Question

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

• A. $(10,10)$
• B. $(15,15)$
• C. $(13,7)$
• D. $Noneofthese$

Answer 1

The correct option is A $(10,10)$

Let the first part be y and other part $=(20-y)$

$f\left(y\right)=(20-y)y^3$

$=20y^3-y^4$

For maximum and minimum, $\frac{dy}{dx}=0$

$\frac{dy}{dx}=60y^2-4y^3$

$4y^2(15-y)=0$

$y=15$

$\frac{d^{2}y}{d{x}^2}=120y-12y^2$

when y=15

$\frac{d^{2}y}{d{x}^2}=120\times 15-12\times {15}^2<0$

20 can be divided into 15 and 5