Maximum With Two Constraints

Calculus Level 3

Let $a$ , $b$ and $c$ be three real positive numbers such that $a+b+c=12$ and $a^2+b^2+c^2=80$ . The maximum value of $abc$ is $\frac{p\sqrt{q}}{r}$ , where $p$ , $q$ and $r$ are positive integers, $p$ and $r$ are coprime and $q$ is square-free. Find $p+q+r$ .

The answer is 140.

1 solution

Kushal Bose
Mar 27, 2017

$ab+bc+ca=(a+b+c)^2 -(a^2+b^2+c^2)=64/2=32 \\ ab +c(a+b)=32 \\ ab + c(12-c)=32 \\ ab=c^2-12c+32 \\ abc=c^3-12c^2+32c$

Now define $f(c)=c^3-12c^2+32c$

Using differentiation we get maximum of $f(c)$ occurs at $c=4- \dfrac{4 \sqrt{3}}{3}$

Putting this we get maximum value of $abc$ is $\dfrac{128 \sqrt{3}}{9}$

The question says r is not divisible by square of any prime. r = 9 is divisible by 3 squared.

Siva Bathula - 4 years, 2 months ago

I donno but I got this as a max value.

Have u got any other flaw in this method ?

Kushal Bose - 4 years, 2 months ago

@Ervin Tronati I got the same answer too, the question is flawed is what I meant.

Siva Bathula - 4 years, 2 months ago

@Siva Bathula – yes the question is ok but the statement is flawed

Kushal Bose - 4 years, 2 months ago

@Kushal Bose – Yes, I don't know why there would be such a restriction on $r$ . I think it should be that $p$ and $r$ are coprime and $q$ is square-free. I could edit it but I would rather that Ervin confirm that that is what he intended.

Nice solution, by the way. I used Lagrange multipliers, which gave me that the maximum occurs when

$b = c = 4 - \dfrac{4\sqrt{3}}{3}$ and $a = 12 - 2c = 4 + \dfrac{8\sqrt{3}}{3}$ .

Brian Charlesworth - 4 years, 2 months ago

Thanks @Siva Bathula and @Kushal Bose for the correction.

Ervin Tronati - 4 years, 2 months ago

Thanks for changing the wording!.

Siva Bathula - 4 years, 2 months ago

Maximum With Two Constraints

The answer is 140.

1 solution

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