Question

The rectangle has area of 50 cm². Complete the following activity to find its dimensions for least perimeter.

Solution: Let x cm and y cm be the length and breadth of a rectangle.

Then its area is xy = 50

∴ `y =50/x`

Perimeter of rectangle `=2(x+y)=2(x+50/x)`

Let f(x) `=2(x+50/x)`

Then f'(x) = `square` and f''(x) = `square`

Now,f'(x) = 0, if x = `square`

But x is not negative.

∴ `x = root(5)(2)   "and" f^('')(root(5)(2))=square>0`

∴ by the second derivative test f is minimum at x = `root(5)(2)`

When x = `root(5)(2),y=50/root(5)(2)=root(5)(2)`

∴ `x=root(5)(2)  "cm" , y = root(5)(2)  "cm"`

Hence, rectangle is a square of side `root(5)(2)  "cm"`

Answer 1

Let x cm and y cm be the length and breadth of a rectangle.

Then its area is xy = 50

∴ `y =50/x`

Perimeter of rectangle `=2(x+y)=2(x+50/x)`

Let f(x) `=2(x+50/x)`

Then f'(x) = `2d/dx(x+50/x)=2[1+50(-1)x^(-2)]`

∴ f'(x)=`bb(2(1-50/x^2))`

and f''(x) = `2d/dx(1-50/x^2)=2[0-50(-2)x^(-3)]`

∴ f''(x) = `bb(200/x^3)`

Now,f'(x) = 0, if `2(1-50/x^2)=0   "i.e if"  1-50/x^2 =0`

i.e. if `50/x^2=1,"i.e. if"  x^2 = 50`

if x = `bb(+-root(5)(2))`

But x is not negative.

∴ `x = root(5)(2)   "and" f^('')(root(5)(2))=bb(200/(root(5)(2))^3)>0`

∴ by the second derivative test f is minimum at x = `root(5)(2)`

When x = `root(5)(2),y=50/root(5)(2)=root(5)(2)`

∴ `x=root(5)(2)  "cm" , y = root(5)(2)  "cm"`

Hence, rectangle is a square of side `root(5)(2)  "cm"`