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 August 24th, 2013, 07:19 AM #1 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 necessary & sufficient condition of utility maximization Q. Given the utility function U= f(x,y) & budget constraint Y=px+p'y. Derive the necessary & sufficient conditions of utility maximisation. [x= quantity of commodity A, y=quantity of commodity B, p=price of A, p'=price of B] I don't know how to derive the necessary & sufficient conditions. I can just solve a problem where the utility function and budget constraint are given( through Lagrange's multiplier and bordered Hessian determinant). But yet, I've tried to derive the necessary condition as follows- DERIVATION OF NECESSARY CONDITION Joining the given utility function & budget constraint with the help of Lagrange's multiplier lambda(i don't know how to write lambda here, so i am writing lam. as lambda), we get Z= f(x,y) + lam.(Y-px-p'y)…(1) Now, ${\partial Z}/{\partial x}={\partial U}/{\partial x}- (lam.p)$ ${\partial Z}/{\partial y}={\partial U}/{\partial y}- (lam.p#39" /> ${\partial Z}/{\partial lam.}=Y-px-p#39;y$ Setting these 3 partial derivatives equal to zero, we get Now, ${\partial U}/{\partial x}- (lam.p)=0$ ${\partial U}/{\partial y}- (lam.p'=0" /> $Y-px-p'y=0$ These three equations are combinedly the necessary condition. Is my answer correct? If no, then please tell me the correct way to write the answer.
 August 29th, 2013, 11:12 AM #2 Newbie   Joined: Aug 2013 From: Porto, Portugal Posts: 4 Thanks: 0 Re: necessary & sufficient condition of utility maximization As far as I can see, you can compute the determinant of the bordered Hessian matrix, which relates the second derivatives of the utility function, and by imposing this determinant to be positive, you have a sufficient condition for a maximum. Possibly some condition on second order derivatives of the function (and concavity normally does exactly this) may make this determinant positive. But the most basic sufficient condition you can state is precisely having a bordered hessian with a positive determinant.
August 29th, 2013, 09:17 PM   #3
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Re: necessary & sufficient condition of utility maximization

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 Originally Posted by Filipe Martins As far as I can see, you can compute the determinant of the bordered Hessian matrix, which relates the second derivatives of the utility function, and by imposing this determinant to be positive, you have a sufficient condition for a maximum. Possibly some condition on second order derivatives of the function (and concavity normally does exactly this) may make this determinant positive. But the most basic sufficient condition you can state is precisely having a bordered hessian with a positive determinant.
Please tell me if my answer for the necessary condition is correct or not.

 August 30th, 2013, 03:06 AM #4 Newbie   Joined: Aug 2013 From: Porto, Portugal Posts: 4 Thanks: 0 Re: necessary & sufficient condition of utility maximization I think your necessary condition is correct. The only thing i'm not sure about are the corner solution, with x=0 or y=0. I need they need to be checked separately.
August 30th, 2013, 03:46 AM   #5
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Re: necessary & sufficient condition of utility maximization

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 Originally Posted by Filipe Martins I think your necessary condition is correct. The only thing i'm not sure about are the corner solution, with x=0 or y=0. I need they need to be checked separately.
Thank u so much for your help.

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# State the necessary and sufficient conditions for utility maximization

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