# Absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$

I need to show absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ is $$\frac{a^ab^bc^c}{(a+b+c)^{a+b+c}}$$

So I I do have $$ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$$

then I went on to equating each of the 2 equations giving

$$y=\frac{b}{a}x, z=\frac{c}{a}x$$

So I have $x+\frac{b}{a}x+\frac{c}{a}x=1$ but I am not so sure how to continue

$$\lambda x = ax^ay^bz^c, \lambda y = bx^ay^bz^c, \lambda z = cx^ay^bz^c$$, then making observation that $x:y:z=a:b:c$, which means

$$x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$$

I don't quite get this ... how do I derive this?

The rest ...

$$(\frac{a}{a+b+c})^a(\frac{b}{a+b+c})^b(\frac{c}{a+b+c})^c=1$$

...

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You probably mean that the constraint is $x+y+z=1$, or maybe $x+y+z \le 1$, and also that everything is non-negative. –  André Nicolas May 1 '12 at 5:43

You were doing fine, the argument from the textbook that you are quoting is more symmetrical, that's all.

You obtained $x+\frac{b}{a}x+\frac{c}{a}x=1$. That almost finishes things! Multiply through by $a$. You get $(a+b+c)x=a$ and therefore $$x=\frac{a}{a+b+c}.$$

From your $y=\frac{b}{a}x$ you can then get $y=\dfrac{b}{a+b+c}$, and similarly $z=\dfrac{c}{a+b+c}$. That gets you to exactly the same place as the solution you quoted.

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great answer @André Nicolas –  dato datuashvili May 1 '12 at 9:38

When you have

$$ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$$

$ax^{a-1}y^bz^c =\lambda \hspace{4pt}$ will give you $\hspace{4pt} \lambda x = a x^a y^b z^c$

Similarly

$bx^ay^{b-1}z^c =\lambda \hspace{4pt} \Rightarrow \hspace{4pt} \lambda y = b x^a y^b z^c$

$cx^ay^bz^{c-1} =\lambda \hspace{4pt} \Rightarrow \hspace{4pt} \lambda z = c x^a y^b z^c$

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

$$\frac{\lambda y}{\lambda z} = \frac{y}{z} = \frac{b x^a y^b z^c}{cx^a y^b z^c} = \frac{b}{c}$$

$$x : y : z = a : b : c$$

What happens to $$\lambda x + \lambda y + \lambda z \hspace{5pt} ?$$

Is it

$$x^a y^b z^c (a+b+c)$$

Now can you relate to the answer you were supposed to get?

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