Chapter 7.4, Problem 97E

To determine

To find: The number of pounds of each grade of paper by solving the system of equations using matrices.

Answer to Problem 97E

Number of pounds of glossy $=50$ .

Number of pounds of semi-glossy $=35$ .

Number of pounds of matte $=15$ .

Explanation of Solution

Given information: 100 pound package contains 3 grades of papers, Glossy which costs $5.50 per pound, semi-gloss costing $4.25 per pound and Matte costs $3.75 per pound. The cost of 100 pound package is $480. Half of 100 pound is less expensive grade papers.

Calculation:

Let,

  $x=$ Number of pounds of glossy.

  $y=$ Number of pounds of semi-glossy.

  $z=$ Number of pounds of matte.

Using the given information we can make the system of equations as,

  $\begin{array}{l}x+y+z=100\\ y+z=50\\ 5.5x+4.25y+3.75z=480\end{array}$

Now the augmented matrix can be written as,

  $\left(\begin{array}{cccc}1& 1& 1& 100\\ 0& 1& 1& 50\\ 5.5& 4.25& 3.75& 480\end{array}\right)$

Using row transformations,

  $\begin{array}{l}{R}_3\to R_3-5.5R_1\\ \left(\begin{array}{cccc}1& 1& 1& 100\\ 0& 1& 1& 50\\ 0& -1.25& -1.75& -70\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_1\to R_1-R_2\\ R_3\to R_3+1.25R_2\\ \left(\begin{array}{cccc}1& 0& 0& 50\\ 0& 1& 1& 50\\ 0& 0& -0.5& -7.5\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_3\to -2R_3\\ \left(\begin{array}{cccc}1& 0& 0& 50\\ 0& 1& 1& 50\\ 0& 0& 1& 15\end{array}\right)\end{array}$

  $\begin{array}{l}{R}_2\to R_2-R_3\\ \left(\begin{array}{cccc}1& 0& 0& 50\\ 0& 1& 0& 35\\ 0& 0& 1& 15\end{array}\right)\end{array}$

The matrix is now in reduced row echelon form.

Converting back to the system of linear equations,

  $x=50,y=35,z=15$ .

Therefore,

Number of pounds of glossy $=50$ .

Number of pounds of semi-glossy $=35$ .

Number of pounds of matte $=15$ .