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
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
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Let the parts be \(x\) and \(28-x\) and minimise \(x^3+(28-x)^2\).
Updated On: Oct 1, 2026
- \(24\)
- \(12\)
- \(8\)
- \(20\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Let the two positive parts be \(x\) and \(y=28-x\). Minimise \(S=x^3+y^2\).
Step 2: Key Formula or Approach
\(S(x)=x^3+(28-x)^2\), so \(S'(x)=3x^2-2(28-x)=3x^2+2x-56\).
Step 3: Detailed Explanation
Set \(S'=0\): \(3x^2+2x-56=0\), giving \(x=\dfrac{-2\pm26}{6}\), so \(x=4\) (the positive root).
\(S''=6x+2>0\), so this is a minimum.
The parts are \(4\) and \(24\), and the absolute difference is \(24-4=20\).
Final Answer:
The difference between the parts is 20, option (D). \[ \boxed{20\ \text{(D)}} \]
Let the two positive parts be \(x\) and \(y=28-x\). Minimise \(S=x^3+y^2\).
Step 2: Key Formula or Approach
\(S(x)=x^3+(28-x)^2\), so \(S'(x)=3x^2-2(28-x)=3x^2+2x-56\).
Step 3: Detailed Explanation
Set \(S'=0\): \(3x^2+2x-56=0\), giving \(x=\dfrac{-2\pm26}{6}\), so \(x=4\) (the positive root).
\(S''=6x+2>0\), so this is a minimum.
The parts are \(4\) and \(24\), and the absolute difference is \(24-4=20\).
Final Answer:
The difference between the parts is 20, option (D). \[ \boxed{20\ \text{(D)}} \]
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