Question

The number 28 is divided into two positive parts such that the sum of the cube of one part and the square of the other part is minimum, then the absolute difference between the two parts is

• A. \(24\)
• B. \(12\)
• C. \(8\)
• D. \(20\)

Hint:

Let the parts be \(x\) and \(28-x\) and minimise \(x^3+(28-x)^2\).

Answer 1

The Correct Option is D

Step 1: Substitute and expand
$S=x^3+x^2-56x+784$.

Step 2: Minimise
$S'=3x^2+2x-56=(3x+14)(x-4)$, so $x=4$ for positive parts. Then $y=24$ and the difference is 20, option (D).

Final Answer:
The parts are 4 and 24, so the difference is 20, option (D). \[ \boxed{20} \]