# Maximum of $f(x) = (45-2x)\cdot (24-2x)\cdot (2x)\;,$ Where $0<x < 12$

How 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

• I don't understand how using A.M $\ge$ G.M leads to a maximum? Can't you simply use calculus to solve this? – Mufasa Dec 27 '14 at 16:55
• I don't get why are you using AM-GM for it.This inequality gives Minima of the function and in no way gives Max possible value. – Devarsh Ruparelia Dec 27 '14 at 17:32
• @PeterWoolfitt Will you suggest how? As I am thinking about it since past 10 minutes. – Devarsh Ruparelia Dec 27 '14 at 17:45
• @DevarshRuparelia Well, here's a toy example, consider maximizing the function $(2-x)(2+x)$ on the range $x\in[0,2]$, then from AM-GM, we have $\dfrac{(2-x)+(2+x)}{2}\ge \sqrt{(2-x)(2+x)}$, so we have $2\ge \sqrt{(2-x)(2+x)}$ and we know equality occurs in AM-GM if and only if each of the functions are equal, so we have the maximum of $(2-x)(2+x)$ occurs where $2-x=2+x$ - that is at $x=0$. Hence by AM-GM we have shown that the maximum value of this function is at $x=0$ and at $0$, the function has value $4$. – Peter Woolfitt Dec 27 '14 at 18:30
• oops, sorry for my earlier post, I meant that the maximum is $70^2$ at $x=5$, which means the problem might me amenable to some AM-GM idea with two functions. – Peter Woolfitt Dec 27 '14 at 18:32

Forget the AM>GM inequality, just use differential calculus. (This is a standard calculus problem, coming from finding the maximum volume of a box made by cutting out the corners of a 24x45 rectangle, although the final term should then be $x$ rather than $2x$.)

$$f(x) = (45-2x)\cdot (24-2x)\cdot (2x)$$

$$=8x^3-276x^2+2160x$$

$$f'(x)=24x^2-552x+2160$$

$$=24(x-5)(x-18)$$

The only critical point in $0<x<12$ is $x=5$. To confirm that is a maximum, we find

$$f''(5)=48x-552|x=5=-312$$

The second derivative is negative, so $x=5$ is indeed the only local maximum. The function trivially is zero at the endpoints, so $f(5)=4900$ is the global maximum value.

We can also use AM-GM as follows:

$$f(x) = \frac{1}{2\cdot 5\cdot 7} [2(45-2x)]\cdot [5(24-2x)] \cdot [7(2x)]$$

$$\leq \frac{1}{2\cdot 5\cdot 7\cdot 27}\left[2(45-2x) + 5(24-2x) + 7(2x)\right]^3 = \frac{210^3}{2\cdot 5\cdot 7\cdot 27} = 70^2$$