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95. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

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Vishal Baghel | Contributor-Level 10

4 months ago

Let x and y in ‘m’ be the length of side of the square the radius of the circle respectily

Then, length of wire = perimeter of square + circumference of circle

$\Rightarrow$ 28 = 4x + 2πy

$\Rightarrow$ 2x + πy = 14

$\Rightarrow y = \frac{14 - 2 x}{x}$

The combine area A of the square and the circle is

A = x² + πy²

$= x^{2} + π {[\frac{14 - 2 x}{π}]}^{2}$

${= x^{2} + \frac{[2 (7 - x)}{x}]}^{2}$

$= x^{2} + \frac{4}{π} {(7 - x)}^{2} .$

So, $\frac{dA}{d x} = 2 x + \frac{4 \cdot 2}{π} (7 - x) + (- 1) = 2 x - \frac{8}{π} (7 - x)$

$\frac{d^{2A}}{d x^{2}} = 2 + \frac{8}{π}$

At, $\frac{dA}{d x} = 0$

$\Rightarrow 2 x - \frac{8 (7 - x)}{π} = 0$

$\Rightarrow 2 x = \frac{8}{π} (7 - x)$

$\Rightarrow 2 π x = 56 - 8 x$

$\Rightarrow (2 x + 8) x = 56$

$\Rightarrow x = \frac{56}{2 π + 8} = \frac{28}{x + 4} .$

At, $x = \frac{28}{π + 4}, \frac{d^{2A}}{d x^{2}} = 2 + \frac{γ}{π} > 0$

$∴ x = \frac{28}{π + 4}$ isa point of minima

Hence, length of square = $4 x = 4 (\frac{28}{1 + 4}) = \frac{112}{π + 4} m$

and length of circle = 2πy

$= 2 π (\frac{14 - 2 x}{π})$

$= 2 π [\frac{14}{π} - \frac{2}{1} [\frac{28}{π + 4}]]$

$= 28 - 4 [\frac{28}{π + 4}] = 28 - \frac{112}{1 + 4}$

$= \frac{28 x + 112 - 112}{x + 4}$

$= \frac{28 π}{π + 4}$

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95. A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum?

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