A rectangle has an area of 50 `"cm"^(2)`. Find its dimensions for least perimeter.

Explanation

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 the rectangle `=2(x+y)`
`=2(x+(50)/(x))`
Let `f(x)=2(x+(50)/(x))" ":.f'(x)=2(1-(50)/(x^(2)))`
and `f''(x)=-2(-(100)/(x^(3)))=(200)/(x^(3))`
Now f'(x)=0, if `1-(50)/(x^(2))=0`, i.e., if `x^(2)=50`, i.e., if `x=+-5sqrt(2)`
But x is not negative. `" ":.x=5sqrt(2)` and
`f^(n)(5sqrt(2))=(200)/((5sqrt(2))^(3))gt 0`
`:.` by the second derivative test, f has a minimum value at `x=5sqrt(2)`
When `x=5sqrt(2),y=(50)/(5sqrt(2))=5sqrt(2)`
`:.x=5sqrt(2)" cm",y=5sqrt(2)" cm"`
`:.` the rectangle is a square of side `5sqrt(2)` cm.