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Let $m,n,s \in \mathbb N$. Let $x,y \in \mathbb R^+$ such that $x+y=s$. Prove that the maximum of the product $$P=x^m y^n$$ is attained when $$x=\frac{ms}{m+n} \textrm{ and }y=\frac{ns}{m+n}.$$

Any help would be great.

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    Have you heard of AM-GM inequality? – 2017-01-08
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    @JyrkiLahtonen how does the AM-GM inequality help here? – 2017-01-08
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    @jack I don't even know where to start. =/ – 2017-01-08
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    Take a look : http://math.stackexchange.com/questions/2086190/find-maximum-of-product – 2017-01-08
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    Saying that you don't know where to start is just lazy. This is obviously a homework problem; what have you been covering in class? What sorts of tools do you have to solve extremal problems in general? What have you tried so far (and if the answer is nothing, then this is not the right time to ask on this board)? – 2017-01-08
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    @anomaly I simply wanted a tip, I really was stomped. With Jack's comment, I should be able to solve it. Edit. Just to be clear, this is not a homework problem and I'm not taking any classes atm. – 2017-01-08

4 Answers 4

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HINT:

Note that $$ P(x)=x^m(s-x)^n. $$ Can you find $P'(x)=0$ for what $x$?

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We want to aximize $P(x,y) = x^my^n$ subject to the constraint that $g(x,y) = x+y-s = 0$. Using Lagrange multipliers, the solution would have to satisfy $$mx^{m-1}y^n = \lambda$$ $$nx^my^{n-1} = \lambda$$ $$x+y=s$$ There are two obvious solutions: $(s,0)$ and $(0,s)$. $P=0$ at these points, so they're clearly not maxima. Now assume $x,y\neq 0$. Dividing the first equation by the second, $$\frac{m}{n} \frac{y}{x} = 1 \implies$$ $$x = \frac{my}{n}$$ Then using the last equation, $$y + \frac{m}{n}y = s$$ $$y = \frac{s}{1+\frac{m}{n}} = \frac{ns}{n+m}$$

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    Way overkill imho – 2017-01-08
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Since $x+y=s\implies y=s-x$. So the function $P$ is

$P=x^m(s-x)^n$. For maximum, you must have $P^{'}(x)=0\implies ms=x(m+n)\implies x=\frac{ms}{m+n}$. $P^{''}(x)$ is negative at that point, hence the point is a maxima. From $x$ we see that $y=x-s$.

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$P=x^my^n$

$\ln P=m\ln x+n\ln y$

Differentiating implicity w.r.t $x$,

$\dfrac1P\dfrac{\mathrm dP}{\mathrm dx} = \dfrac mx+\dfrac ny\dfrac{\mathrm dy}{\mathrm dx}$

$\therefore\ \dfrac{\mathrm dP}{\mathrm dx} = P\left(\dfrac mx-\dfrac ny\right)$ as $y=s-x$ $\implies$ $\dfrac{\mathrm dy}{\mathrm dx}=-1$

Thus $\dfrac{\mathrm dP}{\mathrm dx}=0$ $\implies$ $\dfrac mx=\dfrac ny$ as $P>0$.

$\dfrac{\mathrm d^2P}{\mathrm dx^2} = \dfrac{\mathrm dP}{\mathrm dx}\left(\dfrac mx-\dfrac ny\right)-P\left(\dfrac m{x^2}+\dfrac n{y^2}\right)$; when $\dfrac mx=\dfrac ny$, $\dfrac{\mathrm d^2P}{\mathrm dx^2} = -P\left(\dfrac m{x^2}+\dfrac n{y^2}\right)<0$.

Thus $P$ is maximized when $\dfrac mx=\dfrac ny$, i.e.

$nx = my = m(s-x)$ $\implies$ $x=\dfrac{ms}{m+n}$

and

$y = s-x = s-\dfrac{ms}{m+n} = \dfrac{ns}{m+n}$.