Find the minimum using lagrange

Question

I've got the formula $$x^3+y^3+z^3$$

With the constraint $ax+by+cz = 1$

I tried to solve this using lagrange but every possible way I try to use does not get me to the right answer

Using the lagrange I got to $$3x^2+ka = 0 $$ $$3y^2+kb = 0 $$ $$3z^2+kc = 0 $$

=

$$3x^2bc+kabc = 0 $$ $$3y^2ac+kabc = 0 $$ $$3z^2ab+kabc = 0 $$

$$z = \sqrt{\frac{cy^2}{b}} $$

$$x = \sqrt{\frac{ay^2}{b}} $$

Fill this in the constraint and I then I cannot figure it out

Any got a clue how to solve this properly?

Answer 1

Assuming $x \ge 0, y \ge 0, z \ge 0$ to assure a bounded solution, we have from the stationary conditions

$$ x = \sqrt\frac{\lambda a}{3}\\ y = \sqrt\frac{\lambda b}{3}\\ z = \sqrt\frac{\lambda c}{3}\\ $$

and substituting into the restriction

$$ \frac{\lambda}{3} = \frac{1}{(a\sqrt a+b\sqrt b+c\sqrt c)^2} $$

but

$$ x^3+y^3+z^3 = \left(\frac{\lambda}{3}\right)^{\frac 32}(a\sqrt a+b\sqrt b+c\sqrt c) $$

hence

$$ \left\{\min_{x>0,y>0,z>0}(x^3+y^3+z^3)\ \ \mbox{s. t.}\ \ \ ax+by+cz=1\right\} = \frac{1}{(a\sqrt a+ b\sqrt b+c\sqrt c)^2} $$

Now the bordered Hessian

$$ H_g = \left( \begin{array}{cccc} 0 & -a & -b & -c \\ -a & 6 x & 0 & 0 \\ -b & 0 & 6 y & 0 \\ -c & 0 & 0 & 6 z \\ \end{array} \right) $$

has value

$$ \det(H_g) = -36 \left(a^2 y z+b^2 x z+c^2 x y\right) $$

Which with $x>0,y>0,z>0$ characterizes a minimum.

Answer 2

This is largely a distillation of the comments. As lulu suggested, I am assuming $abc\neq0$. If this is not valid, then you have to assume each is $0$ in turn, and apply Lagrange multipliers.

We have $$\begin{align}3x^2&=-ka\\3y^2&=-kb\\3z^2&=-kc\end{align}$$ We must have $k\neq0$ for otherwise $x=y=z=0$ and the constraint is not satisfied. Therefore, the left-hand sides in these equations are all positive, and $a,b,c$ all have the same sign. If this is the case, then taking square roots as you have done is justified. If not, there is no local minimum.

Your equations $$\begin{align} z &= \sqrt{\frac{cy^2}{b}}\\ x &= \sqrt{\frac{ay^2}{b}}\end{align}$$ are correct, so far as they go. We have $$\begin{align} z &= \pm\sqrt{\frac{c}{b}}|y|\\ x &= \pm\sqrt{\frac{a}{b}}|y|\end{align}$$

and of course $$y=\pm|y|$$ If we express the constraint in terms of $|y|$, that gives $8$ local minima.

There will not, in general, be a global minimum. For example, if $$(a,b,c)=\left(\frac14,\frac12,\frac14\right)$$ then $$(x,y,z)=(0,n,4-2n)$$ satisfies the constaints and $$x^3+y^3+z^3=n^3+(4-2n)^3$$ has no global minimum.