Show that semi-vertical angle of a cone of fixed surface area and max.volume is sin-1(1/3)?

If the height is h and the base radius is r we have

V = ⅓πr²h … (i)

S = πr√(r²+h²) + πr² … (ii) … (S is a constant)

From (ii) (S/π−r²)² = r⁴+ r²h² → S²/π²−2Sr²/π = r²h² … (iii)

Anticipating use in (i) this gives h²r⁴ = S²r²/π²−2Sr⁴/π

Using (i) V² = (π²/9)( S²r²/π²−2Sr⁴/π ) = (S/9)( Sr²−2πr⁴ )

V is greatest when V² is greatest so we just set derivative to zero

(S/9)( 2Sr−8πr³ ) = 0 → r² = S/(4π)

Substituting in (iii) leads to h² = 2S/π

∴ tan(θ) = r/h = 1/√8

From a right-angled triangle of sides 1, √8 and 3, sin(θ) = ⅓

in the beginning, as you have in all hazard already been instructed, * perspective is a geometrical shape; * angel is a spirit being created via God. and that i assume you meant the later. particular, angels can sin. devil and his demons are angels who've sinned. And in the event that they might sin, there is not any longer something retaining the "Godly" angels from sinning too. yet do they? The Bible does no longer straight away answer that. even although, the Bible does exhibit that the saints will choose angels (a million Corinthians 6:3). Is it referring basically to those angels who accompanied devil? There would not seem that disadvantage on that Scripture; in any different case it ought to have suggested we would choose demons. So, it incredibly is available that even the Godly angels can sin; yet, in the event that they did, it can be a count of ideas-set as to what they do as quickly as they sinned -- the demons could proceed questioning that their sins have been suitable and justified, the Godly angels could repent and look for God's justification. .

Constraint Equation:

A = πr² + πrs

πrs = A - πr²

s = (A - πr²) / (πr)

Pythagorean Theorem:

r² + h² = s²

h² = s² - r²

h = √(s² - r²)

Trig Functions:

sinθ = r/s

sinθ = πr²/(A - πr²)

sinθ (A - πr²) = πr²

A sinθ - πr² sinθ = πr²

πr² + πr² sinθ = A sinθ

πr² (1 + sinθ) = A sinθ

πr² = A sinθ / (1 + sinθ)

r² = A sinθ / (π(1 + sinθ))

tanθ = r/h

h = r/tanθ

Objective Equation:

V = (1/3) π r² r / tanθ

V = (1/3) A sinθ / (1 + sinθ) * r / tanθ

V = (1/(3√π)) ((A sinθ) / (1 + sinθ))^(3/2) / tanθ

V = (1/(3√π)) ((A sinθ) / (1 + sinθ))^(3/2) cosθ / sinθ

V = (1/(3√π)) ((A sinθ) / (1 + sinθ))^(3/2) √(1 - sin²θ) / sinθ

V = (1/(3√π)) ((A sinθ) / (1 + sinθ))^(3/2) √((1 + sinθ)(1 - sinθ)) / sinθ

V = (1/(3√π)) A^(3/2) √(sinθ) √(1 - sinθ) / (1 + sinθ)

V = (1/(3√π)) A^(3/2) √[(sinθ)(1 - sinθ) / (1 + sinθ)²]

V = (1/(3√π)) A^(3/2) √[(sinθ - sin²θ) / (1 + sinθ)²]

V' = (1/2) [(sinθ - sin²θ) / (1 + sinθ)²]^(-1/2) * [(cosθ - 2sinθcosθ)(1 + sinθ)² - 2(sinθ - sin²θ)(1 + sinθ)cosθ]/(1 + sinθ)² = 0

(1/2) (1 + sinθ) / √(sinθ - sin²θ) * (-cosθ)(3sin²θ + 2sinθ - 1)/(1 + sinθ)² = 0

(1/2) (-cosθ)(3sin²θ + 2sinθ - 1) / [(1 + sinθ)√((sinθ)(1 - sinθ))] = 0

(-1/2) (3sin²θ + 2sinθ - 1) / √((sinθ)(1 + sinθ)) = 0

3sin²θ + 2sinθ - 1 = 0

(3sinθ - 1)(sinθ + 1) = 0

sinθ = -1, 1/3

θ = 3π/2 (reject), arcsin(1/3)

I get multiple solutions for θ, but θ is restricted to 0 < θ < π/2, so θ = arcsin(1/3) is the only solution.

Also, I get that the maximum volume is A^(3/2)/(6√(2π))