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If $x$ and $y$ are positive numbers such that $x + y = 1$, find the maximum value of $x^4y + xy^4$.

I could do this problem my simplifying the expression to $xy(1-3xy)$ and taking $k=xy$, forming a quadratic equation and got the answer as $1/12.$ But is there another method using calculus to do this problem?

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  • $\begingroup$ Do you know what AM-GM is? How about the rearrangement inequality? $\endgroup$ – Calvin Lin Aug 31 '13 at 4:05
  • $\begingroup$ Yeah, I know the AM-GM inequality. $\endgroup$ – Rajath Krishna R Aug 31 '13 at 4:06
  • $\begingroup$ Related problem. $\endgroup$ – Mhenni Benghorbal Aug 31 '13 at 4:34
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Given that $x+y=1$, you can transform $x^4y+xy^4=x^4(1-x)+x(1-x)^4$ then take the derivative, set to zero, and find the value of $x$. The method you cite works fine for this problem, but it relies on the nature of parabolas. Just substituting in like this will do more problems.

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  • $\begingroup$ Yeah I knew this method but, this would leave terms with powers of 4 which need to be simplified so I thought manipulating the given expression would be best. $\endgroup$ – Rajath Krishna R Aug 31 '13 at 4:22
  • $\begingroup$ Your approach works fine for this problem. I was showing another approach that works on other problems. You are right that it leaves $x^4$ which will be an obstacle on this problem. My approach is more general in that it gives you an equation in one variable, but you may need to use numeric methods to find the solution. It is good to have both in your toolbox. $\endgroup$ – Ross Millikan Aug 31 '13 at 4:27
  • $\begingroup$ Yeah its good to know more than one method and now I also studied Lagrange Multiplier method. $\endgroup$ – Rajath Krishna R Aug 31 '13 at 4:36
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Here is another technique. You can use Lagrange Multiplier method.

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    $\begingroup$ That's a nice method. Thanks for pointing out. $\endgroup$ – Rajath Krishna R Aug 31 '13 at 4:23
  • $\begingroup$ @RajathKrishnaR: You are welcome. $\endgroup$ – Mhenni Benghorbal Aug 31 '13 at 4:35
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You already have got $x^4y + xy^4 = xy(1-3xy)$ in this case.

Now $(3xy) (1-3xy)$ is the product of two terms with fixed sum of $1$, so it achieves its maximum when the terms are equal. Thus for maximum, we need to have $3xy = 1-3xy \implies xy = \frac{1}{6}$.

So $x^4y + xy^4 \ge \frac{1}{12}$.

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