Find the maximum value of $8\cdot27^{\log_6 x}+27\cdot8^{\log_6 x}-x^3$

Question

Find the maximum value of $$8\cdot27^{\log_6 x}+27\cdot8^{\log_6 x}-x^3.$$

If I apply AM${}\ge{}$GM, then I can find the minimum value of this expression, but not sure how to find the max value.

Answer 1

Hint:

$\log_6x=y,x=6^y$

$$8\cdot 27^y+27\cdot8^y-(6^y)^3=216(3^{3(y-1)}+2^{3(y-1)}-3^{3(y-1)}2^{3(y-1)}-1)+216$$

$$=216-216(3^{3(y-1)}-1)(2^{3(y-1)}-1)$$

Answer 2

prove that $$8\cdot 27^{\log_{6}{x}}+27\cdot 8^{\log_{6}{x}}-x^3\le 216$$ and the equal sign holds for $x=6$

Answer 3

For $x=6$ we get a value $216$.

We'll prove that it's a maximal value.

Indeed, we need to prove that $$x^3+216\geq8\cdot27^{\log_6x}+27\cdot8^{\log_6x}$$ or $$\left(6^{\log_6x}\right)^3+216\geq8\cdot27^{\log_6x}+27\cdot8^{\log_6x}$$ or $$27^{\log_6x}\cdot8^{\log_6x}-8\cdot27^{\log_6x}-27\cdot8^{\log_6x}+216\geq0$$ or $$\left(27^{\log_6x}-27\right)\left(8^{\log_6x}-8\right)\geq0,$$ which is obvious for $x\geq6$ and for $0<x\leq6.$

Answer 4

$$y=8\cdot 27^{\log_6{x}}+27\cdot 8^{\log_6{x}}-x^3\Rightarrow \max(y)=8\cdot 27^{\log_6{\max(x)}}+27\cdot 8^{\log_6{\max(x)}}-\max(x)^3\Rightarrow \frac{d}{dx}\bigg[8\cdot 27^{\log_6{x}}+27\cdot 8^{\log_6{x}}-x^3\bigg]=\frac{27\cdot 8^{\frac{\ln(x)}{\ln(6)}}\ln(8)+8\cdot 27^{\frac{\ln(x)}{\ln(6)}}\ln(27)}{x\ln(6)}=0\Rightarrow \max(x)=6\Rightarrow \max(y)=216$$