of the trapezium. Then A ≡ A( x ) = (sum of parallel sides) (height) ( ) ) A ′ ( x ) = ( ) ) Now A ′ ( x ) = gives x + x – = , i.e., x = and x = – . Since x represents distance, it can not be negative. So, x = .
Now A ″ ( x ) = ( ) ( ) ( ) −− ( (on simplification) A ″ ( ) = ( ) ( ) ( ( ) ) Thus, area of trapezium is maximum at x = and the area is given by A( ) = ( ) ( ) cm Example Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. Solution Let OC = r be the radius of the cone and OA = h be its height. Let a cylinder with radius OE = x inscribed in the given cone (Fig . ).
The height QE of the cylinder is given by QE OA = EC OC (since ∆ QEC ~ ∆ AOC) QE h = r QE = h r Let S be the curved surface area of the given cylinder. Then S ≡ S( x ) = xh r h rx S ( ) ) S ( ) h h ′ −π ′′ Now S ′ ( x ) = gives x = . Since S ″ ( x ) < for all x , S ′′ . So x = is a point of maxima of S.
Hence, the radius of the cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. . . Maximum and Minimum Values of a