Problem 56 - College Algebra

Question

For the following exercises, write a system of equations that represents the situation. Then, solve the system using the inverse of a matrix. A sorority held a bake sale to raise money and sold brownies and chocolate chip cookies. They priced the brownies at \(\$ 1\) and the chocolate chip cookies at \(\$ 0.75\) . They raised \(\$ 700\) and sold 850 items. How many brownies and how many cookies were sold?

Step-by-Step Solution

Answer

250 brownies and 600 cookies were sold.

Step 1: Define Variables

Let \( x \) represent the number of brownies sold and \( y \) represent the number of cookies sold.

Step 2: Set Up the System of Equations

From the problem, we know two things: the total number of items sold and the total revenue generated. This translates to the following equations:1. \( x + y = 850 \) - represents the total number of items sold.2. \( 1x + 0.75y = 700 \) - represents the total revenue.

Step 3: Write the System as a Matrix Equation

Express the system of equations in a matrix form. We can write it as:\[\begin{bmatrix} 1 & 1 \ 1 & 0.75 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 850 \ 700 \end{bmatrix}\]

Step 4: Find the Inverse of the Coefficient Matrix

First, find the inverse of the coefficient matrix, \(\begin{bmatrix} 1 & 1 \ 1 & 0.75 \end{bmatrix} \). The inverse, if it exists, can be calculated using:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]For our matrix:\( a=1, b=1, c=1, d=0.75 \).Calculate: \( ad-bc = 1(0.75) - 1(1) = -0.25 \).The inverse is:\[A^{-1} = \frac{1}{-0.25} \begin{bmatrix} 0.75 & -1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} -3 & 4 \ 4 & -4 \end{bmatrix}\]

Step 5: Multiply the Inverse Matrix by the Result Matrix

To find the solution vector \(\begin{bmatrix} x \ y \end{bmatrix}\), multiply the inverse matrix by the result matrix:\[\begin{bmatrix} -3 & 4 \ 4 & -4 \end{bmatrix} \begin{bmatrix} 850 \ 700 \end{bmatrix} = \begin{bmatrix} x \ y \end{bmatrix}\]Perform the multiplication:1. For \( x \): \(-3 \times 850 + 4 \times 700 = -2550 + 2800 = 250\)2. For \( y \): \(4 \times 850 - 4 \times 700 = 3400 - 2800 = 600\)Thus, \( x = 250 \) and \( y = 600 \).

Step 6: Conclusion

The sorority sold 250 brownies and 600 chocolate chip cookies.

Key Concepts

Matrix Inverse

When solving a system of equations using the mathematical approach of matrices, it often involves a key step called finding the matrix inverse. Understanding this concept can greatly simplify the process of solving simultaneous equations, especially when dealing with systems that can be represented in matrix form.

A matrix inverse is somewhat akin to how the number 1/x can reverse the multiplication by x in arithmetic. For a matrix \( A \), the inverse is represented as \( A^{-1} \) and satisfies the relationship \( A \cdot A^{-1} = I \), where \( I \) is the identity matrix (a matrix equivalent to a numeral 1 in scalar terms but applicable in multidimensional contexts).

To find the inverse of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \), one can use the formula:
\[ A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \]

This formula only works if the determinant \( ad-bc \) is not zero. If it is zero, the matrix does not have an inverse, and the system cannot be solved using this method. In the problem given, the specific inverse calculated can be used to multiply against the output matrix to achieve the desired solutions for the variables in question.

Bake Sale Problem

In this example, we are tasked with resolving a real-world scenario in which a group conducts a bake sale with distinct pricing for their brownie and chocolate chip cookie items. This problem can be interpreted and solved as a system of equations that represent the constraints and goals of our problem space.

The key to the bake sale problem lies within the two fundamental equations set up based on given data:

• **Total items sold**: Combining both brownies and cookies, we sold 850 items, which becomes the equation \( x + y = 850 \).
• **Revenue generated**: From the sale of these items with differing prices, we obtained \$700 total, forming the second equation \( 1x + 0.75y = 700 \).

By translating this into a matrix equation, we can later solve it through matrix operations highlighted in the previous section. By precisely defining our product as well as payment equations, such systems enable solutions pertaining to how many of each item were actually sold during the bake sale.

Revenue Calculation

Revenue calculation in practical scenarios often involves ensuring accurate equations reflect all monetary aspects of the problem. In the current bake sale problem, revenue intuition extrapolates from how individual sales of brownies and cookies conflate to a total of \\(700.

Consider:

• Each brownie gained \\)1 per sale.
• Each cookie brought in \$0.75 per sale.

Consequently, we derive the fundamental revenue equation \( 1x + 0.75y = 700 \). This equation accounts precisely for the cumulative revenue, indicating how sales get integrated through varied item pricing. As we weave these monetary constraints seamlessly into our earlier established matrix framework, it guards against oversight or miscalculation while allowing the derivation or adjustment of solutions that definitively allocate numbers to each type of item sold.

Thus, applying strategic revenue calculations empowers accuracy and a definitive solution that precisely satisfies given fiscal goals or expectations postulated within the spectrum of sale-oriented problems.