Question

A local sorority sold hot dogs and bratwursts at the spring fling picnics. The first day they sold 8 dozen hot dogs and 13 dozen bratwursts for ​$282.60. The second day they sold 10 dozen hot dogs and 15 dozen bratwursts for a total of ​$333.00. How much did each​ cost? Solve this system using an inverse matrix.​ (Hint: It's easier to leave the dozens in the problem until the last​ step.)

Answer 1

1 hot dog costs $0.75

1 bratwurst costs $1.35

Explanation:

Let x and y be the price per dozen of hot dogs and bratwursts respectively.

The first day they sold 8 dozen hot dogs and 13 dozen bratwursts for ​$282.60

8x + 13y = 282.60

The second day they sold 10 dozen hot dogs and 15 dozen bratwursts for a total of ​$333.00

10x + 15y = 333

and we have the linear system

8x + 13y = 282.60

10x + 15y = 333

which can be written in matrix form as

`\bf \left(\begin{array}{cc}8&13\\10&15\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}282.60\\333\end{array}\right)`

The solution would be given by

`\bf \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}8&13\\10&15\end{array}\right)^(-1)\left(\begin{array}{c}282.60\\333\end{array}\right)`

We have

`\bf \left(\begin{array}{cc}8&13\\10&15\end{array}\right)^(-1)=\left(\begin{array}{cc}-3/2&13/10\\1&-4/5\end{array}\right)`

hence

`\bf \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{cc}-3/2&13/10\\1&-4/5\end{array}\right)\left(\begin{array}{c}282.60\\333\end{array}\right)=\left(\begin{array}{c}9\\ 16.2\end{array}\right)`

Now,

if a dozen hot dogs cost $9, 1 hot dog costs 9/12 = $0.75

if a dozen bratwursts cost $16.2, 1 bratwurst costs 16.2/12 = $1.35