maximum problem (1 Viewer)

[fannymasta]

Find the maximum volume of a right circular cone whose slant edge has a constant length measure a.

the answer is

(2root3 x pi x a^3) /27

please help

 

[lolokay]

what's the volume of the cone in terms of a, and say, the height of the cone?

 

[fannymasta]

what's the volume of the cone in terms of a, and say, the height of the cone?

umm.. well i know v = 1/3 x pi x r^2 x h

are you saying i need to eliminate r?

 

[Timothy.Siu]

umm.. well i know v = 1/3 x pi x r^2 x h

are you saying i need to eliminate r?

prolly just go a^2=r^2+h^2

and then u can eliminate r and then differentiate find the value for max. and sub it back in and ur done

 

[Trebla]

Let h be the height of the cone, and r be the radius of the cone. We know that r² = a² - h² from Pythagoras' theorem.

V = πr²h/3
= π(a² - h²)h/3
= π(a²h - h³)/3

dV/dh = πa²/3 - πh²
Maximum occurs when dV/dh = 0
i.e. πa²/3 - πh² = 0
=> h = a/√3 (taking positive root as h > 0)
d²V/dh² = - 2πh < 0 for all h > 0, therefore concave down so local maximum occurs at h = a/√3

V_max = π(a²[a/√3] - [a/√3]³)/3
= π(a³/√3 - a³/3√3)/3
= 2πa³/9√3
Rationalising denominator gives
= 2√3πa³/27

 

[fannymasta]

prolly just go a^2=r^2+h^2

what do i do after that ?

do i get r^2 = a^2 - h^2 then try sub it in?

 

[fannymasta]

Let h be the height of the cone, and r be the radius of the cone. We know that r² = a² - h² from Pythagoras' theorem.

V = πr²h/3
= π(a² - h²)h/3
= π(a²h - h³)/3

dV/dh = πa²/3 - πh²
Maximum occurs when dV/dh = 0
i.e. πa²/3 - πh² = 0
=> h = a/√3 (taking positive root as h > 0)
d²V/dh² = - 2πh < 0 for all h > 0, therefore concave down so local maximum occurs at h = a/√3

V_max = π(a²[a/√3] - [a/√3]³)/3
= π(a³/√3 - a³/3√3)/3
= 2πa³/9√3
Rationalising denominator gives
= 2√3πa³/27

THANKs trebla!