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JEE Main 2016

JEE Main 2016 Mathematics question on Application of Derivatives

A wire of length 2 units is cut into two parts which are bent respectively to form a square of side =x= x units and a circle of radius =r= r units. If the sum of the areas of the square and the circle so formed is minimum, then
  1. A.
    2x=(π+4)r2x = (\pi + 4)r
  2. B.
    (4−π)x=πr(4 - \pi)x = \pi r
  3. C.
    x=2rx = 2r
  4. D.
    2x=r2x = r
Show answer and solution

Correct option: C

Solution

The constraint is 4x + 2πr = 2. The total area is A = x² + πr². From the constraint, x = (1 - πr)/2. Minimizing A with respect to r: dA/dr = 2x(dx/dr) + 2πr = 0. Since dx/dr = -π/2, this gives 2x(-π/2) + 2πr = 0, so -πx + 2πr = 0, which simplifies directly to x = 2r.

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