Attachment 28367

in the picture above, shouldn't the last expression be the negative of what it is? Shouldn't it be

(-1)( p1x1 + p2x2 - m ) = -p1x1 - p2x2 + m ?

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- May 14th 2013, 09:22 PMkingsolomonsgraveLagrange method of constrained optimization question
Attachment 28367

in the picture above, shouldn't the last expression be the negative of what it is? Shouldn't it be

(-1)( p1x1 + p2x2 - m ) = -p1x1 - p2x2 + m ? - May 14th 2013, 09:39 PMMarkFLRe: Lagrange method of constrained optimization question
Yes, but since it is equated to zero, they have simplified by multiplying through by negative 1, I am guessing so that there is only 1 negative term, rather than two.

- May 14th 2013, 09:49 PMkingsolomonsgraveRe: Lagrange method of constrained optimization question
Does that mean I could do the same (multiply by -1) to the first two partial derivatives and not change the answer set either?

- May 14th 2013, 10:01 PMMarkFLRe: Lagrange method of constrained optimization question
Yes, algebraically there would be no problem in doing that if you wish. I was taught the equivalent notation:

If given $\displaystyle f\left(x_1,x_2 \right)$ subject to the constraint $\displaystyle g\left(x_1,x_2 \right)=p_1x_1+p_2x_2-m=0$, then we wish to solve the following system:

$\displaystyle f_{x_1}\left(x_1,x_2 \right)=\lambda\cdot g_{x_1}\left(x_1,x_2 \right)$

$\displaystyle f_{x_2}\left(x_1,x_2 \right)=\lambda\cdot g_{x_2}\left(x_1,x_2 \right)$

$\displaystyle p_1x_1+p_2x_2-m=0$

Then we draw an implied relationship between the two independent variables from the first to equations which we then use in the third to get the critical point(s).