Application of Derivatives

Question

For a given current surface area of right circular cone when the volume is maximum. Prove that the semi-vertical angle is $straight theta$ where $sin space straight theta space equals space fraction numerator 1 over denominator square root of 3 end fraction.$

Solution

Let x be radius, h be the vertical height and y be the slant height of cone.
∴ y² = h²+ x² ...(1)
Let V be volume of cone.

$therefore space space space space space straight V space equals space 1 third πx squared straight h space equals space 1 third πx squared square root of straight y squared minus straight x squared end root space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets therefore space space space space space straight V squared space equals space 1 over 9 straight pi squared straight x to the power of 4 left parenthesis straight y squared minus straight x squared right parenthesis therefore space space space space space space fraction numerator 9 straight V squared over denominator straight pi squared end fraction space equals space straight x to the power of 4 left parenthesis straight y squared minus straight x squared right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis$
Now total surface area of cone = constant
$therefore space space space space πx squared plus πxy space equals constant rightwards double arrow space space space space straight x squared plus xy space equals space constant space space rightwards double arrow space space space 2 straight x plus straight x dy over dx plus straight y. space 1 space equals space 0 rightwards double arrow space space space space space dy over dx space equals space minus open parentheses fraction numerator 2 straight x plus straight y over denominator straight x end fraction close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 3 right parenthesis$
Now, V will be maximum when $straight V squared space space or space space fraction numerator 9 straight V squared over denominator straight pi squared end fraction$ is maximum.
$Let space fraction numerator 9 straight V squared over denominator straight pi squared end fraction equals straight Z therefore space space space space space straight z space equals space straight x to the power of 4 left parenthesis straight y squared minus straight x squared right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 2 right parenthesis close square brackets space space space space space space space space space space space dz over dx space equals space 4 straight x cubed space. space left parenthesis straight y squared minus straight x squared right parenthesis space plus space straight x to the power of 4 open parentheses 2 straight y dy over dx minus 2 straight x close parentheses space space space space space space space space space space space space space equals space 4 straight x cubed left parenthesis straight y squared minus straight x squared right parenthesis plus straight x to the power of 4 open square brackets 2 straight y open parentheses negative fraction numerator 2 straight x plus straight y over denominator straight x end fraction close parentheses minus 2 straight x close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 3 right parenthesis close square brackets$
$equals 4 straight x cubed left parenthesis straight y squared minus straight x squared right parenthesis space minus space 2 straight x cubed straight y left parenthesis 2 straight x plus straight y right parenthesis space minus 2 straight x to the power of 5 space equals space 4 straight x cubed straight y squared minus 4 straight x to the power of 5 minus 4 straight x to the power of 4 straight y minus 2 straight x cubed straight y squared space minus 2 straight x to the power of 5 equals space 2 straight x cubed straight y squared minus 6 straight x to the power of 5 minus 4 straight x to the power of 4 straight y$
$Now comma space dz over dx space equals space 0 space space gives space us space space space 2 straight x cubed straight y cubed space minus space 6 straight x to the power of 5 space minus space 4 straight x to the power of 4 straight y space equals space 0 space space space space therefore space space space space straight x cubed left parenthesis straight y squared minus 3 straight x squared minus 2 xy right parenthesis space equals space 0 space space space rightwards double arrow space space space space straight x cubed left parenthesis straight y plus straight x right parenthesis thin space left parenthesis straight y minus 3 straight x right parenthesis space equals space 0 space space space space therefore space space space space straight x space equals space 0 comma space space minus straight y comma space space space straight y over 3$
$Now space straight x space equals space 0 comma space minus straight y space are space not space possible comma space space space therefore space space space straight y space equals space 3 straight x space space When space straight y space is space slightly space less than space 3 straight x comma space space dz over dx space equals plus ve When space straight y space is space slightly greater than 3 straight x comma space space dz over dx space equals space minus ve therefore space space space space space space space straight z space or space space straight V space is space maximum space for space straight y space equals space 3 straight x. Let space space straight theta space be space the space semi minus vertical space angle therefore space space space space space space sin space straight theta space equals space straight x over straight y space equals space fraction numerator straight x over denominator 3 straight x end fraction space equals space 1 third space space space$

For Daily Free Study Material Join wiredfaculty