A wire of length 2 units is cut into two parts which are bent respectively to form a square of side = x units and a circle of radius = r units. If the sum of the areas of the square and the circle so formed is minimum, then

Engineering

Mathematics

Maxima and Minima

Question

2x = r

(4 – π)x = πr

2x = (π + 4) r

x = 2r

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4x + 2πr = 2 ⇒ 2x + πr = 1

A = x2 + πr² = x² + π ${(\frac{1 - 2 x}{π})}^{2} = \frac{1}{π} ({πx}^{2} + (4 x^{2} - 4 x + 1)) = \frac{1}{π} ((π + 4) x^{2} - 4 x + 1)$

$\frac{dA}{dx} = \frac{1}{π}$ (2x (π + 4) – 4) = 0

$x = \frac{2}{π + 4}$ (point of minimum)

2x + πr = 1.

$\frac{4}{π + 4} + πr = 1$ ⇒ $πr = 1 - \frac{4}{π + 4} = \frac{π}{π + 4}$

$∴ r = \frac{1}{π + 4} \Rightarrow x = 2 r .$