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一个长方形的周长为20cm,围成圆柱体积最大为
题目内容:
一个长方形的周长为20cm,围成圆柱体积最大为优质解答
将长方形围成圆柱体后,长方形的两条边分别为圆柱体的高和底面周长:
设所围成的圆柱体的高为x,其中0<x≤10
则底面周长=(20÷2)-x=10-x
底面半径为 = (10-x)/(2π)
圆柱体积V = π*[(10-x)/(2π)]^2*x = x(x-10)^2 /(4π) = (x^3-20x^2+100x) /(4π)
V' = 1/π (3x^2-40x+100) = 3/(4π) * (x-10/3)(x-10)
当x∈(0,10/3)时,V'>0,V单调增;x∈(10/3,10)时,V'<0,V单调减
当x=10/3时,V取最大值:
Vmax = x(x-10)^2 /(4π) = 10/3*(10/3-10)^2 /(4π) = 1000/(27π) - 追问:
- V' = 1/π (3x^2-40x+100) = 3/(4π) * (x-10/3)(x-10)怎么算的
- 追答:
- 对V = (x^3-20x^2+100x) /(4π) 求导数: V' = 1/(4π) (3x^2-40x+100) = 1/(4π) (3x-10)(x-10) 【分解因式】 = 3/(4π) (x-10/3)(x-10)
优质解答
设所围成的圆柱体的高为x,其中0<x≤10
则底面周长=(20÷2)-x=10-x
底面半径为 = (10-x)/(2π)
圆柱体积V = π*[(10-x)/(2π)]^2*x = x(x-10)^2 /(4π) = (x^3-20x^2+100x) /(4π)
V' = 1/π (3x^2-40x+100) = 3/(4π) * (x-10/3)(x-10)
当x∈(0,10/3)时,V'>0,V单调增;x∈(10/3,10)时,V'<0,V单调减
当x=10/3时,V取最大值:
Vmax = x(x-10)^2 /(4π) = 10/3*(10/3-10)^2 /(4π) = 1000/(27π)
- 追问:
- V' = 1/π (3x^2-40x+100) = 3/(4π) * (x-10/3)(x-10)怎么算的
- 追答:
- 对V = (x^3-20x^2+100x) /(4π) 求导数: V' = 1/(4π) (3x^2-40x+100) = 1/(4π) (3x-10)(x-10) 【分解因式】 = 3/(4π) (x-10/3)(x-10)
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