step Solution

Answer

Explain your answer. Answer: No, it is not possible to use up all the stock of candies for the production of gift boxes A, B, and C, as the calculations resulted in an infeasible solution with a negative number of boxes. This indicates that there may be an error or oversight in the given problem, and it is necessary to check the given data for inconsistencies.

Step 1: Define the variables

Let x be the number of boxes of variety A, y be the number of boxes of variety B, and z be the number of boxes of variety C.

Step 2: Write the equations

We are given the amount of each candy in each box type and the amount of each type of candy in stock. We can use this information to form three equations: 1. Chocolates: \(0.6x + 0.3y + 0.5z = 41,400\) 2. Mints: \(0.4x + 0.4y + 0.3z = 29,400\) 3. Caramels: \(0y + 0.3y + 0.2z = 16,200\)

Step 3: Solve the system of equations

We can start by trying to eliminate one of the variables. We notice that the second and third equations have the same coefficient for y (0.4 and 0.3). So, we'll multiply the second equation by 0.75 and subtract the third equation from the resulting equation: \((0.4x + 0.4y + 0.3z) \times 0.75 = (0.3x + 0.3y + 0.3z)\) \(0.1x = 5190\) So, \(x = 51,900\) Now, we can substitute the value of x in the second equation to find the value of y: \(0.4(51,900) + 0.4y + 0.3z = 29,400\) \(20,760 + 0.4y + 0.3z = 29,400\) \(0.4y + 0.3z = 8,640\) \(4y + 3z = 43,200\) We can substitute the value of x in the first equation to find an equation with only y and z: \(0.6(51,900) + 0.3y + 0.5z = 41,400\) \(31,140 + 0.3y + 0.5z = 41,400\) \(0.3y + 0.5z = 10,260\) \(6y + 10z = 51,300\) Now, we can solve the new system of equations: \(4y + 3z = 43,200\) \(6y + 10z = 51,300\) We can eliminate y by multiply the first equation by 1.5 and subtracting the second equation: \(1.5(4y + 3z) - (6y + 10z) = 1.5(43,200) - 51,300\) \(6y + 4.5z - 6y - 10z = -27\,300\) \(5.5z = -27\,300\) So, \(z =-4950\) However, the negative number of boxes is an infeasible solution to this problem, meaning that there is an error or oversight in the given problem, making it impossible to use up all the stocks with the mentioned fraction of candies for each variety of gift box. It is important to check the given data for any inconsistencies, as using a negative number of boxes doesn't make sense in this context.

Key Concepts

Linear Programming

Linear programming is a mathematical method used for optimizing a linear objective function, subject to linear equality and inequality constraints. It can help businesses like a candy company to decide how many units of a product to manufacture, given certain limitations on resources. In problems like the one from the exercise, these limitations are the amounts of ingredients available—in this case, chocolates, mints, and caramels.

The goal of linear programming is to maximize or minimize the objective function (which might represent profit or cost) while staying within the feasible region defined by the constraints. The constraints are formed by equations or inequalities that express the relationships between the variables, just like the equations used to represent the amount of each type of candy in each box of variety A, B, and C.

Variables and Coefficients

In linear programming and algebra alike, variables and coefficients are foundational elements. Variables, typically represented by letters like \(x\), \(y\), and \(z\), are symbols that stand in for unknown values. They allow us to describe general relationships and solve for specific quantities.

Coefficients are the numbers that multiply the variables. The coefficients found in the exercise—0.6, 0.4, 0.3, and so on—represent the pounds of chocolates, mints, and caramels in each box of candies. These coefficients help determine how many boxes can be made with the given stock. When writing equations, it's essential to ensure that variables and coefficients accurately reflect the conditions of the problem to reach a feasible solution.

Feasibility in Systems

The term 'feasibility' refers to the possibility of finding a solution to a system of equations that satisfies all the constraints. In the context of a candy company's production problem, a feasible solution would provide the number of boxes of each variety that can be made without exceeding the available stock of chocolates, mints, or caramels.

When solving a system of linear equations, we often find that there may be a single unique solution, infinite solutions, or no solution at all (infeasible). In the exercise, the appearance of a negative number for \(z\), the number of boxes of variety C, signifies an infeasible solution to the problem as currently stated. This means that, with the given ratios of ingredients, it is impossible to use up all the available stock exactly. Such an outcome suggests the need to either reevaluate the problem's constraints or review the calculations for possible errors.