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# Examples of finding Maxima and Minima

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