Problem 56 - Precalculus: Mathematical for Calculus

Question

A roadside fruit stand sells apples at \(75 \notin\) a pound, peaches at \(90 €\) a pound, and pears at \(60 \notin\) a pound. Muriel buys 18 pounds of fruit at a total cost of \(\$ 13.80 .\) Her peaches and pears together cost \(\$ 1.80\) more than her apples. (a) Set up a linear system for the number of pounds of apples, peaches, and pears that she bought. (b) Solve the system using Cramer's Rule.

Step-by-Step Solution

Answer

Muriel bought 18 pounds of apples, 0 pounds of peaches, and 0 pounds of pears.

Step 1: Define Variables

Let \( x \) be the number of pounds of apples, \( y \) be the number of pounds of peaches, and \( z \) be the number of pounds of pears. The equation for the total weight is: \( x + y + z = 18 \).

Step 2: Set Up Cost Equation

The cost equation for all fruits is given by: \(0.75x + 0.90y + 0.60z = 13.80\).

Step 3: Set Up Comparative Cost Equation

The peaches and pears cost \(1.80\) more than apples, so the equation is: \(0.90y + 0.60z = 0.75x + 1.80\).

Step 4: Simplify Comparative Cost Equation

Rearrange the third equation to \(0.90y + 0.60z - 0.75x = 1.80\).

Step 5: Construct the Augmented Matrix

The linear system is represented in matrix form as follows: \[\begin{align*}& \begin{bmatrix}1 & 1 & 1 \0.75 & 0.90 & 0.60 \-0.75 & 0.90 & 0.60\end{bmatrix}\begin{bmatrix}x \y \z\end{bmatrix} =\begin{bmatrix}18 \13.80 \1.80\end{bmatrix}\end{align*}\]

Step 6: Calculate Determinant of Coefficient Matrix

Calculate the determinant of the coefficient matrix, \( D \), which is: \[D = \begin{vmatrix} 1 & 1 & 1 \0.75 & 0.90 & 0.60 \-0.75 & 0.90 & 0.60\end{vmatrix} = 0.75(0.90 - 0.60) - 1(0.90(-0.75) - 0.60(-0.75)) + 1(-0.75*0.90 + 0.75*0.90)\] This simplifies to \( D = 0.225 + 0.225 = 0.45 \).

Step 7: Solve for x Using Cramer's Rule

Replace the first column in the determinant with the constants vector and calculate \( D_x \): \[D_x = \begin{vmatrix} 18 & 1 & 1 \13.80 & 0.90 & 0.60 \1.80 & 0.90 & 0.60\end{vmatrix}\] Calculate \( D_x \) to get \( D_x = 8.1 \). Now, use Cramer’s rule: \( x = \frac{D_x}{D} = \frac{8.1}{0.45} = 18 \).

Step 8: Solve for y Using Cramer's Rule

Replace the second column in the determinant with the constants vector and calculate \( D_y \): \[D_y = \begin{vmatrix} 1 & 18 & 1 \0.75 & 13.80 & 0.60 \-0.75 & 1.80 & 0.60\end{vmatrix}\] Calculate \( D_y \) to get \( D_y = 0 \). Now, use Cramer’s rule: \( y = \frac{D_y}{D} = \frac{0}{0.45} = 0 \).

Step 9: Solve for z Using Cramer's Rule

Replace the third column in the determinant with the constants vector and calculate \( D_z \): \[D_z = \begin{vmatrix} 1 & 1 & 18 \0.75 & 0.90 & 13.80 \-0.75 & 0.90 & 1.80\end{vmatrix}\] Calculate \( D_z \) to get \( D_z = 0 \). Now, use Cramer’s rule: \( z = \frac{D_z}{D} = \frac{0}{0.45} = 0 \).

Key Concepts

Cramer's Rule

Cramer's Rule is a powerful mathematical tool used to solve linear systems of equations with as many equations as unknowns. It works by relating the determinants of matrices. When you have a system of linear equations, you can express these using matrices. Cramer's Rule gives you a way to find solutions for each variable by using the determinant of the matrix of coefficients and the determinant of matrices obtained by replacing one column of the coefficient matrix with constants. To apply Cramer's Rule, follow these steps:

• Write the system of equations in matrix form, where the coefficient matrix is derived from the equations.
• Calculate the determinant of the coefficient matrix, called \(D\).
• Replace one column of this matrix at a time with the constants from the equations and calculate the determinant of this new matrix.
• The solutions for variables \(x, y, z, \) etc., are calculated as the ratio of these new determinants to \(D\): \( x = \frac{D_x}{D}, y = \frac{D_y}{D}, z = \frac{D_z}{D}\), assuming \(D eq 0\).

If the determinant \(D\) is zero, the system may have no solution or infinitely many solutions, hence Cramer's Rule can't be applied in such cases.

Matrix Determinants

A matrix determinant is a special number calculated from square matrices. It's a fundamental concept that has applications in solving linear systems, finding the inverse of a matrix, and understanding properties like eigenvalues and eigenvectors. For a 3x3 matrix such as:\[A = \begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}\]The determinant, \( \det(A) \), is computed using the formula:\[\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)\]In simpler terms, the determinant can be thought of as a scalar value that simplifies solving systems of linear equations. When using Cramer's Rule, calculating determinants correctly is crucial for finding solutions to linear systems.

Augmented Matrix

An augmented matrix is an intuitive way to represent a system of linear equations, incorporating both the coefficients of the variables and the constants from the equations into a single matrix. This representation is beneficial when using techniques such as Gaussian elimination or Cramer's Rule.The layout of an augmented matrix includes:

• The left section (before the separator line) contains the coefficient matrix. The matrix is formed using the coefficients of the variables from the equations.
• The right section after the separator contains the constants from each equation.

For the system of equations related to Muriel’s fruit purchases, we have:\[ \begin{bmatrix} 1 & 1 & 1 & | & 18 \ 0.75 & 0.90 & 0.60 & | & 13.80 \ -0.75 & 0.90 & 0.60 & | & 1.80 \end{bmatrix} \]This matrix combines the coefficients of apples, peaches, and pears with their total costs, enhancing the application of matrix solutions for linear systems.