Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

L(x,λ)=1/(3+x₁)-λx₁+7/(2+x₂)-λx₂+4λ

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

share|improve this question
1  
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
add comment

2 Answers

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).

share|improve this answer
add comment

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 $g(y_1)$.)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.