Problems from 4.7
7.) A model for the yield Y of an agricultural crop as a function of the nitrogen level N in
the soil (measured in appropriate units) is
Y =
kN
1 + N2
where k is a positive constant. What nitrogen level gives the best yield?
In our procedure for analyzing maximum and minimums we are already given the
equation as a function of one variable so we do not need to do that part of the
problem, however we should observe the domain of interest. The domain of Y as a
function of N is [0, ∞) (as there can not be a negative level of Nitrogen in the soil).
There is also no diagram that we could draw in this situation.
To find the critical numbers we take the derivative with respect to N giving,
Y 0 (N ) =
(1 + N 2 )(k) − kN (2N )
k(1 − N 2 )
=
(1 + N 2 )2
(1 + N 2 )2
The critical numbers are 1 and -1, but only 1 is in the range of possible N values.
If one tests the derivative for N between 0 and 1 one will see that Y 0 > 0 and then
for N > 1, Y 0 > 0 thus the critical number at 1 corresponds to the location of a
maximum and it must be the absolute maximum by the First Derivative Test for
Absolute Extremes. Thus the maximum yield occurs when the nitrogen level is 1
1
12.) A box with a square base and an open top must have a volume of 32000 cm3 . Find
the dimensions of the box that minimize the amount of material used.
h
w
w
Let, w= length of one side of the base of the box in cm.
h=height of the box in cm.
A=Surface area of the box in cm2
The relations that we have are that:
A = 4wh + w2
This comes from the fact that the four rectangular sides all have area wh and the
base has area w2 . We also know the volume is 32000, but this is also the length times
the width times the height, thus:
32000 = hw2
Solving this relation for h we get that h = 32000/w2 . We substitute this into the
surface area formula above to get:
A=
128000
+ w2 .
w
2
Thinking of A as a function of w we know that it has a domain of (0, ∞). Taking the
derivative we have:
A0 (x) = −128000/w2 + 2w
Setting this equal to 0 we get:
2w =
128000
,
w2
or
w3 = 64000,
or
w = 40.
Also, there is a critical point at 0, but we know that that point is not in our allowable
interval.
If we test the first derivative for w between 0 and 40 we see that A0 (w) is negative here
and then for w > 40, A0 (w) > 0 Thus by the first derivative test for absolute extremes
we know the surface area must be minimized when x = 40. Thus the dimensions of
the box that minimize the materials are when the base is 40 by 40 cm, the height is
then 20 cm.
3
18.) Find the point on the line 6x + y = 9 that is closest to the point (−3, 1).
Which point on the line makes the distance the smallest?
æ
30
25
æ
20
æ
15
10
5
æ
-4
-3
-2
-1
1
To solve this problem let:
(x, y)= be the coordinates of the point on the line.
D= be the distance between (x, y) and (−3, 1).
Using the distance formula we have that:
p
D = (x + 3)2 + (y − 1)2 .
Also we have that x and y satisfy the equation of the line:
y = 9 − 6x
If we put this into the distance formula we have:
p
D(x) = (x + 3)2 + (8 − 6x)2 .
Now we could proceed to take the derivative of this function and find the critical
points but we could save ourselves a little work by considering the square of the
distance (if you think about it doesn’t it make sense that if the distance is minimized
then the square of the distance should be minimized and vice versa? Anyway, let
S(x) = (x + 3)2 + (8 − 6x)2 be the squared distance, the domain of this function is
all real numbers. Taking the derivative,
S 0 (x) = 2(x + 3) − 12(8 − 6x) = 74x − 90.
Setting this equal to 0 we get that x = 90/74. Also S 0 (x) < 0 for x < 90/74 and
S 0 (x) > 0 for x > 90/74. Thus by the first derivative test for Abs. extremes, there
is a local minimum at x=90/74.
The corresponding value of S is 676/37 and the
√
corresponding distance is 26/ 37.
4
24.) Find the dimensions of the rectangle of largest area that has its base on the x-axis and
its other two vertices above the x-axis and lying on the parabola y = 8 − x2 .
10
8
6
(x, 8 - x2 L
4
8 - x2
2
0
x
-4
x
0
-2
2
4
Let x and y be the coordinates of the upper right corner of the rectangle. Using the
fact that they are on the parabola we can write this point at the upper right-most
corner as (x, 8 − x2 ) then the upper left corner is at (−x, 8 − x2 ) and the width of
the rectangle is 2x and the height of the rectangle is 8 − x2 . Thus the area of the
rectangle as a function of x is
A(x) = 2x(8 − x2 ) = 16x − 2x3 .
√
The domain of this function is [0, 8] Taking the derivative we obtain
Setting this function to be 0 we have:
6x2 = 16 → x2 =
dA
dx
= 16 − 6x2 .
√
16
→ x = ±4/ 6
6
Consequently x = ± √46 , but we only need to consider the positive value and the end
√
√
points x = 0 and
x
=
8.
Evaluating
the
original
function
A(0)
=
0,
A(4/
6) ≈
√
17.4185 and A( 8) = 0. Consequently, the dimensions are
2
√
4
4
8
w = 2 ∗ √ = 8/ 6, h = 8 − √
= 8 − = 16/3
3
6
6
5
44.) A boat leaves a dock at 2:00PM and travels due south at a speed of 20 km/h. Another
boat has been heading due east at 15 km/hr and reaches the same dock at 3:00PM.
At what time were the two boats closest together?
Variables:
y= the distance of the boat heading south (boat A) from the dock.
x=the distance of the boat heading east (boat B) from the dock.
dsq=the squared distance between the boats (we will minimize this instead of the
distance, but the time which that happens is the same). t = time in hours since 2PM.
Relationships.
dsq = x2 + y 2
Since the first boat starts at the dock and is moving at a speed of 20 mph then,
y = 20t
Since boat B is an hour away, then it must be 15 miles away, thus its distance starts
at 15 at t=0 and then decreases by 15 mph, thus:
x = 15 − 15t.
Replacing these into the equation for dsg gives:
dsq(t) = (15 − 15t)2 + (20t)2 .
The domain of dsq(t) is [0, 1]. The critical points are places where dsq 0 (x) = 0, so:
dsq 0 (t) = −30(15 − 15t) + 800t = 1250t − 450.
This is zero when t = 450/1250 = .36. If we test the distance at the endpoints and
this critical point we see that:
dsq(0) = (15)2 , dsq(1) = (20)2 dsq(.36) = (12)2 .
Thus the boats are closest together at .36 hours after 2 PM, and their minimal distance
is 12 miles.
6
47.) An oil refinery is located on the north bank of a straight river that is 2 km wide. A
pipeline is to be constructed from the refinery to storage tanks located on the south
bank of the river 6 km east of the refinery. The cost of laying the pipe is $400000/km
over land to a point P on the north bank and $800.000/km under the river to the
tanks. To minimize the cost of the pipeline, where should P be placed?
The set up to this problem is very similar to the running/boat problem we did in
class.
Variable:
x= the distance from the refinery to the point P on the north bank in km.
y=the distance from P to the storage tanks in km
C=the total cost of the pipeline.
The total cost of the pipeline is the cost of the part on land which would be $400000x
plus the cost of the amount underwater which would be $800000y. Thus,
C = 400000x + 800000y.
Using the geometry of the problem similar to what we did in class we can see that,
p
y = 4 + (6 − x)2 .
Thus,
C(x) = 400000x + 800000
p
4 + (6 − x)2 .
The domain of C(x) is [0, 6]. The derivative of C(x) is:
−2(6 − x)
C 0 (x) = 400000 + 400000 ∗ p
4 + (6 − x)2
Setting this equal to zero we get:
400000 = 400000 p
2(6 − x)
4 + (6 − x)2
Dividing out the 400000 and multiplying the square root to the other side we obtain:
p
4 + (6 − x)2 = 2(6 − x)
7
Squaring both sides we get:
4 + (6 − x)2 = 4(6 − x)2 .
or
4 = 3(6 − x)2
or
or
4
= (6 − x)2
3
2
±√ = 6 − x
3
or x = 6 ± √23 . However, the number with the plus sign is not in our domain so we
can disregard it. The minus sign gives 6 − √23 ≈ 4.8453. If we test this and the end
points in C we get:
√
√
C(0) = $800000 24 ≈ $39200000, C(6) = $4000000 C(6 − 2/ 3) = $2400000
Thus the point should be placed approximately 4.8453 km to the east of the refinery
to minimize the cost, which will be $2,400, 000.
8