Question

Divide $12$ into two parts such that the product of the square of one part and the fourth power of the second part is maximum is

• A. $6,6$
• B. $5,7$
• C. $4,8$
• D. $3,9$

Answer 1

The correct option is C

$4,8$

The explanation of the correct option:

Step1. Construct the function :

Let the parts are $x$and $(12–x)$

let the function be $f\left(x\right)=x^4{(12-x)}^2,$where $0<x<12$

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

Now, differentiate with respect to $x$,

$f^{\text{'}}\left(x\right)=6x^5–120x^4+576x^3$

$=6x^3(x–12)(x–8)$

Put $f\text{'}\left(x\right)=0$

$$\begin{gathered} \Rightarrow 6x^3(x–12)(x–8)=0 \\ \Rightarrow x=12,8,0 \end{gathered}$$

Step2. Finding the numbers:

Now, find $f\text{'}\text{'}\left(x\right)$

$f\text{'}\text{'}\left(x\right)=30x^4-480x^3+1728x^2$

Put $x=8$

$$\begin{gathered} \Rightarrow f\text{'}\text{'}\left(x\right)=30{\left(8\right)}^4-480{\left(8\right)}^3+1728{\left(8\right)}^2 \\ =-12288<0 \end{gathered}$$

So, when $x=8$ $f\text{'}\text{'}\left(x\right)$ is negative.

Therefore, $f\left(x\right)$ is maximum when $x=8$

Thus,$x=8$ is the first part and $12–x=4$ is the second part such that $f\left(x\right)$maximum.

Hence, Option(C) is the correct answer.