1-9
(a) Point vehicles
vxQ = cars
hour = v
x = 42 . 1 v − v 2
0 . 324
Seek stationary point maximum
d Q
d v = 0 = 42 . 1 − 2 v
0 . 324 ∴ v * = 21 . 05 mph
Q* = 42 . 1(21 . 05) − 21 . 05 2
0 . 324 = 1367 . 6 cars/h Ans .
(b)
ll x2 − 1
0 . 324
Q = v
v (42 . 1) − v 2 + l
v
x + l =
Maximize Q with l = 10/5280 mi
v Q
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