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've been struggling with the following problem for hours:

Consider the expression $p^2\frac{x}{y+z}+q^2\frac{y}{x+z}+r^2\frac{z}{x+y}$, where $p,q,r>0$ are parameters. Choose $x,y,z\ge0$ so that the value of the expression is a minimum.

Any ideas to begin with? :/

share|improve this question
    
Are you sure $x = y = z = 0$ is allowed? Hint: the order of $p, q, r$ i.e. which is higher and lower is important.. –  Macavity Mar 9 '13 at 15:18
    
Ofc if more than one of them is zero, then the expression is undefined. However, from that statement, eg. x=0 is permitted. –  Healpowah Mar 9 '13 at 15:21
    
Ok - it's implicit I suppose. Still the hint is valid. –  Macavity Mar 9 '13 at 15:23
    
I somehow get the impression that the expression is a minimum if the 3 additives are equal. Is this right? –  Healpowah Mar 9 '13 at 15:51
1  
@Macavity see my answer here: math.stackexchange.com/questions/318982/… –  Ivan Loh Mar 10 '13 at 4:28

1 Answer 1

up vote 0 down vote accepted

Converting to a new set of variables, we have
$p^2 u + q^2 v + r^2 w$ to be minimised, where $\dfrac{1}{1+u} + \dfrac{1}{1+v}+ \dfrac{1}{1+w} = 2$

From the constraint, we have $w = \dfrac{1- uv}{2 uv + u + v}$, so we can write the unconstrained problem as:
$$\text{Minimise} \quad M = p^2 u + q^2 v + r^2 \dfrac{1- uv}{2 uv + u + v}$$

This should be possible to address easily by taking derivatives w.r.t. $u, v$ and setting them to zero. Using this approach, I get one solution which could be positive: $$ u = \frac{-p + q + r}{2p} \quad \text{and} \quad v = \frac{p-q + r}{2q} \quad \text{and} \quad w = \frac {p + q - r}{2 r}$$

(Note these need not be always positive, unless $p, q, r$ can form a triangle. Also notable is that the weights are smaller, larger the square.)

With this, the minimum value of $M$ is $\quad \dfrac{(p+q+r)^2}{2} - (p^2 + q^2 + r^2)$ which suggests there should be a nice inequality way of getting here as well.

share|improve this answer
    
Thanks. :) But what happens if p,q,r don't form a triangle? eg. $p=1000$, $q=2$, $r=1$. I find that with $x,y,z\ge 0$, the minimum is $4$, with $x=0$, $y=1/2$, $z=1$. There also seems to be a minimum when p is relatively big to q and r. How can you deal with that case? :/ –  Healpowah Mar 9 '13 at 20:54
    
Well, as this is really a convex optimisation problem in $u, v, w$ with convex constraints, in case the global minimum is not in the feasible region, the optimum is on the boundary. In this case, if $p$ is too large, the corresponding weight then has to be on the boundary - viz. the weight is zero, taking $p$ out of the objective function and the rest follows as mentioned earlier in comments with a minimum of $2qr$. –  Macavity Mar 10 '13 at 3:27

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.