Question

$20$ is divided in two parts so that product of cube of one quantity and square of other quantity is maximum.The parts are-

• A. $10,10$
• B. $16,4$
• C. $12,8$
• D. $6,14$

Answer 1

The correct option is C $12,8$

Let the parts be $x,y$
So, $x+y=20$and $\left(x^3\right)\left(y^2\right)=$maximum
Hence,$f\left(x\right)=\left(x^3\right)(20-x{)}^2$ should be maximum
So, $f^{\prime }\left(x\right)=3x^2(20-x{)}^2+x^3(-2(20-x))$
$\Rightarrow f^{\prime }\left(x\right)=x^2(20-x)(60-5x)=0$
$\Rightarrow x=0,12,20$
$f^{\prime \prime }\left(x\right)=2x(20-x)(60-5x)-x^2(60-5x)-5x^2(20-x)$
$\Rightarrow f^{\prime \prime }\left(12\right)<0$
$\therefore x=12,y=8$
$\{\because x=0,20\text{will make the product}0\}$