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My question is how to prove that $f(x) = 4x(16 - x)(6 - x)$ has its maximum at point $x_0 = \frac{8}{3}$.

I am trying to use arithmetic-geometric inequality, but I am unable to solve it.

Mathsource
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  • Assuming that you mean the maximum value on the interval $[0,16]$. You know that it is negative on the interval $(6,16)$ and positive on $(0,6)$ so you could say you want to find the value of $x\in(0,6)$ for which $f$ has a maximum value. – John Wayland Bales Jan 12 '21 at 19:07
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    you can't. it has a local maximum at ypur $8/3$ and a local minimum at $12.$ For your purpose, you can simply translate with $x = t +\frac{8}{3}$ – Will Jagy Jan 12 '21 at 19:07
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    https://www.researchgate.net/publication/234612942_Optimization_of_Cubic_Polynomial_Functions_without_Calculus – Amaan M Jan 12 '21 at 19:10

3 Answers3

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The problem can be greatly simplified by considering $f(8/3+t)-f(8/3)$.

This is $4(t^3-14t^2)$. The graph of this cubic has a repeated root at $0$ and a single root at $14$ and is negative at $7$. It therefore clearly has a local maximum at $t=0$ i.e. $x=8/3$.

The same idea can be used for other polynomial functions.

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If AM-GM is allowed then when $x \in (0,16)$ we have $$4x(16-x)(6-x) = \frac 15 (5x) (16-x) (4(6-x))\\ \le \frac 15 \left( \frac{5x+16-x+24-4x}{3}\right)^3 = \frac 15 \cdot \left(\frac{40}{3} \right)^3$$ The equality holds when $5x=16-x=24-4x$, or equivalently $x=\frac 83$.

Neat Math
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$f(x)=4x(16-x)(6-x)=4x^3 - 88x^2 +384x$

Now calculate it's derivative $f'(x)=12x^2 -176x + 384 = 12(x-6)(x - \frac{8}{3})$

Now just find increasing and decreasing intervals

$f(x)$ is increasing on $ x< \frac{8}{3} $ and $ x>6$

And decreasing on ($ \frac{8}{3} ; 6)$ so after that you can say $x_0 = \frac{8}{3} $ is a local maximum point

NightEye
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    I think the reason your answer is attracting downvotes is that the question specifically states that no calculus should be used. – A-Level Student Jan 12 '21 at 20:49