Application of Derivatives

Question

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $tan to the power of negative 1 end exponent square root of 2.$

Solution

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

$therefore space space space space straight V space equals space 1 third πx squared straight h space equals space 1 third straight pi left parenthesis straight y squared minus straight h squared right parenthesis space straight h space space space space space space space space space space space space space space left square bracket because space space of space left parenthesis 1 right parenthesis right square bracket therefore space space space space space space straight V space equals space 1 third straight pi left parenthesis straight y squared straight h minus straight h cubed right parenthesis space space space space space space dV over dh space equals straight pi over 3 left parenthesis straight y squared minus 3 straight h squared right parenthesis$
For V to be maximum or minimum, $dV over dh space equals space 0$
$therefore space space space straight pi over 3 left parenthesis straight y squared minus 3 straight h squared right parenthesis space equals space 0 space space space space space space space rightwards double arrow space space space space space straight y squared minus 3 straight h squared space equals space 0 space space space space space rightwards double arrow space space space space straight h space equals space fraction numerator straight y over denominator square root of 3 end fraction space space space space space space space space fraction numerator straight d squared straight V over denominator dh squared end fraction space equals space straight pi over 3 left parenthesis 0 minus 6 straight h right parenthesis space equals space minus 2 space straight pi space straight h At space space space straight h equals fraction numerator straight y over denominator square root of 3 end fraction space fraction numerator straight d squared straight V over denominator dh squared end fraction space equals space minus 2 straight pi fraction numerator straight y over denominator square root of 3 end fraction less than 0$
$therefore space space space space space space space straight V space is space max. space when space straight h space equals space fraction numerator straight y over denominator square root of 3 end fraction space straight i. straight e. comma space space space space when space straight h squared space equals space straight y squared over 3 straight i. straight e. comma space when space 3 straight h squared space equals space straight y squared space space space space space space space space straight i. straight e. comma space space space when space 3 straight h squared space equals space straight x squared plus straight h squared space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets straight i. straight e. comma space when space 2 straight h squared space equals space straight x squared space space space space space space space space straight i. straight e. space space space space when space square root of 2 space straight h space equals space straight x straight i. straight e. comma space when space straight x over straight h equals space square root of 2 space space space space space space space straight i. straight e. space space space when space tanθ space equals square root of 2 space space space space straight i. straight e. comma space when space tanθ space equals space tan to the power of negative 1 end exponent square root of 2$

For Daily Free Study Material Join wiredfaculty

Application Of Derivatives

Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is $tan to the power of negative 1 end exponent square root of 2.$

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction