Math Prob

Volume of a cone = πr^2h/3 = 36 => h = 108/(πr^2) ... (1)
The slant height L = √(h^2 + r^2)
L^2 = r^2 + [108/(πr^2)]^2
L^2 = r^2 + (11664/π^2)r^(-4) ...(2)
The circumference of the circle (cup mouth) = 2πr
When open up the cup, the slant height becomes the radius
The arc length = Lθ = 2πr => θ = 2πr/L
Area of the paper A = (L^2)θ/2 = (L^2)(2πr/L)/2 = πrL ... (3)
dA/dr = πL + πr(dL/dr)
dA/dr = 0 => dL/dr = -L/r ... (4)
Differentiate (2) wrt x,
2L(dL/dr) = 2r - (46656/π^2)r^(-5)
Use (4), 2L(-L/r) = 2r - (46656/π^2)r^(-5)
-2L^2 = 2r^2 - (46656/π^2)r^(-4)
-L^2 = r^2 - (23328/π^2)r^(-4)
Use (2),
- r^2 - (11664/π^2)r^(-4) = r^2 - (23328/π^2)r^(-4)
2r^2 = (11664/π^2)r^(-4)
r^6 = 5832/π^2
r = 2.897 => h = 4.097 => L = 5.017 => A = 45.66
r = 3 => h = 3.82 => L = 4.857 => A = 45.78
r = 2.8 => h = 4.385 => L = 5.203 => A = 45.76
Hence when r = 2.897 and h = 4.097, the paper consumed is a minimum

2009-11-23 22:08:18 補充：
Area of paper = (πL^2)(θ/2π)
θ is the angle of the sector in radian
Complete circle = 2π radian
Hence area of paper = (L^2)θ/2
And θ has been obtained in previous step to be 2πr/L
okay?

2009-11-23 22:11:50 補充：
See this site with diagram : http://www.mathsteacher.com.au/year10/ch14_measurement/18_cone/20cone.htm

What about i use s=pi*r*sqrt(r^2+h^2) to solve the problem. Is it duable?

2009-11-23 20:56:58 補充：
Area of the paper A = (L^2)θ/2 = (L^2)(2πr/L)/2 = πrL ...
I don't get this step.