Question

Divide $20$ into two parts such that the product of one part and the cube of the other is maximum. The two parts are

• A. $(12,8)$
• B. $(15,5)$
• C. $(10,10)$
• D. $(2,18)$

Answer 1

Answer: (B)

Let the first part = y, then other part = $(20-y)$.
$f(y)=(20-y)y^3$ or $f(y)=20y^3-y^4$
For maximum or minimum, put $f^{\prime }(y)=0$
$\Rightarrow f^{\prime }\left(y\right)=60y^2-4y^3=0$
$\Rightarrow 4y^2\left(15-y\right)=0$
$\Rightarrow y=0ory=15$
Now, $f^{\prime \prime }\left(y\right)=120y-12y^2$
$\Rightarrow f^{\prime \prime }\left(15\right)=120\left(15\right)-12{\left(15\right)}^2<0$
$\therefore f\left(y\right)$ is maximum at y=15
Hence, 20 can be divided in 15 and 5