Question

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

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

Answer 1

The correct option is C

$4,8$

Explanation for the correct options:

Step1: Forming $f\left(x\right)$ for the given condition:

Let the parts be $x$ and $12–x$

$f\left(x\right)=x^4{(12–x)}^2$, where $0<x<12$ [Given]

$=x^4(144–24x+x^2)$

$=x^6–24x^5+144x^4$

Step2: Differentiating $f\left(x\right)$:

$f’\left(x\right)=6x^5–120x^4+576x^3$

$=6x^3(x^2–20x+96)$

$=6x^3(x–12)(x–8)$

$f\text{'}\left(x\right)=0$ when $x=12,8,0$

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

Thus$x=8$ and $12–x=4$ are the two parts

Hence, Option(C) is correct.