Show that the volume of largest cone that can be inscribed in a sphere of radius R is 8 over 27 of the volume of the sphere.

Asked by Topperlearning User | 21st Aug, 2014, 01:22: PM

Expert Answer:

Let space straight r space be space the space radius space of space the space inscribed space cone space and space straight h space be space the space height.
Let space straight R space be space the space radius space of space the space sphere.
Using space pythagoras space theorem
left parenthesis straight h minus straight R right parenthesis squared plus straight r squared equals straight R squared
rightwards double arrow straight h squared minus 2 hR plus straight R squared plus straight r squared equals straight R squared
rightwards double arrow straight h squared minus 2 hR plus straight r squared equals 0
rightwards double arrow straight r squared equals 2 hR minus straight h squared
Let space straight V subscript straight s space be space the space volume space of space sphere space and space straight V subscript straight c space be space the space volume space of space the space cone.
straight V subscript straight s equals 4 over 3 πR cubed
Also comma space straight V subscript straight c equals 1 third πr squared straight h
straight V subscript straight c equals 1 third straight pi left parenthesis 2 hR minus straight h squared right parenthesis straight h
space space space space equals space 1 third straight pi left parenthesis 2 straight h squared straight R minus straight h cubed right parenthesis
Differenetiating space straight w. straight r. straight t. space straight h comma space we space get comma
dV subscript straight c over dh equals 1 third straight pi left parenthesis 4 hR minus 3 straight h squared right parenthesis
Now comma
dV subscript straight c over dh equals 0
rightwards double arrow 1 third straight pi left parenthesis 4 hR minus 3 straight h squared right parenthesis equals 0
rightwards double arrow straight h equals fraction numerator 4 straight R over denominator 3 end fraction
For space maxima space or space minima comma
fraction numerator straight d squared straight V subscript straight c over denominator dh squared end fraction equals 1 third straight pi left parenthesis 4 straight R minus 6 straight h right parenthesis
space space space space space space space space space space space space equals 1 third straight pi left parenthesis 4 straight R minus 6 cross times fraction numerator 4 straight R over denominator 3 end fraction right parenthesis
space space space space space space space space space space space space equals fraction numerator negative 4 straight R over denominator 3 end fraction straight pi less than 0
therefore Volume space is space maximum.
straight r squared equals 2 hR minus straight h squared
space space space equals 2 open parentheses fraction numerator 4 straight R over denominator 3 end fraction close parentheses straight R minus open parentheses fraction numerator 4 straight R over denominator 3 end fraction close parentheses squared
space space space equals fraction numerator 8 straight R squared over denominator 3 end fraction minus fraction numerator 16 straight R squared over denominator 9 end fraction
space space space equals fraction numerator 8 straight R squared over denominator 9 end fraction
straight V subscript straight c equals 1 third straight pi left parenthesis 2 straight h squared straight R minus straight h cubed right parenthesis
space space space space space equals 1 third straight pi open parentheses 2 open parentheses fraction numerator 4 straight R over denominator 3 end fraction close parentheses squared straight R minus open parentheses fraction numerator 4 straight R over denominator 3 end fraction close parentheses cubed close parentheses
space space space space space equals 1 third straight pi open parentheses fraction numerator 32 straight R cubed over denominator 9 end fraction minus fraction numerator 64 straight R cubed over denominator 27 end fraction close parentheses
straight V subscript straight c space equals 1 third straight pi fraction numerator 32 straight R cubed over denominator 27 end fraction....... left parenthesis straight i right parenthesis
straight V subscript straight s equals 4 over 3 πR cubed....... left parenthesis ii right parenthesis
left parenthesis straight i right parenthesis divided by left parenthesis ii right parenthesis comma
straight V subscript straight c over straight V subscript straight s equals 8 over 27
rightwards double arrow straight V subscript straight c equals 8 over 27 straight V subscript straight s
Hence space Proved

Answered by Renu Varma | 21st Aug, 2014, 03:22: PM