So,
x = 5 Now
A″(x) =
2
2
2
2
( 2 )
100
( 4
10)
( 2
10
100)
2 100
100
x
x
x
x
x
x
x
−
−
−
−
− −
−
+
−
−
=
3
3
2
2
2
300
1000
(100
)
x
x
x
−
−
−
 (on simplification)
or
A″(5) =
3
3
2
2
2(5)
300(5)
1000
2250
30
0
75 75
75
(100
(5) )
−
−
−
−
=
=
<
−
Thus, area of trapezium is maximum at x = 5  and the area is given by
A(5) =
2
2
(5
10) 100
(5)
15 75
75 3 cm
+
−
=
=
Example 26 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 6