美国微积分问题（懂英语的来）The cost of fuel to propel a boat through the water（in dollars per hour)is proportional to the cube of the speed.A certain ferry boat uses $100 worth of fuel per hour when crusing at 10 miles per hour.Apart

美国微积分问题（懂英语的来）The cost of fuel to propel a boat through the water（in dollars per hour)is proportional to the cube of the speed.A certain ferry boat uses $100 worth of fuel per hour when crusing at 10 miles per hour.Apart
美国微积分问题（懂英语的来）
The cost of fuel to propel a boat through the water（in dollars per hour)is proportional to the cube of the speed.A certain ferry boat uses $100 worth of fuel per hour when crusing at 10 miles per hour.Apart from fuel,the cost of running this ferry(labor,maintenance,and so on) is $675 per hour.AT what speed should it travel so as to minimize the cost per mile traveled?

美国微积分问题（懂英语的来）The cost of fuel to propel a boat through the water（in dollars per hour)is proportional to the cube of the speed.A certain ferry boat uses $100 worth of fuel per hour when crusing at 10 miles per hour.Apart
Solution:
Let the cost of fuel be x
x ∝ v³
Let x = kv³,where k is the proportionality.
Sub v = 10 mi/hr,x = 100 into x = kv³,
we have k = 100/1000 = 1/10
So,x = v³/10
Let the total cost per mile be y.
The time taken for one mile = 1/v.
y = (1/v)×v³/10 + (1/v)×675
= v²/10 + 675/v
dy/dv = v/5 - 675/v²
Let dy/dv = 0
so v³ = 675×5 = 3375,
v = 15 (mi/hr)
d²y/dv² = 1/5 + 1350/v > 0
So,when v = 15 (mi/hr)
the minimized cost per mile is
ymin = 15²/10 + 675/15 = 67.5($)

驱动船在水中前进的燃料费（按美元/小时计算）与船速的立方成正比。某船每小时消耗100美元以达到10英里/小时的速度。除了燃料消耗费以外，其他花销（人力，保养等）为675美元/小时。那么使得此船每英里消耗资金最低的速度是多少？
根据已知，燃料费F=a*v^3
v是速度，a是常数
代入100 = a*10^3 得出a=0.1
所以燃料费 F = 0.1*v^3
...

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驱动船在水中前进的燃料费（按美元/小时计算）与船速的立方成正比。某船每小时消耗100美元以达到10英里/小时的速度。除了燃料消耗费以外，其他花销（人力，保养等）为675美元/小时。那么使得此船每英里消耗资金最低的速度是多少？
根据已知，燃料费F=a*v^3
v是速度，a是常数
代入100 = a*10^3 得出a=0.1
所以燃料费 F = 0.1*v^3
所以每小时的消耗为
675+0.1v^3
因为每小时行驶v*1=v英里
所以每英里消耗资金Y为
Y = 675/v + 0.1v^2
也就是求这个函数的最小值
这个函数的导数是
y' = 0.2v - 675/v^2
另y' = 0 => v = 15
知道在v=15时导数为0
在v<15时导数小于0，说明此时y递减
在v>15时导数大于0，说明此时y递增
所以v=15时取得最小值
ymin = 0.2*15^2 + 675/15 = 90

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请楼主追加楼下分数，15分一般找不到这种答案。