Application of Derivatives

Question

Prove that a conical tent of given capacity will require the least amount of convas when the height is $square root of 2$ times the radius of the base.

Solution

Let x be the radius of the base, h be the height and y the slant height of the conical tent.
Let V be the given capacity (i.e., volume)
$therefore space space space space straight V space equals space 1 third πx squared straight h rightwards double arrow space space space space space space straight h space equals space fraction numerator 3 straight V over denominator πx squared end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$
Let S be the curved surface.
$therefore space space space straight S space equals space πxy space equals space πx square root of straight h squared plus straight x squared end root space space space space rightwards double arrow space space space space space straight S squared space equals space straight pi squared straight x squared open square brackets straight h squared plus straight x squared close square brackets$
$equals space straight pi squared straight x squared space open square brackets fraction numerator 9 straight V squared over denominator straight pi squared straight x to the power of 4 end fraction plus straight x squared close square brackets space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 1 right parenthesis close square brackets$
$therefore space space space straight S squared space equals space fraction numerator 9 straight V squared over denominator straight x squared end fraction plus straight pi squared space straight x to the power of 4$
$Let space straight z space equals space straight S squared space space space space space space space space space space space space space space space rightwards double arrow space space space space space straight z space equals space fraction numerator 9 straight V squared over denominator straight x squared end fraction plus straight pi squared straight x to the power of 4 space space space space space dz over dx space equals space minus fraction numerator 18 space straight V squared over denominator straight x cubed end fraction plus 4 space straight pi squared space straight x cubed space space space space space$
$dz over dx space equals space 0 space space space space rightwards double arrow space space space space space fraction numerator negative 18 space straight V squared over denominator straight x cubed end fraction plus 4 straight pi squared straight x cubed space equals space 0 space space space space rightwards double arrow space space space space 9 straight V squared space equals space 2 space straight pi squared space straight x to the power of 6 space... left parenthesis 2 right parenthesis space space fraction numerator straight d squared straight z over denominator dx squared end fraction space equals space fraction numerator 54 space straight V squared over denominator straight X to the power of 4 end fraction plus 12 straight pi squared space straight x squared greater than 0 space for all space straight x therefore space space space space straight z space straight i. straight e. comma space straight S squared space and space hence space straight S space is space minimum space when space straight x space is space given space by space 9 straight V squared space equals space 2 space straight pi squared space straight x to the power of 6 straight i. straight e. comma space space space 9 1 over 9 straight pi squared space straight x to the power of 4 space straight h squared space equals space 2 straight pi squared straight x to the power of 6 space space space space space straight i. straight e. comma space space space straight h squared space equals space 2 straight x squared space space straight i. straight e. comma space straight h space equals space square root of 2 straight x therefore space space space space space straight S space is space minimum space when space height space is space square root of 2 space times space the space radius space of space the space base.$

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