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    				| 1. | A rectangle has an area of 50 `"cm"^(2)`. Find its dimensions for least perimeter. | 
| Answer» 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 | |