Let us look at two examples of finding maxima and minima of functions.
Example 1: A juice manufacturer wants to minimise the amount of plastic needed to make a cylindrical bottle of 500 ml capacity. Given 500 ml = 500 cm3, find the radius of the bottle with minimum surface area.
Volume of a cylinder = 
Therefore = 500
Rearranging the above, we get h = 
Surface area of a cylinder S = 
+ 
S = 
+
= 
+
= 
+ 
For minimum value
= 0 and 
> 0
When = 0, we have
=
r3 = 
r = 
= 4.30
= 
+ 
= 37.72 > 0 (r = 4.30 from above)
So at r = 4.3 cm, the surface area will be minimum.
And that surface area S = 
+ 
= 348.73 cm2
Example 2: Sand is tipped from a truck onto a pile. The rate, R kg/sec at which the sand is flowing is given by the expression
R = 100t – t3, for 0 ≤ t ≤ T where t is the time in seconds after the sand begins to flow.
What is the largest value of T for which the expression R is physically possible, and find the maximum rate of flow of the sand?
R ≥ 0 since you can’t untip the sand, so let us examine the value of t for which 100t – t3 ≥ 0
t(100 − t²) ≥ 0
t(10 − t)(10 + t) ≥ 0
This is a cubic curve. By substituting the values between -10, 0 and 10, we get the curve
The curve is above x-axis in the range 0 ≤ t ≤ 10
Since t cannot be negative, 0 ≤ t ≤ 10 is the solution, and the largest value of t is 10 seconds.
Maximum rate of flow of sand = 
when 
and 
= 100 – 3t2 = 0
3t2 = 100
t = 
= 
(t > 0)
= -6t = -6 x < 0 hence maximum
The maximum rate of flow is when t =
R = 100 x – 
= 
– 
= 
= 
= 384.90 kg/sec