# A non-linear maximisation

We know that $x+y=3$ where x and y are positive real numbers. How can one find the maximum value of $x^2y$? Is it $4,3\sqrt{2}, 9/4$ or $2$?

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Do you know what AM-GM is? Do you know what calculus is? – Calvin Lin Dec 29 '12 at 13:53

## 4 Answers

Lagrange multipliers: Let $f(x,y)=x^2y$ and $g(x,y,\lambda)=x^2y-\lambda (x+y-3)$. Then: $$\partial _y g=0\iff x^2-\lambda=0\iff x=\pm \sqrt{\lambda}$$ $$\partial _x g=0\iff 2xy-\lambda=0\iff xy=\frac{\lambda}2\iff y=\pm \frac{\sqrt{\lambda}}2$$ $$\partial _{\lambda} g=0\iff x+y=3$$

with $\lambda>0$. We want $x+y=3$ and so $\sqrt{\lambda}=2\iff \lambda=4$ or $-\sqrt{\lambda}=2$ which is impossible. Thus, $$(x,y)=(2,1)$$ and $f(2,1)=4$ is a possible maximum value. To verify that it is the maximum value you will have to determine the eigenvalues of the Hessian (which is simple)

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@BabakSorouh You are welcome. – Nameless Dec 29 '12 at 15:57
Thank you all guys here. – Basil R Dec 31 '12 at 6:00

By AM-GM

$$\sqrt[3]{2x^2y} \leq \frac{x+x+2y}{3}=2$$

with equality if and only if $x=x=2y$.

Second solution

This one is more complicated, and artificial (since I needed to know the max)$. $$x^2y=3x^2-x^3=-4+3x^2-x^3+4=4- (x-2)^2(x+1)\leq 4$$ since$(x-2)^2(x+1) \geq 0$. - I alredy upvoted this answer ! – Amr Dec 29 '12 at 14:22$x^2y=x^2(3-x)=3x^2-x^3$So,$\frac{x^2y}{dx}=3(2x)-3(x^2)=3x(2-x)$For the extreme values of$x^2y,\frac{x^2y}{dx}=0\implies x=0$or$x=2$Now,$\frac{d^2(x^2y)}{dx^2}=6-3(2x)=6(1-x)$At$x=0,\frac{d^2(x^2y)}{dx^2}=6>0$so$x^2y$will have minimum value at$x=0$At$x=2,\frac{d^2(x^2y)}{dx^2}=-6<0\implies x=2$will make$x^2y$the maximum - Is there a solution that does not use calculus ? – Amr Dec 29 '12 at 14:03 @Amr, find in N.S.'s answer – lab bhattacharjee Dec 29 '12 at 14:20 @lab, although it didn't matter in this case, shouldn't you have also checked the edge cases as well? You already have$x=0$as a minimum but$y=0, x=3$should have been explicitly checked since the variables were constrained to be positive. – half-integer fan Dec 29 '12 at 14:24 @half-integerfan,$x=3$does not purvey any extreme value of$x^2y$, at least by Calculus. – lab bhattacharjee Dec 29 '12 at 14:27 @lab, if the constraint had been$x \gt -10$instead of$x \gt 0$then the maximum value would be at$x = -10$even though it does not have slope zero, because the function continues to$+ \infty$as$x$goes to$- \infty$. – half-integer fan Dec 29 '12 at 14:33 Besides' to @lab's answer, the following fact may help you: If$x$and$y$be positive numbers such that$ax+by=k$,$k$is constant, then$x^{\alpha}y^{\beta}$has maximum value if$\frac{ax}{\alpha}=\frac{by}{\beta}=\frac{k}{\alpha+\beta}\$.

Now find the value.

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Dear Babak, your solution is elegant and simple. But the fact you are using is probably not to the OP's knowledge. Not that my answer is (as it seems), but I think Lagrange multipliers are more well known than this fact. The AM-GM answer is probably the most elementary out of all these answers. – Nameless Dec 29 '12 at 15:44