Chapter 7.3, Problem 94E

Solution

To calculate: The number of each solution should be used if Stephanie's wants to use as little of the 50% solution as possible.

The amounts of the solutions are 0 L of 10% acid solution, $28.8$ L of 25% acid solution, $11.2$ L of 40% acid solution

Given Information:

Stephanie's Drugstore needs to prepare a 40-L mixture that is 32% acid from three solutions: a 10% acid solution, a 25% acid solution, and a 50% acid solution.

Calculation:

Consider the given information,

Suppose that x is representing amount of 10% acid solution, in L, and y is representing amount acid 25% acid solution, in L, and z is representing amount of 50% acid solution, in L.

Now, write the equation, as Stephane’s Drugester needs to prepare a 40-L mixture.

  $x+y+z=40$   ...... (1)

And, the acid solution must be 32%.

  $\begin{array}{c}10\%x+25\%y+50\%z=32\%\left(40\right)\\ 0.10x+0.25y+0.50z=12.8\end{array}$   ...... (2)

It can be observed that combining a 10% and 25% acid solution will never yield a 32% acid solution. Since the both are less than 32% this implies that to not use 10% acid solution at all. So,

  $x=0$

Now, write the system of the matrix as $AX=B$ .

  $A=\left[\begin{array}{ccc}1& 1& 1\\ 0.10& 0.25& 0.50\\ 1& 0& 0\end{array}\right],X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right],\text{ and }B=\left[\begin{array}{c}40\\ 12.8\\ 0\end{array}\right]$

The equation can be solved as,

  $\begin{array}{c}{A}^{-1}AX=A^{-1}B\\ X=A^{-1}B\end{array}$

Solve for  X by multiplying both sides by  $A^{-1}$ . That is,  $X=A^{-1}B$. Using a graphing calculator (as mention in the book), find the solution of the system as shown below:

  

The obtained solutions are defined as,

  $\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 28.8\\ 11.2\end{array}\right]$

Hence, the amounts of the solutions are 0 L of 10% acid solution, $28.8$ L of 25% acid solution, $11.2$ L of 40% acid solution.