# Constrained optimisation question using Lagrange multiplierWatch

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#1
Hi guys,

I am really struggling to find an answer to this question (the answer is that the max utility = 1849). What am I doing wrong? I think my budget constraint is correct and it seems like I have also done the partial differentiation correctly. However, I think I am not solving the system of equations correctly, what am I doing wrong? I have shown some of my working out.
0
#2
Here is the actual question
0
10 months ago
#3
(Original post by Logolept)
Hi guys,

I am really struggling to find an answer to this question (the answer is that the max utility = 1849). What am I doing wrong? I think my budget constraint is correct and it seems like I have also done the partial differentiation correctly. However, I think I am not solving the system of equations correctly, what am I doing wrong? I have shown some of my working out.
Your final bit of solving the linear equations looks a bit complicated?
you have
lambda = x_1
so
2 x_2 - x_1 = -3
x_1 + 2 x_2 = 83
so
x_2 = 20, x1= 43
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#4
(Original post by mqb2766)
Your final bit of solving the linear equations looks a bit complicated?
you have
lambda = x_1
so
2 x_2 - x_1 = -3
x_1 + 2 x_2 = 83
so
x_2 = 20, x1= 43
Yeah, I was trying to solve the system of equations using another method. Funny enough, when I put these equations into Wolfram I get my answer-- thanks for your explanation, I did not spot the simplification.
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#5
Can I also ask you, would there be a general way of solving such equations or do you need to look for these simplifcations.
0
10 months ago
#6
(Original post by Logolept)
Can I also ask you, would there be a general way of solving such equations or do you need to look for these simplifcations.
Without being funny, if you put it into wolfram and got the wrong answer, you must have put the wrong equation in. There is no special trick here, you can solve a system of equations (linear in this case) by elimination (which is what you were trying to do) or substitution (which is what I did first). Elimination is generally used, but here you have a single variable lambda in both equations, and in the second equation you have only x_1 as well. It reduces the problem by one dimension to note x_1 = lambda, even if you then do elimination on this reduced problem, after the substitution. With a simple lagrange problem like this, you'll always have a single lambda appearing the equations (dL/dx), so substitution to get rid of it seems an obvious thing to do and then work in x-space for the remaining equations.
0
#7
(Original post by mqb2766)
Without being funny, if you put it into wolfram and got the wrong answer, you must have put the wrong equation in. There is no special trick here, you can solve a system of equations (linear in this case) by elimination (which is what you were trying to do) or substitution (which is what I did first). Elimination is generally used, but here you have a single variable lambda in both equations, and in the second equation you have only x_1 as well. It reduces the problem by one dimension to note x_1 = lambda, even if you then do elimination on this reduced problem, after the substitution. With a simple lagrange problem like this, you'll always have a single lambda appearing the equations (dL/dx), so substitution to get rid of it seems an obvious thing to do and then work in x-space for the remaining equations.
Much appreciated.
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