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)

$(15,5)$

Let the numbers be $x$ and $y$

Given, $x+y=20\to \left(1\right)$
Product $P=x{y}^3=(20-y)y^3$ using (1)

$\Rightarrow P=20y^3-y^4$
$\frac{dP}{dy}=60y^2-4y^3$
$\frac{d^{2}P}{d{y}^2}=120y-12y^2$
for extrema, $\frac{dP}{dy}=0\Rightarrow 60y^2=4y^3\Rightarrow y=15\therefore x=5\therefore$ two parts are $(15,5)$