Chapter 4.6, Problem 26PPS

To determine

To find: the milliliters of blue and red food coloring is used to make purple food coloring.

Answer to Problem 26PPS

The red food coloring is 6 milliliters.

The blue food coloring is 19 milliliters.

Explanation of Solution

Given information:

The purple food coloring require 25 milliliters of a 44% concentration food coloring.

The concentration of red and blue concentration of food coloring is 25% and 50%.

Formula used:

Inverse of a $2\times 2$ matrix $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is $A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ is, where $ad-bc\ne 0$ .

The value of a second order determinant is the difference of the products of the products of the two diagonals.

Steps to solve the equations are:-

Step1. Find the inverse of the coefficient matrix.

Step 2. Multiple each side of the matrix equation by the inverse matrix.

Calculation:

According to the question ,

Let x be the red food coloring .

And y be the blue food coloring .

Thus, the system of equation formed is

  $\begin{array}{l}x+y=25\\ 0.25x+0.50y=0.44(25)\\ 0.25x+0.50y=11\end{array}$

The matrix equation is

  $\left[\begin{array}{cc}1& 1\\ 0.25& 0.50\end{array}\right].\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}25\\ 11\end{array}\right]$ .

Step 1.

  $=A^{-1}=\frac{1}{(1)(0.50)-(1)(0.25)}\left[\begin{array}{cc}0.50& -1\\ -0.25& 1\end{array}\right]$

  $\begin{array}{l}=\frac{1}{0.50-0.25}\left[\begin{array}{cc}0.50& -1\\ -0.25& 1\end{array}\right]\\ =\frac{1}{0.25}\left[\begin{array}{cc}0.50& -1\\ -0.25& 1\end{array}\right]\end{array}$

Step2.

  $\begin{array}{l}=\frac{1}{0.25}\left[\begin{array}{cc}0.50& -1\\ -0.25& 1\end{array}\right]\left[\begin{array}{cc}1& 1\\ 0.25& 0.50\end{array}\right].\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{0.25}\left[\begin{array}{cc}0.50& -1\\ -0.25& 1\end{array}\right]\left[\begin{array}{c}25\\ 11\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{0.25}\left[\begin{array}{c}(0.50\times 25)+(-1\times 11)\\ (-0.25\times 25)+(1\times 11)\end{array}\right]\\ =\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{0.25}\left[\begin{array}{c}12.5-11\\ -6.25+11\end{array}\right]\\ =\left[\begin{array}{c}x\\ y\end{array}\right]=\frac{1}{0.25}\left[\begin{array}{c}1.5\\ 4.75\end{array}\right]\end{array}$

Or

  $=\left[\begin{array}{c}x\\ y\end{array}\right]=\left[\begin{array}{c}6\\ 19\end{array}\right]$

Thus, 6 milliliters of red food coloring and 19 milliliters of blue food coloring is required for purple food coloring.