Gauss-Jordan help

tony77

New member
Hi all, I could use some help on setting up and solving (mainly setting up) two word problems involving Gauss-Jordan for linear Algebra.

  1. "a coffee dealer has 3 kinds of beans costing $3.00, $3.75, and $4.00 per pound. If the dealer wants 100 ln of a blend that costs $3.60 per lb, what amounts of the 3 beans can they use?"
  2. UC browser SHAREit Appvn
2) "A company has on hand 60 lb of chemical A and 50 lb of chemical B that it wants to use up. Each unit of product 1 uses 3 lb of A and 2 lb of B, which each unit of product 2 uses 1 lb of A and 1 lb of B. Can the company use up all of A and B in making the two products, if so how many units of each should it make? (# of each units of each product need not be an integer)"

Any help will be appreciated!
 
Last edited:

MarkFL

La Villa Strangiato
Math Helper
1.) Let's let \(x\) be the amount of the $3.00 per pound bean, \(y\) be the amount of the $3.75 per pound bean and \(z\) be the amount of the $4.00 per pound bean used to make the blend. The dealer wants to have 100 pounds of the blaned, so we have:

[MATH]x+y+z=100[/MATH]
Because the dealer wants the blend to cost $3.60/pound, we have:

[MATH]3.00x+3.75y+4.00z=100\cdot3.60[/MATH]
Or:

[MATH]12x+15y+16z=1440[/MATH]
Because we have 3 variables, and only two constraints, we can let \(z=t\), and set up our augmented matrix as follows:

[MATH]\left[\begin{array}{ccc|c}1 & 1 & 1 & 100 \\ 12 & 15 & 16 & 1440 \\ 0 & 0 & 1 & t \end{array}\right][/MATH]
The first operation I would perform is [MATH]R_2-12R_1[/MATH] to get:

[MATH]\left[\begin{array}{ccc|c}1 & 1 & 1 & 100 \\ 0 & 3 & 4 & 240 \\ 0 & 0 & 1 & t \end{array}\right][/MATH]
Next [MATH]\frac{1}{3}R_2[/MATH]
[MATH]\left[\begin{array}{ccc|c}1 & 1 & 1 & 100 \\ 0 & 1 & \dfrac{4}{3} & 80 \\ 0 & 0 & 1 & t \end{array}\right][/MATH]
[MATH]R_1-R_2[/MATH]
[MATH]\left[\begin{array}{ccc|c}1 & 0 & -\dfrac{1}{3} & 20 \\ 0 & 1 & \dfrac{4}{3} & 80 \\ 0 & 0 & 1 & t \end{array}\right][/MATH]
Finally [MATH]R_1+\frac{1}{3}R_3[/MATH] and [MATH]R_2-\frac{4}{3}R_3[/MATH]
[MATH]\left[\begin{array}{ccc|c}1 & 0 & 0 & 20+\dfrac{1}{3}t \\ 0 & 1 & 0 & 80-\dfrac{4}{3}t \\ 0 & 0 & 1 & t \end{array}\right][/MATH]
Since no amount can be negative, we require:

[MATH]0\le t[/MATH]
[MATH]0\le80-\frac{4}{3}t\implies t\le60[/MATH]
And so we have:

[MATH](x,y,z)=\left(\frac{60+t}{3},\frac{4(60-t)}{3},t\right)[/MATH] where \(0\le t\le60\)
 

Jason

Well-known member
HA. And @MarkFL said he didn't take linear algebra? ;)
 

MarkFL

La Villa Strangiato
Math Helper
Gaussian elimination (aka the Gauss-Jordan method) is something I learned in College Algebra, and then again in Precalculus. Learning to work with a parameter is something I picked up along the way on the forums. :)
 
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