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Finding Maxima and Minima of a Function

Differentiation has many applications in our day to day lives, and one of them is in finding maxima and minima of a given function, i.e. maximum and minimum values of a function.

This has many applications in real life, like finding the maximum volume of a given shape, etc.


Let’s look at an example of finding maxima and minima of a function:

Example 1: A rectangular prism with a square base is to have a surface area of 300 cm2.

maximum volume of a rectangular prism

a) Show that the volume of the prism is given by V = {150x~-~x^3)/2

b) Find the dimensions of the prism that will have the maximum volume.


a) Given surface area of prism = 300 cm2,

SA y  = 2x2 + 4xy  =  300. Hence

y = x2 + 2xy  =  150, or

y = {150~-~x^2)/{2x}

Volume of the prism V = x2y

V = x2 {150~-~x^2)/{2x}

= x{150~-~x^2)/2

= {150x~-~x^3)/2


b) For the volume to be maximum

dV/dx = 0   and    {d^2V}/{dx^2} < 0

dV/dx150/2 – 3x^2/2

= 75 – 3x^2/2 = 0

75 = 3x^2/2


x = pm{sqrt{50}}

Since x cannot be -{sqrt{50}}, x is {sqrt{50}}

{d^2V}/{dx^2}  = {-3/2}2x  = -3x

When x = {sqrt{50}},   {d^2V}/{dx^2}  = -3 {sqrt{50}}  < 0 (maximum point)

Maximum volume V = {150~*~{sqrt{50}}~-~({sqrt{50}})^3}/{2}



50{sqrt{50}} cm3  =  100{sqrt{2}} cm3


Substituting the value of x = {sqrt{50}} in y = {150~-~x^2)/{2x}, we get

y = {150~-~50}/{2 {sqrt{50}}}

y = {100}/{2 {sqrt{50}}}{50}/{sqrt{50}}

{50}/{5{sqrt{2}}} = {10}/{sqrt{2}}

{10{sqrt{2}}}/2 = 5{sqrt{2}}


Maximum volume of V = 100{sqrt{2}} cm3 has dimensions of the prism to be x = 5{sqrt{2}} cm and y = 5{sqrt{2}} cm.

In other words, maximum volume of a prism is achieved when it is a cube.


Here are some more applications of the maxima and minima of a function.