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Ok.so I do understand up to step 9. But then it gets all confusing to me...Normally what I would do for these sort of problems is to isolate the lambda symbols in equations 8 and 9 and then equal them, so I can isolate lets say x1. I worked it out like that : 2x1+b=4x2 so x1=2x2-b and x2=(x1+b)/2 I would plug this x1 in equation ( 9 ) in order to get the minimazing cost function for firm one, I would proceed equally for x2. finally I would plug this cost minimazing functions into the objective function inorder to get the final function . And all that is totally different from the way that this problem is solved. Could anyone explain to me why?

Maximilian1988
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2 Answers2

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First : 2 x1 + b = 4 x2 leads to x1 = 2 x2 - b /2. Second : you can handle the problem the manner you want as long as you accept the Lagrangian and equations 7, 8 and 9 which are just the derivatives wtith respect to x1, x2 and lambda. I hope this helps. If not, post a question to me.

Claude Leibovici
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  • (Ill call x2 "y" ) ok , so what I do is isolate lagrange multiplier in equations 7 and 8, then I equal them and I have 2x + b = 4y, I then isolate y and I have y=(2x+b)/4 . I plug this y in equation 9 and what I get at the end is x=(400-b)/3. So what am I doing qring? why this x doesnt match equation 13 from the book? – Maximilian1988 Oct 24 '13 at 02:27
  • As Michael wrote, you have three equations (7-8-9) to solve for the three unknowns x1, x2 and lambda. You must get rid of lambda and finally express x1 and x2 which depend on b. The solution in your book is correct. – Claude Leibovici Oct 24 '13 at 03:34
  • well, yes but as I write above I also get rid of Lambda and express x1 in terms of b, why then my answer differs from the one in the book? – Maximilian1988 Oct 25 '13 at 02:47
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There are three things we don't know - $x_1$, $x_2$ and $\lambda$.
We have three equations, which are usually enough information. Two equations are not. The book gives one of many ways to solve the three equations.
If you substitute $x_1=2x_2-b/2$ back into the cost minimizing function, it still has $x_2$ in it, which you don't know. Then, if you put $x_2=x_1/2+b/4$ in, it now has $x_1$ which you don't know. Somehow, you have to get rid of both at the same time. So you need all three equations 7, 8 and 9.

Empy2
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  • (Ill call x2 "y" ) ok , so what I do is isolate lagrange multiplier in equations 7 and 8, then I equal them and I have 2x + b = 4y, I then isolate y and I have y=(2x+b)/4 . I plug this y in equation 9 and what I get at the end is x=(400-b)/3. So what am I doing qring? why this x doesnt match equation 13 from the book? – Maximilian1988 Oct 24 '13 at 02:20