volume (v) of a cone of base radius (r) and verticl height (h) is?

Question

V=1/3 X pi X r^2 X h

What is maximum volume of a cone with SLOPE HEIGHT 3m????

Answer 1

v = pi x r^2 x h x 1/3

Pythagorous for slope gives:
r^2 + h^2 = 3^2 = 9
substituting for r^2 gives
v=pi(9-h^2)h/3
= (9 x pi x h)/3 - (pi x h^3)/3

To get max volume, differentiate dv/dh and solve for 0 which gives

dv/dh = (3 x pi) - (pi x h^2) = 0
which gives h = sqrt 3

Substituting this value back to the opriginal volume equation, we get

v =( 3 x pi x sqrt 3) - (pi x (sqrt 3)^3)/3
= pi x sqrt3 x (3 - 3/3)
= 2 x pi x sqrt 3

So, in answer to your question, max volume = 2 x pi x sqrt 3

Answer 2

max volume of cone with slope height 3 is when radius of base is sqrt(6) and height is sqrt(3)
max V = 2XpiXsqrt(3)

Answer 3

use pythagoras to work out the vertical height, then the equation.
do u know the radius? or the angle of the slope?

is there any other formula u can use to work out either the angle of the slope or the radius/height?
(sorry, im not much help. im only in year 10 at school :)

Answer 4

V = ⅓πr²h
s = slope height
s = √(r² + h²)
s² = r² + h²
r² = s² - h²

V = ⅓πr²h = ⅓π(s² - h²)h = ⅓πhs² - ⅓πh³
dV/dh = ⅓πs² - πh² = 0
⅓πs² = πh²
s²/3 = h²
h² = s²/3
h = s/√3
r² = s² - h² = s² - s²/3 = (2/3)s²
r = √(2/3)s

h = s/√3 = 3/√3 = √3
r = √(2/3)(3) = √6

V = ⅓πr²h = ⅓π(√6)²(√3) = ⅓π*6*√3 = 2π√3

Answer 5

question the quantity V of a amazing around cone varies quickly with the sq. of its base radius r and its top h.? The consistent of proportionality is (a million/3)pie. that's a real assertion so what's your question?