Hope N.

asked • 07/11/18# A sector with central angle θ is cut from a circle of radius R = 24 inches, and the edges of the sector are brought together to form a cone.

A sector with central angle θ is cut from a circle of radius R = 24 inches (see figure), and the edges of the sector are brought together to form a cone. Find the magnitude of θ such that the volume of the cone is a maximum.

More

## 1 Expert Answer

PROBLEM

A sector with central angle θ is cut from a circle of radius R = 24 inches, and the edges of the sector are brought together to form a cone.

A sector with central angle θ is cut from a circle of radius R = 24 inches (see figure), and the edges of the sector are brought together to form a cone. Find the magnitude of θ such that the volume of the cone is a maximum.

A sector with central angle θ is cut from a circle of radius R = 24 inches (see figure), and the edges of the sector are brought together to form a cone. Find the magnitude of θ such that the volume of the cone is a maximum.

SOLUTION

Consult the information on Building a Cone at http://mathforum.org/dr.math/faq/formulas/BuildCone.html

Use Case 9 since we know the radius, s = 24 inches and we are solving for the central angle, T = θ, which will maximize the volume, V = πr

^{2}h/3.From Case 9, we know for the cone, r = sT/(2π) and h = √[s

^{2}- r^{2}].Therefore, V = (π/3)[(sT)/(2π)]

^{2}(√[s^{2}- ((sT)/(2π))^{2}])To maximize V find dV/dT and set it equal to zero.

Differentiate using the product rule, resulting ultimately after simplification (and substituting θ for T) as:

(2θs

^{3}/4π)√[(4π^{2}-θ^{2})/(4π^{2})] - (θ^{3}s^{3})/(4π√[(4π^{2}-θ^{2})/(4π^{2})]) = 0So, (2θs

^{3}/4π)√[(4π^{2}-θ^{2})/(4π^{2})] = (θ^{3}s^{3})/(4π√[(4π^{2}-θ^{2})/(4π^{2})])^{2}= 4π

^{2}/(4π

^{2}+1) θ = √[4π

^{2}/(4π

^{2}+1)] θ = 0.98757049215 radians θ = 56.5836211719 degrees

## Still looking for help? Get the right answer, fast.

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

#### OR

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

Andy C.

07/11/18