Question

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

• A. 13 and 7
• B. 14 and 6
• C. 15 and 5
• D. 16 and 4

Answer 1

Answer: (C)

15 and 5

Let the two parts be $x$ and $20-x$
Let $y=(20-x)x^3$
$\Rightarrow y=20x^3-x^4$
For maximum or minimum,
$\frac{dy}{dx}=0$
$\Rightarrow 60x^2-4x^3=0$
$\Rightarrow 4x^2(15-x)=0$
$\Rightarrow x=0,x=15$
$\frac{d^{2}y}{d{x}^2}=120x-12x^2$
At $x=15$, $\frac{d^{2}y}{d{x}^2}<0$
Hence, $y$ has a maximum at $x=15$
So, the two numbers are 15, 5