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I am using the lagrangian to find the minmum of the function:

min f(x₁,x₂)=1/(3+x₁)+7/(2+x₂)

s.t    x₁+x₂=4

Then the lagrangian is


I am stuck at this point because the standard method for solving grad L is tricky.

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Do you have to use the Lagrangian? What if you just use the constraint to eliminate $x_2$? – littleO Nov 3 '12 at 6:29

First, the problem has no solution in that it is unbounded below. Let $x_\epsilon = -3 -\epsilon, y_\epsilon = 7+\epsilon$, with $\epsilon>0$. Then $(x_\epsilon, y_\epsilon)$ satisfies the constraint and the cost goes to $-\infty$ as $\epsilon \to 0$.

However, finding stationary points of the Lagrangian is pretty straightforward, yielding $$\pmatrix{-\frac{1}{(3+x_1)^2} \\ - \frac{7}{(2+x_2)^2}}+ \lambda \pmatrix{1 \\ 1 } = 0$$ This gives $\frac{1}{(3+x_1)^2} = \lambda = \frac{7}{(2+x_2)^2}$, from which we obtain $|3+x_1| = \frac{1}{\sqrt{7}}|2+x_2|$. There are only two possibilities to consider, (1) $3+x_1 = \frac{1}{\sqrt{7}}(2+x_2)$ and (2) $3+x_1 = -\frac{1}{\sqrt{7}}(2+x_2)$.

Solving these results in (1) $x = \frac{-9+3 \sqrt{7}}{2}, y = \frac{17-3 \sqrt{7}}{2}$, and (2) $x = \frac{-9-3 \sqrt{7}}{2}, y = \frac{17+3 \sqrt{7}}{2}$. Substituting these values in gives (1) $\frac{2\sqrt{7}}{4\sqrt{7}-7}$ and (2) $\frac{2\sqrt{7}}{4\sqrt{7}+7}$. However, only (1) could qualify as a minimum in the sense that it does minimize the cost on the set $(-3,\infty)\times (-2,\infty)$ (subject to the constraint, of course).

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Maybe the substitution $y_1=3+x_1$ and $y_2=2+x_2$ makes this a little tidier.

After the substitution you want to minimize $f(y_1,y_2)=\frac1{y_1}+\frac7{y_2}$ with the condition $y_1+y_2=7$.

This is the same as minimizing the function $g(y_1)=\frac1{y_1}+\frac7{7-y_1}$ of one variable - so you just need to find critical points.

WolframAlpha: extrema 1/t+7/(7-t)

If you want to use Lagrange multipliers anyway, you get $L=\frac1{y_1}+\frac7{y_2}+\lambda(7-y_1-y_2)$, which leads to the equations $$ \begin{align*} -\frac1{y_1^2}-\lambda&=0\\ -\frac7{y_2^2}-\lambda&=0\\ y_1+y_2&=7 \end{align*} $$ From the first two equations you get $-\lambda=\frac1{y_1^2}=\frac7{y_2^2}$, which implies $y_2^2=7y_1^2$. Using the third equation you can modify this to $(7-y_1)^2=7y_1^2$. (BTW this is precisely the same equation you get when you look for critical points of $g(y_1)$.)

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