Let the length of rectangle be x cm and breadth of rectangle be y cm .
Given, Area = `50cm^(2)`
`thereforexy=50` . . . (i)
Perimeter of rectangle, P =2 (x+y)
`=2(x+(50)/(x))` [ From (i)]
`=(2(x^(2)+50))/(x)`
Now, `(dP)/(dx)=2(1-(50)/(x^(2)))`
Put, `(dP)/(dx)=0`,
`rArr2(1-(50)/(x^(2)))=0`
`rArr1-(50)/(x^(2))=0`
`rArrx^(2)=50`
`rArrx=5sqrt(2)m`
Put the value of x in equation (i),
`5sqrt(2)y=50`
`rArry=(50)/(5sqrt(2))`
`rArry=5sqrt(2)`
Again, `(d^(2)P)/(dx^(2))=(100)/(x^(3))=` Positive
`therefore` P is minimum.
`therefore` Dimensions when its perimeter is the least are `5sqrt(2)` m and `5sqrt(2)` m. lt
