Question

Show that of all the rectangles of given area, the square has the smallest perimeter.​

Answer 1

ANSWER:-

Given:

Show that of all the rectangles of given area, the square has the smallest perimeter.

Solution:

Let x be the length and y be the breadth of rectangle whose area is A and perimeter is P.

P= 2(x+y)

$= > p = 2(x + \frac{A}{x} ) \: \: \: \: \: \: \: \: a = x.y \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y = \frac{A}{x}$

For maximum or minimum value of perimeter P.

$\frac{dP}{dx} = 2(1 - \frac{A}{ {x}^{2} }) = 0 \\ \\ = > 1 - \frac{A }{ {x}^{2} } = 0 \\ \\ = > {x}^{2} - A= 0 \\ \\ = > {x}^{2} = A \\ \\ = > x = \sqrt{A} \: \: \: [Dimensions \: of \: rectangle \: is \: always \: positive]$

Now,

$\frac{ {d}^{2} P}{d {x}^{2} } = 2(0 - A \times \frac{ - 1}{ {x}^{3} } ) = \frac{2A}{ {x}^{3} } \\ \\ = > {}^{ \frac{ {d}^{2} P}{d {x}^{2} } } x = \sqrt{a} = \frac{2a}{( \sqrt{A} ) {}^{3} } > 0 \\ Hence, \\ = > x = \sqrt{A} \: P (Perimeter \: of \: rectangle) \: is \: smallest. \\ Therefore ;\\ = > y = \frac{A}{x} = \frac{a}{ \sqrt{A} } = \sqrt{A}$

Hence,

For the smallest perimeter, length & breadth of rectangle are equal (x=y= √A)

So,

rectangle is square.

Hope it helps ☺️