Question

$\sf \: class \: 12 \\ \sf \: maths$
$\large\color{purple}\underline { \tt{\underline{ \tt{ \boxed { Question }}}}}$
The wire of length 28m is to be cut into two pieces. One of the two pieces is to be made into a square and the other into a circle. What should be the lengths of the two pieces, so that the combined area of circle and square is minimum?
$\tt \: no \: spams\:pls \\ \tt \: right \: answer \: = 10 \: thanks \:$
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$\sf \: @WildCat7083$​

Answer 1

Concept Used :-

Maxima or Minima by Differentiation.

HOW TO FIND MAXIMUM AND MINIMUM VALUE OF A FUNCTION

• Differentiate the given function f(x).

• let f'(x) = 0 and find critical points.

• Then find the second derivative, i.e. f''(x).

• Apply those critical points in the second derivative.

• The function f (x) is maximum when f''(x) < 0.

• The function f (x) is minimum when f''(x) > 0.

$\large\underline{\bf{Solution-}}$

Let the wire be AB which is 28 m.

Let the length of wire is cut in to two pieces at point P such that AP is converted in to square of side 'x' meter and PB is converted in to circle of radius 'y'.

It implies,

AP = Perimeter of square = 4x meter

PB = Perimeter of circle = 2πy

Since, AP + PB = 28

$\rm :\implies\:4x + 2\pi \: y = 28$

$\rm :\implies\:2x + \pi \: y = 14$

$\bf\implies \:y = \dfrac{14 - 2x}{\pi} - - (1)$

Let suppose that combined area is represented by A.

$\rm :\longmapsto\:A = Area_{(circle)} + Area_{(square)}$

$\rm :\longmapsto\:A = \pi \: {y}^{2} + {x}^{2}$

Now substituting the value of y, from equation (1), we get

$\rm :\longmapsto\:A = \pi \: {\bigg(\dfrac{14 - 2x}{\pi} \bigg) }^{2} + {x}^{2}$

$\rm :\longmapsto\:A = \dfrac{1}{\pi} {(14 - 2x)}^{2} + {x}^{2}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} A = \dfrac{d}{dx} \bigg(\dfrac{1}{\pi} {(14 - 2x)}^{2} + {x}^{2} \bigg)$

$\rm :\longmapsto\:\dfrac{dA}{dx} = \dfrac{1}{\pi}(2) {(14 - 2x)}\dfrac{d}{dx}(14 - 2x) + 2x$

$\rm :\longmapsto\:\dfrac{dA}{dx} = \dfrac{1}{\pi}(2)(14 - 2x)( - 2) + 2x$

$\rm :\longmapsto\:\dfrac{dA}{dx} = \dfrac{ - 4}{\pi}(14 - 2x) + 2x - - - (2)$

For maxima or minima,

$\rm :\longmapsto\:\dfrac{dA}{dx} = 0$

$\rm :\longmapsto\: \dfrac{ - 4}{\pi}(14 - 2x) + 2x = 0$

$\rm :\longmapsto\: \dfrac{4}{\pi}(14 - 2x) = 2x$

$\rm :\longmapsto\: \dfrac{2}{\pi}(14 - 2x) = x$

$\rm :\longmapsto\:28 - 4x = \pi \: x$

$\rm :\longmapsto\:28 = 4x + \pi \: x$

$\rm :\longmapsto\:28 = (4 + \pi) \: x$

$\bf\implies \:x = \dfrac{28}{4 + \pi}$

Again Differentiate equation (2), w. r. t. x, both sides,

$\rm :\longmapsto\:\dfrac{ {d}^{2} A}{ {dx}^{2} } = \dfrac{ - 4}{\pi}( - 2) + 2$

$\rm :\longmapsto\:\dfrac{ {d}^{2} A}{ {dx}^{2} } = \dfrac{8}{\pi} + 2$

$\bf\implies \:\:\dfrac{ {d}^{2} A}{ {dx}^{2} } > 0$

$\bf\implies \:A \: is \: minimum \: when \: x = \dfrac{28}{4 + \pi}$

So,

Length of two pieces are

$\sf \: AP = 4x = 4 \times \dfrac{28}{4 + \pi} = \dfrac{112}{4 + \pi} \: meter$

$\sf \: PB = 2\pi \: y = 2\pi \: \bigg(\dfrac{14 - 2x}{\pi}\bigg) = 2(14 - 2x)$

Please find the attachment after this

Answer 2

Lets take dA/dx = 0

$Thus \: \frac{x}{2\pi} - \frac{28 - x}{8} = 0 \\ 4x = 28\pi - \pi \: x \\ 4x + \pi \: x = 28\pi \\ x(4 + \pi) = 28\pi \\ x = \frac{28\pi}{4 + \pi}$

$other \: part \: = 28 - x \\ where \: x \: = \frac{28\pi}{4 + \pi} \\ = 28 - \frac{28\pi}{4 + \pi} \\ = \frac{112 + 28\pi - 28\pi}{4 + \pi} \\ = \frac{112}{4 + \pi}$

$now \: again \: differentiating \: we \: get \: \\ \frac{d ^{2}A }{d {x}^{2} } = \frac{1}{2\pi} + \frac{1}{8} = + ve$

$A \: is \: minimum \: \\ where \: x \: = \frac{28\pi}{4 + \pi} \: and \: 28 - x = \frac{112}{4 + \pi}$