Question

Divide 12 into two parts so that the product of the square of one part and the fourth power of the other is maximum.

Answer 1

We have,

Total number $=12$

Let the first part be $=x$

And IInd part be $=12-x$

According to given question,

Maximize $f\left(x\right)=x^4\times {(12-x)}^2(\therefore 0<x<12)$

$=x^4(144-24x+x^2)$

$f\left(x\right)=x^6-24x^5+144x^4$

On differentiating and we get,

$f^{\prime }\left(x\right)=6x^5-120x^4+576x^3$

$=6x^3(x^2-20x+96)$

Then,

For maximum and minimum

$f^{\prime }\left(x\right)=0$

$\Rightarrow 6x^3(x-12)(x-8)=0$

Then,

$x=12$ and $x=8$ and $x=0$

If $x=8$ then,

$12-x$

$\Rightarrow 12-8$

$=4$

Hence, $f\left(x\right)$ is maximized.