Show that the rectangle of maximum area that can be inscribed in a circle of radius r is a square of side square root of 2 straight r

Asked by Topperlearning User | 19th Aug, 2014, 09:25: AM

Expert Answer:

Let space ABCD space be space the space rectangle space inscribed space in space straight a space circle space of space radius space straight r.
AC space and space BD thin space are space diameters space of space length space 2 straight r space as space angle space in space semicircle space is space always space 90 degree.
Let space straight x space be space length comma space straight y space be space breadth space of space rectangle.
straight x squared plus straight y squared equals left parenthesis 2 straight r right parenthesis squared
rightwards double arrow straight y squared equals 4 straight r squared minus straight x squared
rightwards double arrow straight y equals square root of 4 straight r squared minus straight x squared end root
Area space of space reactangle comma space straight A equals xy
straight A equals straight x square root of 4 straight r squared minus straight x squared end root
rightwards double arrow dA over dx equals square root of 4 straight r squared minus straight x squared end root minus fraction numerator 2 straight x squared over denominator 2 square root of 4 straight r squared minus straight x squared end root end fraction
rightwards double arrow dA over dx equals fraction numerator 4 straight r squared minus straight x squared minus straight x squared over denominator square root of 4 straight r squared minus straight x squared end root end fraction
rightwards double arrow dA over dx equals fraction numerator 4 straight r squared minus 2 straight x squared over denominator square root of 4 straight r squared minus straight x squared end root end fraction
For space maximum space area comma
dA over dx equals 0
rightwards double arrow fraction numerator 4 straight r squared minus 2 straight x squared over denominator square root of 4 straight r squared minus straight x squared end root end fraction equals 0
rightwards double arrow 4 straight r squared minus 2 straight x squared equals 0
rightwards double arrow straight x equals square root of 2 straight r
Also comma space straight y equals straight r square root of 2
Since space the space length space and space breadth space are space same comma space therefore space it space is space straight a space figure.
So space required space rectangle space of space length space and space breadth space is space straight a space straight a space square space of space side space straight r square root of 2.

Answered by  | 19th Aug, 2014, 11:25: AM