3
$\begingroup$

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

The answer instead had

$\lambda x = ax^ay^bz^c, \lambda y = bx^ay^bz^c, \lambda z = cx^ay^bz^c$, 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 ...

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

...

  • 0
    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.2012-05-01

2 Answers 2

4

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.

  • 0
    great answer @André Nicolas2012-05-01
2

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?