一道微积分应用题

对不起，电脑出了问题，图形出不来，还是语言叙述吧，不好意思。

解答：

设：圆柱体的半径是r；
则：圆柱体的高 H = 2×√(R²-r²)
圆柱的体积 V = πr²H = 2πr²√(R²-r²)
dV/dr = 4πr√(R²-r²) + 2πr²(-r)/[√(R²-r²)]
= 4πr√(R²-r²) - 2πr³/[√(R²-r²)]
令 dV/dr = 0
得 2√(R²-r²) = r²/[√(R²-r²)]
2(R²-r²) = r²， 3r² = 2R² r = R√(2/3)

d²V/dr²= 4π√(R²-r²)-4πr²/[√(R²-r²)]-6πr²/[√(R²-r²)]-6πr⁴/[√(R²-r²)^(3/2)]
= 4π√(R²-r²)-10πr²/[√(R²-r²)]-6πr⁴/[√(R²-r²)^(3/2)]

当 r = R√(2/3)
d²V/dr²= (4√3)πR/3-(20√3)πR/3-(8√3)πR = -(40√3)πR/3 < 0

∴Vmax = 2πr²√(R²-r²)|(r=R√(2/3)) = (4√3)πR³/9

取得极大值的条件是：H = 2×√(R²-r²) = (2√3)R/3