Finding the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$

Question

Given $a,b,c,k > 0$, find the maximum of $f(x,y,z)=x^ay^bz^c$ where $x,y,z\in [0,\infty)$ and $x^k+y^k+z^k=1$

The subject is Lagrange multipliers, thus that is what I tried to use, where the gradients are

$$\triangledown f=\begin{pmatrix}ax^{a-1}y^{b}z^{c}\\ bx^{a}y^{b-1}z^{c}\\ cx^{a}y^{b}z^{c-1} \end{pmatrix} $$

and if $g$ is the function of the constraint, then

$$\triangledown g=\begin{pmatrix}kx^{k-1}\\ ky^{k-1}\\ kz^{k-1} \end{pmatrix}$$

but then I'm having troubles finding the actual maximum, as I'm interested in $x,y,z,\lambda$ satisfying

$$ \begin{pmatrix}ax^{a-1}y^{b}z^{c}\\ bx^{a}y^{b-1}z^{c}\\ cx^{a}y^{b}z^{c-1} \end{pmatrix}=\lambda\begin{pmatrix}kx^{k-1}\\ ky^{k-1}\\ kz^{k-1} \end{pmatrix} $$

and ignoring for a second the possibilities of one of the constants being $1$, this gives some equations which really don't seem all that right, as extracting $x$ from the first equality for example gives

$$x=\left(\frac{ay^{b}z^{c}}{\lambda k}\right)^{\frac{a-1}{k-1}}$$

which then setting into the second equality to extract $y$ gives

$$b\left(\frac{ay^{b}z^{c}}{\lambda k}\right)^{\frac{a-1}{k-1}}y^{b-1}z^{c}=\lambda ky^{k-1}$$

where I'm not sure how to even approach extracting $y$, and even if I will be able to seems to convoluted to be able to help me in the future..

Any better approaches to the question?

Answer 1

Just played with it a little and got: $$ \left(\begin{matrix}x^{a}y^{b}z^{c}\\ x^{a}y^{b}z^{c}\\ x^{a}y^{b}z^{c} \end{matrix}\right)=\lambda k\left(\begin{matrix}x^{k}/a\\ y^{k}/b\\ z^{k}/c \end{matrix}\right) $$ Therefore $x^k/a=y^k/b=z^k/c$. By putting it in the constraint, we get $$ 1=x^k+y^k+z^k=(1+b/a+c/a)x^k \\ x=\sqrt[k]{\frac{a}{a+b+c}} $$ and similarly $y=\sqrt[k]{\frac{b}{a+b+c}}$ and $z=\sqrt[k]{\frac{c}{a+b+c}}$.

Answer 2

$$ax^{a-1}y^bz^c=\lambda kx^{k-1}$$

$$ax^ay^bz^c=af(x,y,z)=\lambda k x^k$$

Similarly

$$bf(x,y,z)=\lambda ky^k$$

And one with $z$

Add all three together

$$(a+b+c)f(x,y,z)=\lambda k $$

$$f(x,y,z)=\frac {\lambda k}{a+b+c}$$

However after this part, I think it is quite impossible to find exactly how $\lambda$ depends on $a,b,c,k$. I have tried using wolfram alpha to check the results for different values of constants but I don't see any patterns. Most of the cases, using the above formula to trace back the value of $\lambda$, it is usually in a very messy surd expression. Maybe if given specific cases for $a,b,c,k$, one can find the value.

Answer 3

For those interested I have asked for help in a different place and was able to reach a solution. Both @Olorin and @Ido have found the same solution. The hint that allowed me to solve this was to try and maximize $\ln f$ instead of $f$. The complete solution is:

Let $g\left(x,y,z\right)=x^{k}+y^{k}+z^{k}-1$. We will search for the maximum of $\ln f$ constrained by $g\left(x,y,z\right)=0$. That is we are interested in $x,y,z$ where for some $\lambda$, $$\nabla\left(\ln f\right)=\lambda\nabla g$$

As $$\ln f\left(x,y,z\right)=a\ln x+b\ln y+c\ln z$$ we can calculate the derivative: $$\nabla\left(\ln f\right)=\begin{pmatrix}a/x\\ b/y\\ c/z \end{pmatrix}$$ and $$\nabla g=\begin{pmatrix}kx^{k-1}\\ ky^{k-1}\\ kz^{k-1} \end{pmatrix}$$

hence we want to find $x,y,z$ and $\lambda$ such that $$\begin{pmatrix}a/x\\ b/y\\ c/z \end{pmatrix}=\lambda\begin{pmatrix}kx^{k-1}\\ ky^{k-1}\\ kz^{k-1} \end{pmatrix}$$

which gives us the equations $$\frac{a}{x} = \lambda kx^{k-1} \frac{b}{y} = \lambda ky^{k-1} \frac{c}{z} = \lambda kz^{k-1}$$

together with $$x^{k}+y^{k}+z^{k}=1$$

Solving the first set of equations for $x,y,z$ we see that $$x = \left(\frac{a}{\lambda k}\right)^{1/k} y = \left(\frac{b}{\lambda k}\right)^{1/k} z = \left(\frac{c}{\lambda k}\right)^{1/k}$$ hence $$\frac{a}{\lambda k}+\frac{b}{\lambda k}+\frac{c}{\lambda k}=1$$ which shows us that $$\lambda=\frac{a+b+c}{k}$$ and that a critical point is at $$x=\left(\frac{a}{a+b+c}\right)^{1/k},\ y=\left(\frac{b}{a+b+c}\right)^{1/k}\text{ and }z=\left(\frac{c}{a+b+c}\right)^{1/k}$$ where as $\ln$ is a monotone function, this must also be a critical point of $f$ on the set. Apart from this point all extrema must appear on the boundary, where one of $x,y,z$ is equal $0$, and thus $f\left(x,y,z\right)=0$ , where the set being compact shows that both a maximum and a minimum exist, and specifically this point is indeed a maximum, and the constrained maximum is $$f\left(\left(\frac{a}{a+b+c}\right)^{1/k},\left(\frac{b}{a+b+c}\right)^{1/k},\left(\frac{c}{a+b+c}\right)^{1/k}\right)=\left(\frac{a^{a}b^{b}c^{c}}{\left(a+b+c\right)^{a+b+c}}\right)^{1/k}$$