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

  • 1
    $\begingroup$ I don't understand how using A.M $\ge$ G.M leads to a maximum? Can't you simply use calculus to solve this? $\endgroup$ – Mufasa Dec 27 '14 at 16:55
  • $\begingroup$ 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. $\endgroup$ – Devarsh Ruparelia Dec 27 '14 at 17:32
  • $\begingroup$ @PeterWoolfitt Will you suggest how? As I am thinking about it since past 10 minutes. $\endgroup$ – Devarsh Ruparelia Dec 27 '14 at 17:45
  • $\begingroup$ @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 $4$. $\endgroup$ – Peter Woolfitt Dec 27 '14 at 18:30
  • $\begingroup$ 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. $\endgroup$ – 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 $2x$.)

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




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


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.

| cite | improve this answer | |

We can also use AM-GM as follows:

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

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

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.