Question

Solve it lazy persons (Q 21) ,in attachment.​

Answer 1

Answer:

( b ) x = 2r

Step-by-step explanation:

ATQ , wire is cut in to two parts and by one part square is formed and by second part circle is formed and side of square and radius of circle are x and r respectively and length of wire is 2 unit

Perimeter of square = 4 × side

= 4 x

Circumference of circle = 2 π × radius

= 2 π r

Length of wire = perimeter of square + circumference of circle

2 = 4 x + 2π r

dividing equation by 2 , we get

=> 1 = 2 x + π r

=> 2 x = 1 - π r

=> x = (1 - πr ) / 2 ........................(1)

Let area of square and circle be A₁ and A₂ respectively

A₁ = side²

= x²

Putting x = ( 1 - πr ) / 2

=> A₁ = { ( 1 - π r ) / 2 }²

=> A₁ = ( 1 - πr )² / 4

Now

A₂ = π × (radius)²

=> A₂ = π r²

Let sum of areas of square and circle is A , then

A = A₁ + A₂

=> A = ( 1 - π r )² / 4 + π r²

Differentiating with respect to r we get

=> dA/dr =(1/4)d/dr (1- πr)² + π d/dr(r²)

= 1/4 {2 (1 - πr )¹} d/dr (1-πr) +π (2r )

= {( 1 - π r ) / 2 } ( 0 - π × 1 ) + 2πr

=> dA / dr = - π ( 1 - πr ) / 2 + 2 π r

Differentiting it again with respect to x

=> d²A/dr² = - {π ( 0 - π × 1 )} / 2 + 2π × 1

=> d²A/dr² = (π² / 2) + 2π ( positive )

For minimum value of A

dA / dr = 0

=> - π (1 - πr ) / 2 + 2 π r = 0

=> - π ( 1 - πr ) / 2 = - 2π r

=> 1 - πr = 4 r

=> πr + 4r = 1

=> r ( π + 4 ) = 1

=> r = 1 / ( π + 4 )

[ d²A / dr² ] at r = 1 /( π +4 )

=( π² / 2 ) + 2π ( positive )

So A is minimum at r = 1 / (π + 4 )

Now r = 1 /( π + 4 )

=> π + 4 = 1 / r

=> π = ( 1 / r ) - 4

Now from ( 1 )

x = ( 1 - πr ) / 2

=> 2x = 1 - πr

Now putting π = (1/r ) - 4

=> 2x = 1 - r { (1/r ) - 4 }

= 1 - r × (1/r ) + 4r

= 1 - 1 + 4r

=>2x = 4r

=> x = 2r

Answer 2

$\color {red}\huge\bold\star\underline\mathcal{Question:-}$ A wire of length 2 units is cut into two parts which are bent respectively to form a square of side and a circle . we have to find relationship between them ?

$\huge\underline\blue{\sf Given:}$ side of square = x

radius of circle = r

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\:Answer}}}}}}}}}}$

Let Length of Two parts = a & (2-a) where a is perimeter of square and (2-a) will be perimeter of circle .

we know that perimeter of square = 4 × side

and,

circumfernce of circle = 2πr

so,

4 × x = a

x = $\huge{\frac{</strong><strong>a</strong><strong>}{</strong><strong>4</strong><strong>}}$

2πr = (2-a)

r = $\huge{\frac{</strong><strong>(</strong><strong>2</strong><strong>-</strong><strong>a</strong><strong>)</strong><strong>}{</strong><strong>2</strong><strong>π</strong><strong>}}$

Now, we know that ,

area of square = (side)²

area of circle = πr²

so,

Area of square = x² = (a/4)² = a²/16

Area of circle =

$\pi( \frac{2 - a}{2\pi} )^{2} = \frac{ {a}^{2} - 4a + 4 }{4\pi}$

Now , sum of both area =

$\huge{</strong><strong>(\frac{ {a}^{2} }{16})}$ + (a²-4a+4)/4π

Now, for minimum area of their sum, differentiate for a , we get,

f(a) = $\huge{(\frac{ {a}^{2} }{16})}$ + $\huge{\frac{</strong><strong>(</strong><strong>a^</strong><strong>{</strong><strong>2</strong><strong>}</strong><strong>-4a</strong><strong>+</strong><strong>4</strong><strong>}{</strong><strong>4</strong><strong>\</strong><strong>p</strong><strong>i</strong><strong>}}$

f'(a) = =$\huge{\frac{1}{16\pi}}$($\huge{2a\pi+8a-16}$)

Now, for minimum value putting this equal to 0 we get,

2aπ+8a = 16

$a = \frac{16}{2\pi + 8} = \frac{8}{\pi + 4} \\ \\ \\ now \: we \: have \: x = \frac{a}{4} \\ \\ so \: x \: = \frac{2}{\pi + 4}$

also ,

$r = \frac{2 - a}{\pi}$

Putting value of a here we get,

$r = \frac{2 - \frac{8}{\pi - 4} }{2\pi} = \frac{1}{\pi + 4}$

so,

$x = \frac{2}{\pi + 4} \: and \: r = \frac{1}{\pi + 4}$

$\large\red{\boxed{\sf </strong><strong>x</strong><strong>=</strong><strong>2</strong><strong>r</strong><strong>}}$