# Confounding Lagrange multiplier problem

Optimize $f(x,y,z) = 4x^2 + 3y^2 + 5z^2$ over $g(x,y,z) = xy + 2yz + 3xz = 6$

According to the theorem the gradients must be parallell, $\nabla f = \lambda \nabla g$, so their cross product must equal zero i.e. $\nabla f \times \nabla g = \mathbf{0}$. This results in the following system of equations:

$$\begin{cases} 3y(2y+3x) = 5z(x+2z) \\ 5z(y + 3z) = 4x(2y + 3x) \\ 4x(x+ 2z) = 3y(y+3z) \end{cases}$$

A system of equations which got me stumped, as I am unable to derive anything useful/sensible from it. Yes I did attempt to incorporate the constraint, albeit unsuccessfully.Now since Lagrange multipliers are failing me (?), I have no clue as to how I can progress. Any help?

• That's not the typical route one takes when applying Lagrangian multipliers. Instead, equate the components of $\nabla f$ and $\lambda \nabla g$ while keeping in mind the $g(x,y,z)=6$ constraint. – Semiclassical Aug 12 '14 at 19:55
• That yields the same set of equations :/ – thelionkingrafiki Aug 12 '14 at 20:31

Let $$L(x,y,z,\lambda)=f(x,y,z)-\lambda[g(x,y,z)-6]$$ Then your optimization problem is: $$\left\{\begin{array}{l}0\stackrel{!}{=}\frac{\partial L}{\partial x}=8x-\lambda(y+3z)\\0\stackrel{!}{=}\frac{\partial L}{\partial y}=6y-\lambda(x+2z)\\0\stackrel{!}{=}\frac{\partial L}{\partial z}=10z-\lambda(2y+3x)\\0\stackrel{!}{=}\frac{\partial L}{\partial \lambda}=xy+2yz+3xz-6\end{array}\right.$$ where the first three equations are equivalent to $\nabla f=\lambda\nabla g$, and the last one to the condition $g(x,y,z)=6$. This gives you a much simpler system of equations than the one you proposed above. Can you take it from here?

• Thanks for your reply, but I am afraid that led me to the same set of equations as before. Care to share a clue of some sort? Sidenote: What's up with the exclamation marks above the equal signs? – thelionkingrafiki Aug 12 '14 at 20:32
• @thelionkingrafiki: It only leads you to the same equations if you eliminate $\lambda$ first. Don't! Try to eliminate the terms $xy$, $yz$, $zx$ instead. – Semiclassical Aug 12 '14 at 20:37
• Semiclassical: Oh ok! I've been trying for a while and I am not getting anywhere. It's that $\lambda$ that confuses me as to how I am to eliminate those terms... – thelionkingrafiki Aug 12 '14 at 21:16
• Any help anyone? – thelionkingrafiki Aug 13 '14 at 16:13

To continue from Daniel's answer, add the first three equations together to get:

$$(2-\lambda)(4x+3y+5z)=0$$

This gives two cases:

$$\lambda=2\\ 4x+3y+5z=0$$

Using $\lambda=2$, we can easily solve the others: $x=\pm 1, y=\pm 1, z=\pm 1$. This gives a minimum.

The condition $4x+3y+5z=0$ will lead to contradiction from the first three equations.

• I have made a small edit to correct evident typoes in this note. – colormegone Jul 26 '15 at 7:48

An interesting challenge arises in working with the Lagrange-multiplier method in that all but the simplest problems (generally those involving linear functions and linear constraint equations) can easily lead to systems of multi-linear or non-linear equations, which often do not have general methods of solution. I find that one must draw on an assortment of possible approaches in looking for reasonable routes to a solution.

This problem turns out not to be one where we would solve the set of "Lagrange equations" for expressions for $\ \lambda \$ , which we could then equate and proceed from there. The system that Daniel Robert-Nicoud presents can be arranged in this fashion, which is suggestive:

$$\begin{array}{cccc}8 \ x&- \ \lambda \ y& - \ 3\lambda \ z& = 0\\\lambda \ x& \ -6 \ y& + \ 2\lambda \ z& = 0\\ 3\lambda \ x&+ \ 2 \lambda \ y& \ - 10 \ z& = 0\\\end{array} \ \ .$$

We know that the solution to this system will be non-trivial if the determinant of the coefficients is singular. This is reminiscent of looking for eigenvalues, but we are instead solving the determinant polynomial $\ -12 \ \lambda^3 \ - \ 96 \ \lambda^2 \ + \ 480 \ = \ 0 \$ for values of the Lagrange-multiplier that will allow the values of the variables to be other than all zeroes. We find $\ \lambda \ = \ 2, \ -5 \ \pm \sqrt{5} \$ .

For $\ \lambda \ = \ 2 \$ , we obtain the result indicated by KittyL , $\ x \ = \ y \ = \ z \$ . For the other values of the multiplier, we find (with the acknowledged help of WolframAlpha)

$$\lambda \ = \ -5 \ + \ \sqrt{5} \ \ : \ \ y \ = \ \frac{(5 - \sqrt{5}) \ (10 - 3 \sqrt{5})}{10 \ (2 \sqrt 5 - 3)} \ x \ \approx \ 0.61803 \ x \ \ , \ \$$ $$z \ = \ -\frac{(5 - \sqrt{5}) \ (4 + \sqrt{5})}{10 \ (2 \sqrt 5 - 3)} \ x \ \approx \ -1.17082 \ x \ \ ;$$

$$\lambda \ = \ -5 \ - \ \sqrt{5} \ \ : \ \ y \ = \ -\frac{(5 + \sqrt{5}) \ (10 + 3 \sqrt{5})}{10 \ (2 \sqrt 5 + 3)} \ x \ \approx \ -1.61803 \ x \ \ , \ \$$ $$z \ = \ \frac{(5 + \sqrt{5}) \ (4 - \sqrt{5})}{10 \ (2 \sqrt 5 + 3)} \ x \ \approx \ 0.17082 \ x \ \ .$$

Inserting the result for $\ \lambda \ = \ 2 \$ into the constraint equation produces the simple result $\ 6 \ x^2 \ = \ 6 \ \ \Rightarrow \ \ x \ = \ y \ = \ z \ \pm 1 \$ , for which the value of the function at either point is $\ 4 \ + \ 3 \ + \ 5 \ = \ 12 \$ .

Is there anything else worth mentioning? If we apply the values of the variables for the $\ \lambda \ = \ -5 \ + \ \sqrt{5} \$ case, the constraint equation yields (and I hope I may be forgiven using the decimal approximations)

$$0.61803 \ x^2 \ + \ 2 \ (0.61803) \ (-1.17082) \ x^2 \ + \ 3 \ (-1.17082) \ x^2 \ \approx \ 6$$ $$\Rightarrow \ \ -4.34164 \ x^2 \ = \ 6 \ \ ,$$

which admits no solution. Finally, the third value $\ \lambda \ = \ -5 \ - \ \sqrt{5} \$ leads to

$$-1.61803 \ x^2 \ + \ 2 \ (-1.61803) \ (0.17082) \ x^2 \ + \ 3 \ (0.17082) \ x^2 \ \approx \ 6$$ $$\Rightarrow \ \ -1.65836 \ x^2 \ = \ 6 \ \ .$$

So there is only the one pair of extremal points $\ (1, \ 1, \ 1) \$ and $\ (-1, \ -1, \ -1 ) \$ , and it is easily checked (even just by trying simple integer triples that solve the constraint equation) that the value found for the function is minimal. Because the function and the constraint equation have symmetry about the origin, it would be expected that extremal solutions of the same value would be found in pairs observing this symmetry.

[I reported this approach because it does prove useful in solving such systems of equations. This one had somewhat complicated values for the multiplier, so it is not simply to work through entirely "by hand". (I also thought I had found another extremum of interest, but in typing this out, I uncovered a sign-error in a late stage of my work...) It is also curious that decimal approximations of the "golden ratio" appear; I've no idea whether this is just coincidental. ]