Question

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

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

Answer 1

Answer: (D)

$12,8$

Let one part be $x$

Then other will be $20-x$

Now $f\left(x\right)=x^3\times {\left(20-x\right)}^2$

We have to maximizer $f\left(x\right)$

$f^{\prime }\left(x\right)=3x^2{\left(20-x\right)}^2-2x^3\left(20-x\right)$

$=x^2\left(20-x\right)\left(60-5x\right)$

$f^{\prime \prime }\left(x\right)=2x\left(20-x\right)\left(60-5x\right)-x^2\left(60-5x\right)-5x^2\left(20-x\right)$

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

$\Rightarrow x^2\left(20-x\right)\left(60-5x\right)=0$

$\Rightarrow x=0$ or $x=20$ or $x=12$

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

$f^{\prime \prime }\left(20\right)=16000>0$

$f^{\prime \prime }\left(12\right)=144\times -40<0$ (maxima)

Therefore $f\left(x\right)$ is maximum when one part is $12$ and other part is 8