Linear Algebra: Linear Economic Model-Help Please

yj456

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May 3, 2016
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I am stuck with this one problem: Find annual production (output) of coal and steel if total annual production is $20 million.

Know:
  • coal and steel industries form a closed economy
  • Coal Industry:
    • For every $1 coal produced, $0.30 of coal needed and $0.70 of steel needed
  • Steel Industry:
    • For every $1 steel produced, $0.80 of coal needed and $0.20 of steel needed

1st: I created a table to help me understand this problem more
csTotal Produced
coal, c$0.30$0.70$1
steel, s$0.80$0.70$1






2nd: I wrote 2 systems of equations and then rewrote them into homogeneous linear system of equations

c= 0.30c + 0.70s --> 0= -0.70c + 0.70s
s= 0.80c + 0.20s --> 0= 0.80c - 0.80s

3rd: Placed equations into an augmented matrix (imagine this table written as a matrix in augmented form)
-0.700.70 0
0.80-0.80 0

4th: I reduced the matrix into row echelon form

1-10
000

5th: After seeing the matrix, I noticed that there was a free variable. I made s as the free variable and defined it as t because I wanted to make the equation into vertex form. (s=t)
c-s = 0 --> c-t = 0

6th: From there, I found that c equals t. So this means that c also equals s.
c=t
c=s

I know that if I find c, I can find s. This is where I get stuck on. How do I incorporate $20 million to find c?
-------

Wait, now that I think about it, is the answer $10 million because both coal c and steel s are equal to each other. Both c and s need to add up to $20 million.

Is this correct?
 
Last edited:
I am stuck with this one problem: Find annual production (output) of coal and steel if total annual production is $20 million.

Know:
  • coal and steel industries form a closed economy
  • Coal Industry:
    • For every $1 coal produced, $0.30 of coal needed and $0.70 of steel needed
  • Steel Industry:
    • For every $1 steel produced, $0.80 of coal needed and $0.20 of steel needed

1st: I created a table to help me understand this problem more
csTotal Produced
coal, c$0.30$0.70$1
steel, s$0.80$0.70$1






2nd: I wrote 2 systems of equations and then rewrote them into homogeneous linear system of equations

c= 0.30c + 0.70s --> 0= -0.70c + 0.70s
s= 0.80c + 0.20s --> 0= 0.80c - 0.80s

3rd: Placed equations into an augmented matrix (imagine this table written as a matrix in augmented form)
-0.70
0.70 0
0.80-0.80 0

4th: I reduced the matrix into row echelon form

1-10
000

5th: After seeing the matrix, I noticed that there was a free variable. I made s as the free variable and defined it as t because I wanted to make the equation into vertex form. (s=t)
c-s = 0 --> c-t = 0

6th: From there, I found that c equals t. So this means that c also equals s.
c=t
c=s

I know that if I find c, I can find s. This is where I get stuck on. How do I incorporate $20 million to find c?
-------

Wait, now that I think about it, is the answer $10 million because both coal c and steel s are equal to each other. Both c and s need to add up to $20 million.

Is this correct?
Correct. c = s = $10M
 
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