1. Find the points of local maximum (or) local minimum for the function x3 - 9x2 + 24x - 12
Sol: Let f(x) = x3 - 9x2 + 24x - 12
⇒ f' (x) = 3x2 - 18x + 24
⇒ f" (x) = 6x - 18
For maximum (or) minimum f' (x) = 0
⇒ 3x2 - 18x + 24 = 0
⇒ x2 - 6x + 8 = 0
⇒ x2 - 2x - 4x + 8 = 0
⇒ x (x - 2) - 4 ( x - 2) = 0
⇒ (x - 2) (x - 4) = 0
∴ x = 2 ; x = 4
At x = 2; f"(2) = 6(2) - 18 = - 6 < 0
At x = 2, f(x) has maximum value
∴ Local maximum value f(2) = 8 - 36 + 48 - 12
⇒ f(2) = 8
At x = 4: f"(4) = 24 - 18 = 6 > 0
At x = 4, f(x) has minimum value
∴ Local minimum value f(4) = 64 - 144 + 96 - 12
f(4) = 4
2. Show that the function 
has a minimum value at x = e
sol: Given :
for minimum value, f' (x) = 0
⇒ log x = 1
∴ x = e ; At x = e;
⇒ f" (e) = 1/e > 0
∴ At x = e, f(x) has minimum value.
3. Show that y = sin3 x cos x has a maximum value at x = 
and find its value.
Sol : Given: y = sin3 x cos x
for maximum value = 0
⇒ 3 sin2 x cos2 x - sin4x = 0
⇒ sin2 x (3cos2x - sin2 x) = 0
⇒ sin2x = 0 ; ⇒ 3 cos2x - sin2x = 0
⇒ sin x = 0 ; ⇒ 3 cos2x = sin2x
∴ x = 0 ; ⇒ tan2x = 3
⇒ tan x = 
∴ x =
At x = ;
∴ At x = , y = sin3x cos x has maximum value.
∴ Maximum value: sin3 cos
4. Show that the semi-vertical angle of the cone of maximum volume and of given slant height is tan-1(
)
Sol : Let '' be the semi-vertical angle of the cone.
Let r, h, l be the radius, height and slant height of the cone respectively.
Volume of the cone: V = 
π r2h
For maximum value of V : 
= 0
⇒ 2 sin α cos2α - sin3 α = 0
⇒ 2 cos2 α - sin2 α = 0
⇒ 2 cos2 α = sin2 α
= tan2 α = 2
⇒ tan α = 
∴ α = tan-1 ()
∴ If V is maximum then semi-vertical angle is tan-1 ().
5. Show that the maximum rectangle inscribed in a circle is a square.
Sol: Let 'l' be the length and 'b' be the breadth of the rectangle.
Let ABCD be the rectangle inscribed in a circle of radius 'r'.
∴ l2 + b2 = (2r)2
For maximum value : 
= 0
4 r2 - 2 l2 = 0
2 l2 = 4r2
l2 = 2r2
∴ l = r
∴ A is maximum when l = b
Hence the maximum rectangle inscribed in a circle is a square.
6. Show that the height of the cylinder of maximum volume that can be inscribed in a sphere of radius 'a' is 
Sol: Let R be the radius and H be the height of the cylinder.
From ∆OAB, OA2 + AB2 = OB2
Volume of the cylinder : V = πR2 H
For maximum value : 
= 0
7. A cylinder is inscribed in a sphere of radius R. Show that the height of the cylinder is R when its surface area is maximum.
Sol: Let 'O' be the centre of the sphere of radius R.
Let 'r' be the radius and 'h' be the height of the cylinder inscribed in a sphere.
For maximum (or) minimum : 
= 0
⇒ h2 = 2R2
∴ h = R
Hence the surface area of the cylinder is maximum when h = R
Writer Sayyad Anwar
