Q. 85 - Precalculus Enhanced with Graphing Utilities

Question

Computing the Cost of Production The Acme Steel    Company is a producer of stainless steel and aluminum containers. On a certain day, the following stainless steel containers were manufactured: 500 with 10-gallon (gal) capacity, 350 with 5-gal capacity, and 400 with 1-gal capacity. On the same day, the following aluminum containers were manufactured: 700 with 10-gal capacity, 500 with 5-gal capacity, and 850 with 1-gal capacity.

(a) Find a 2 by 3 matrix representing these data. Find a 3 by 2 matrix to represent the same data. 

(b) If the amount of material used in the 10-gal containers is 15 pounds (lb), the amount used in the 5-gal containers is 8 lb, and the amount used in the 1-gal containers is 3 lb, find a 3 by 1 matrix representing the amount of material used.

(c) Multiply the 2 by 3 matrix found in part (a) and the 3 by 1 matrix found in part (b) to get a 2 by 1 matrix showing the day’s usage of material.

(d) If stainless steel costs Acme \(0.10 lb and aluminum costs \)0.05 lb, find a 1 by 2 matrix representing cost.

(e) Multiply the matrices found in parts (c) and (d) to determine the total cost of the day’s production.

Step-by-Step Solution

Answer

(a)  A 2 by 3 matrix $=\left[\begin{array}{ccc}500& 350& 400\\ 700& 500& 850\end{array}\right]$

      A 3 by 2 matrix $=\left[\begin{array}{cc}500& 700\\ 350& 500\\ 400& 850\end{array}\right]$

(b) A 3 by 1 matrix $=\left[\begin{array}{c}15\\ 8\\ 3\end{array}\right]$

(c) The 2 by 1 matrix $\left[\begin{array}{c}11500\\ 17050\end{array}\right]$ 

(d) The 1 by 2 matrix $\left[\begin{array}{cc}0.10& 0.05\end{array}\right]$

(e)  The total cost of the day's production is $\$2002.5$

Step 1. Given information

On certain day stainless steel containers were manufactured 500 with 10-gallon (gal) capacity, 350 with 5-gal capacity, and 400 with 1-gal capacity 

On the same day aluminum containers were manufactured: 700 with 10-gal capacity, 500 with 5-gal capacity, and 850 with 1-gal capacity. 

Step 2. (a) Find a 2 by 3 matrix representing these data. Find a 3 by 2 matrix to represent the same data.

The table represents the given data

Containers(stainless steel)	500	350	400
Containers(aluminium)	700	500	850
Capacity	10	5	1

A 2 by 3 matrix formed from this data will be $\left[\begin{array}{ccc}500& 350& 400\\ 700& 500& 850\end{array}\right]$

A 3 by 2 matrix representing the same data will be $\left[\begin{array}{cc}500& 700\\ 350& 500\\ 400& 850\end{array}\right]$

Step 3. (b) If the amount of material used in the 10-gal containers is 15 pounds (lb), the amount used in the 5-gal containers is 8 lb, and the amount used in the 1-gal containers is 3 lb, find a 3 by 1 matrix representing the amount of material used.

The table represents the amount of material used for each container

Amount of material used	15	8	3
Capacity	10	5	1

A 3 by 1 matrix formed from this  data will be $\left[\begin{array}{c}15\\ 8\\ 3\end{array}\right]$

Step 4. (c) Multiply the 2 by 3 matrix found in part (a) and the 3 by 1 matrix found in part (b) to get a 2 by 1 matrix showing the day’s usage of material.

For finding the day's ge of materials,multiplay the two matrices $\left[\begin{array}{c}15\\ 8\\ 3\end{array}\right] and \left[\begin{array}{ccc}500& 350& 400\\ 700& 500& 850\end{array}\right]$

$$\begin{gathered} \left[\begin{array}{ccc}500& 350& 400\\ 700& 500& 850\end{array}\right]\left[\begin{array}{c}15\\ 8\\ 3\end{array}\right]=\left[\begin{array}{c}500x15+350x8+400x3\\ 700x15+500x8+850x3\end{array}\right] \\ 2x3 3x1 2x1 \\ =\left[\begin{array}{c}11,500\\ 17,050\end{array}\right] \end{gathered}$$

The 1 by 2 matrix representing the cost per pound for the material used is $\left[\begin{array}{cc}0.10& 0.05\end{array}\right]$

Step 5. (d) If stainless steel costs Acme $0.10 lb and aluminum costs $0.05 lb, find a 1 by 2 matrix representing cost.

The table represents the cost per pound for the material used

Cost per pound

0.10              0.05

The 1 by 2 matrix representing the cost per pound for the material used is $\left[\begin{array}{cc}0.10& 0.05\end{array}\right]$

step 6. (e) Multiply the matrices found in parts (c) and (d) to determine the total cost of the day’s production.

For finding the total cost of the day's production,multiplay the matrices $\left[\begin{array}{c}11500\\ 17050\end{array}\right] and \left[\begin{array}{cc}0.10& 0.05\end{array}\right]$

$$\begin{gathered} \left[\begin{array}{c}11500\\ 17050\end{array}\right]\left[\begin{array}{cc}0.10& 0.05\end{array}\right]=11500(0.10)+17050(0.05) \\ =1150+852.5 \\ =2002.5 \end{gathered}$$

Therefore,the total cost of the day's production is $\$2002.5$