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NCERT Class 12 · Ch 6 Exercise 6.3 · Q22

NCERT Class 12 Maths · Chapter 6 Applications Of Derivatives · Exercise 6.3 Question 22

Cut 28 m wire into a square perimeter and circle circumference to minimize their combined area.
Optimum wire allocation: square side = circle diameter
Optimum wire allocation: square side = circle diameter

💡 Concept Hints

  1. Identify the domain and the inner and outer functions, or the relevant one-sided limits.
  2. Apply the appropriate differentiation or limit rule, keep all chain factors, and check exceptional points.

✅ Step-by-Step Solution

Final Answer: Square piece 112/(π+4) m; circle piece 28π/(π+4) m.
  1. Let x be the square piece, 0≤x≤28. The square area is x²/16 and the circle area is (28−x)²/(4π).
  2. A′=x/8−(28−x)/(2π). Set it to zero: πx=4(28−x), hence x=112/(π+4).
  3. A″=1/8+1/(2π)>0, so this unique feasible stationary point is the global minimum. The remaining length is 28π/(π+4).

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