Proving Maximum Volume of a Right Circular Cone: Optimization Problem Solution

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SUMMARY

The discussion centers on proving the maximum volume of a right circular cone given a fixed surface area. The established equation is 9V² = r²(S² - 2πr²S), where V is the volume, r is the base radius, and S is the surface area. The maximum volume occurs when the semi-vertical angle θ satisfies tan θ = 1/(2√2). The differentiation of the volume function with respect to r leads to critical values, and the second derivative test is recommended to confirm whether these values correspond to a maximum.

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  • Understanding of calculus, specifically differentiation and critical points.
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  • Knowledge of surface area and volume formulas for cones.
  • Ability to apply the second derivative test for optimization problems.
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Homework Statement



A right circular cone of base radius r and height h has a total surface area S and volume V . Show that 9V2=r2(S2-2pir2S) . (i can do this part) . Hence or otherwise , show that for a fixed surface area S , the maximum volume of the cone occurs when its semi-vertical angle , theta is given by tan theta=1/2(root 2)

The Attempt at a Solution



From the proven equation ,

9V2=r2(S2-2pir2S)

Differentiate this wrt to r ,

dV/dr=(2S2r-8pi Sr3)/(18V)

dV/dr=0 , S=4pi r2 , substitue S with the area of cone , then
tan theta=r/h=1/(2 root 2)

This is my question , how do i prove that its a maximum ?
 
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when you took the derivative of the Volume function and set it equal to zero you are finding a critical value.
There are a couple ways to test whether it is a local min or max. The second derivative test is one of them.

note: I have to check your derivation; not that I am doubting it or anything.
 

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