A grain silo has the shape of a right circular cylinder surmounted by a hemisphere. If the silo is to have a volume of 495π ft^3, determine the radius and height of the silo that requires the least amount of material to build. Hint: The volume of the silo is
πr^2 h + 2/3πr^3,
and the surface area (including the floor) is π(3r^2 + 2rh)
DWRead
total volume = ⅔πr³ + πr²h = 495π
⅔r³ + r²h = 495
r²h = 495-⅔r³
h = 495/r² - ⅔r
total surface area A = 3πr² + 2πrh
= 3πr² + 2πr(495/r² - ⅔r)
= 3πr² + 2·495π/r - 4πr²/3
= (5/3)πr² + 2·495π/r
dA/dr = 2(5/3)πr - 2·495π/r² = 0
(5/3)πr = 495π/r²
(5/3)r³ = 495
r³ = 297
r = ∛297 = 3∛11 ≅ 6.67 ft
h = 495/r² - ⅔r ≅ 6.67 ft
radius is 6.67 ft
cylinder is 6.67 ft high
total height = 13.34 ft
ted s
so 495 = r² h + 2 r³ / 3 ===> h = [ 495 - 2 r³ / 3 ] / r² = 495 r^(-2) - 2 r / 3 ====> SA = π ( 3r² + 990/r - 4 r / 3 )
minimize SA.....................r ≈ 5.56