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Solution :

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