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($)