JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Application of Derivatives

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A wire of length 36 m is cut into two pieces, one of the pieces is bent to form a square and the other is bent to form a circle. If the sum of the areas of the two figures is minimum, and the circumference of the circle is

k

(meter), then

(\frac{4}{π} + 1) k

is equal to ______.

Answer: 36

Solution

Let the square take

x

metres of wire, leaving

36 - x

metres for the circle. The square has side

\frac{x}{4}

and the circle has radius

\frac{36 - x}{2 π}

, so the total area is

A = \frac{x ^{2}}{16} + \frac{( 36 - x ) ^{2}}{4 π} .

Setting

\frac{d A}{d x} = \frac{x}{8} - \frac{36 - x}{2 π} = 0

gives

π x = 4 (36 - x)

, so

x = \frac{144}{π + 4}

, and

\frac{d ^{2} A}{d x ^{2}} = \frac{1}{8} + \frac{1}{2 π} > 0

makes this the minimum.
The circumference of the circle is

k = 36 - \frac{144}{π + 4} = \frac{36 π}{π + 4},

(\frac{4}{π} + 1) k = \frac{π + 4}{π} \cdot \frac{36 π}{π + 4} = 36.

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Numerical answer

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JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Application of Derivatives

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