<s>[INST] Maximum of $f(x) = (45-2x)\cdot (24-2x)\cdot (2x)\;,$ Where $0How Can I Maximise $f(x) = (45-2x)\cdot (24-2x)\cdot (2x)\;,$ Where $0<x < 12$
Using Inequality
$\bf{My\; Try::}$ In $0<x<12\;,$ The value of $(45-2x)\;,(24-2x)\;,2x>0$
and we can write $\displaystyle f(x) = \frac{1}{2}\bigg[(45-2x)\cdot (24-2x)\cdot (4x)\bigg]$
So Applying $\bf{A.M\geq G.M\;,}$ We get
$\displaystyle \frac{(45-2x)+(24-2x)+4x}{3}\geq \bigg[(45-2x)\cdot (24-2x)\cdot (4x) \bigg]^{\frac{1}{3}}$
and equality hold when $(45-2x)= (24-2x)=(4x)\;,$ But this is wrong bcz no common
value of $x$ for which equality holds.
Help required, Thanks
 [/INST] We can also use AM-GM as follows:
$$f(x) = \frac{1}{2\cdot 5\cdot 7} [2(45-2x)]\cdot [5(24-2x)] \cdot [7(2x)]$$
$$\leq \frac{1}{2\cdot 5\cdot 7\cdot 27}\left[2(45-2x) + 5(24-2x) + 7(2x)\right]^3 = \frac{210^3}{2\cdot 5\cdot 7\cdot 27} = 70^2$$
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