Math 217
Name:
HW3 Answer
02/18/ 2014
TA: Jiyuan Han
Problem 11, Page 108 If a locomotive is traveling at v miles per hour, the cost per hour
of fuel v 2 /25 dollars, and other costs are 100 dollars per hour regardless of speed. What is the
speed that minimizes cost per mile.
[ 1]Solution: In order to find the money cost per mile, we need to find the money cost per
hour, and the distance the locomotive moves per hour.
Some notations: let s be the total money cost per hour, let m be the money cost per mile. So
we want to minimize m
2
Since the speed is v, the distance the locomotive moves in 1 hour is v mile;s = v25 + 100. Then
we have:
s
v
100
m(v) = =
+
v
25
v
First we want to find the stationary points:
m0 (v) =
1
100
− 2 =0
25
v
⇒ v = 50
Then, we need to check v = 50 is a minimum point. When v < 50, m0 (v) < 0; when v > 50,
m0 (v) > 0. Thus v = 50 is a global minimum point. Thus the speed v = 50 mile/hour would
minimize the cost per mile.
Problem 24, page 109 : In the xy-plane, an observer stands at the origin (0,0). A car
travels along the curve y = x12 . How close does the car come to the observer?
[ 2]Solution:
The location of the car is: (x, y). The distance from the car to the observer is:
p
2
2
d = x + y . We want to minimize d. But this is equivalent to minimize d2 , which makes
the calculation simpler.
Don’t forget we also have: y = x12 .
To find the stationary point:
(d2 )0 (x) = (x2 +
1 0
4
)
=
2x
−
=0
x4
x5
1
1
⇒ x = (2) 6 , −(2) 6
1
1
Still we need to check if x = (2) 6 , −(2) 6 are minima points.
Since the path of the car is symmetric along the y-axis, we only need to check the point
1
1
1
x = (2) 6 . When 0 < x < (2) 6 , (d2 )0 (x) < 0; when x > (2) 6 , (d2 )0 (x) > 0. Thus x =
1
1
6 , −(2) 6 are global minimal points. Then the shortest distance from the car to the observer
(2)q
is
1
1
(2) 3 + (4)− 3 .