Question

find the maximum volume of a right circular cone whose slant height is 3 cm

Answer 1

Answer:

$V_{max}= 10.89\:{cm}^{3}$

Solution:

we know that volume of a right circular cone is$V= \frac{1}{3} \pi \: {r}^{2} \: h$

here r: radius of cone

h height of cone

Relation between radius,height and slant height(l) of cone is given by

${l}^{2} = {h}^{2} + {r}^{2}$

let height of cone is x cm

so we can write radius in terms of x

$( {3)}^{2} = {x}^{2} + {r}^{2} \\ \\ {r}^{2} = 9 - {x}^{2}$

put this value in Formula of volume and differentiate with respect to x,to find maximum volume

$V= \frac{1}{3} \pi(9 - {x}^{2} )x \\ \\ \frac{dV}{dx} = 0 \\ \\ \frac{dV}{dx} = \frac{\pi}{3} (9 - {x}^{2} ) + \frac{x\pi}{3} ( 0- 2x) \\ \\ \frac{\pi}{3} (9 - {x}^{2} ) + \frac{x\pi}{3} ( 0- 2x) = 0 \\ \\ 9 - {x}^{2} - 2 {x}^{2} = 0 \\ \\ 9 - 3 {x}^{2} = 0 \\ \\ 3 {x}^{2} = 9 \\ \\ {x}^{2} = 3 \\ \\ x = \sqrt{3}$

So, Volume of cone will be maximum when it's height is √3 cm, and
it is

$V_{max} = \frac{1}{3} \pi(9 - 3)( \sqrt{3} ) \\ \\V_{max} = 2\pi \sqrt{3} \: {cm}^{3}$

Hope it helps you.