868
Chapter 29
◆
More Applications of the Derivative
Fluid Flow
22. Water is running from a vertical cylindrical tank 3.00 m in diameter at the rate of 3兹h
m3/min, where h is the depth of the water in the tank. How fast is the surface of the water
falling when h 9.00 m?
23. Water is flowing into a conical reservoir 20.0 m deep and 10.0 m across the top, at a rate
of 15.0 m3/min. How fast is the surface rising when the water is 8.00 m deep?
24. Sand poured on the ground at the rate of 3.00 m3/min forms a conical pile whose height is
one-third the diameter of its base. How fast is the altitude of the pile increasing when the
radius of its base is 2.00 m?
25. A horizontal trough 5.00 m long has ends in the shape of an isosceles right triangle (Fig.
29–32). If water is poured into it at the rate of 0.500 m3/min, at what rate is the surface of
the water rising when the water is 1.00 m deep?
0.500 m3/min
5.00 m
FIGURE 29–32
Gas Laws
Use Boyle’s law, pv constant.
26. A tank contains 30.0 m3 of gas at a pressure of 35.0 kPa. If the pressure is decreasing at the
rate of 0.350 kPa/h, find the rate of increase of the volume.
27. The adiabatic law for the expansion of air is _pv1.4 C. If at a given time the volume is
observed to be 2.00 m3, and the pressure is 300 kPa, at what rate is the pressure changing
if the volume is decreasing 0.100 m3/s?
Miscellaneous
x
FIGURE 29–33
y
28. The speed v m/s of a certain bullet passing through wood is given by v 300兹 1 3x ,
where x is the depth in metres. Find the rate at which the speed is decreasing after the bullet
has penetrated 10.0 cm. (Hint: When substituting for dx/dt, simply use the given expression for v.)
29. As a man walks a distance x along a board (Fig. 29–33), he sinks a distance of y in., where
Px3
y 3EI
Here, P is the person’s weight, 165 lb.; E is the modulus of elasticity of the material in the
board, 1 320 000 lb./sq. in.; and I is the modulus of elasticity of the cross section, 10.9 (in.4).
If he moves at the rate of 25.0 in./s, how fast is he sinking when x 75.0 in.?
30. A stone dropped into a calm lake causes a series of circular ripples. The radius of the outer
one increases at 0.600 m/s. How rapidly is the disturbed area changing at the end of 3.00 s?
29–4
Optimization
In Chapter 28, we found extreme points, the peaks and valleys on a curve. We did this by finding
the points where the slope (and hence the first derivative) was zero. We now apply the same idea
to problems in which we find, for example, the point of minimum cost, or the point of maximum
efficiency, or the point of maximum carrying capacity.
Section 29–4
◆
869
Optimization
Suggested Steps for Maximum-Minimum Problems
1. Locate the quantity (which we will call Q) to be maximized or minimized, and locate the
independent variable, say, x, which is varied in order to maximize or minimize Q.
2. Write an equation linking Q and x. If this equation contains another variable as well, that
variable must be eliminated by means of a second equation. A graph of Q f(x) will show
any maximum or minimum points.
3. Take the derivative dQ/dx.
4. Locate the values of x for which the derivative is zero. This may be done graphically by
finding the zeros for the graph of the derivative or analytically by setting the derivative to
zero and solving for x.
5. Check any extreme points found to see if they are maxima or minima. This can be done
simply by looking at your graph of Q f(x), by applying your knowledge of the physical
problem, or by using the first- or second-derivative tests. Also check whether the maximum or minimum value you seek is at one of the endpoints. An endpoint can be a maximum or minimum point in the given interval, even though the slope there is not zero.
A list of general suggestions such as these is usually so vague as to be useless without
examples. Our first example is one in which the relation between the variables is given verbally
in the problem statement.
◆◆◆
Example 17: What two positive numbers whose product is 100 have the least possible sum?
Solution:
1. We want to minimize the sum S of the two numbers by varying one of them, which we call
x. Then
100
x
other number
2. The sum of the two numbers is
100
S x x
3. Taking the derivative yields
dS
dx
100
x2
1 4. Setting the derivative to zero and solving for x gives us
x2 100
x 10
Since we are asked for positive numbers, we discard the 10. The other number is
100/10 10.
5. But have we found those numbers that will give a minimum sum, as requested, or a maximum sum? We can check this by means of the second-derivative test. Taking the second
derivative, we have
200
S x3
When x 10,
200
S(10) 0.2
1000
which is positive, indicating a minimum point.
The sum of two positive numbers
whose product is a constant
is always a minimum when the
numbers are equal, as in this
example. Can you prove the
general case?
870
◆
Chapter 29
More Applications of the Derivative
Graphical Solution: A graph of S versus x, shown in Fig. 29–34(a), shows a minimum at
x 艐 10. To locate the minimum more accurately, we graph the derivative as shown in Fig. 29–34(b).
Note that it has a zero at x 10.
S
dS
dx
40
0.5
Min S
20
0
5
10
15
x
−0.5
5
0
10
x
15
(b)
(a)
◆◆◆
FIGURE 29–34
In the next example, the equation linking the variables is easily written from the geometrical relationships in the problem.
x
◆◆◆
Example 18: An open-top box is to be made from a square of sheet metal 40 cm on
a side by cutting a square from each corner and bending up the sides along the dashed
lines (Fig. 29–35). Find the dimension x of the cutout that will result in a box of the
greatest volume.
x
40 cm
Solution:
1. We want to maximize the volume V by varying x.
2. The equation is
V length width depth
(40 2x)(40 2x)x
x(40 2x)2
40 cm
FIGURE 29–35
3. Taking the derivative, we obtain
dV
dx
x(2)(40 2x)(2) (40 2x)2
4x(40 2x) (40 2x)2
(40 2x)(4x 40 2x)
(40 2x)(40 6x)
4. Setting the derivative to zero and solving for x, we have
V(cm3)
40 2x 0
Max
x 20 cm
4000
V
8
0
4
FIGURE 29–36
6.67
10
40 6x 0
20
x 6.67 cm
3
We discard x 20 cm, because it is a minimum value and results in the entire sheet
of metal being cut away. Thus we keep x 6.67 cm as our answer.
x(cm)
dV
dx
Graphical Check: The graph of the volume equation, shown in Fig. 29–36, shows a
maximum at x ⬵ 6.67, so we don’t need a further test of that point. The graph of the
◆◆◆
derivative has a zero at x 6.67, verifying our solution.
Our next example is one in which the equation is given.
Section 29–4
◆
871
Optimization
Example 19: The deflection y of a simply supported beam with a concentrated load P
(Fig. 29–37) at a distance x from the left end of the beam is given by
◆◆◆
Pbx
y (L2 x2 b2)
6LEI
where E is the modulus of elasticity, and I is the moment of inertia of the beam’s cross section.
Find the value of x at which the deflection is a maximum for a 8.00-m-long beam with a concentrated load 2.00 m from the right end.
This equation is actually valid
only for points to the left of the
load—the region of interest
to us. The equation is slightly
different for deflections to the
right of the load.
L
Load P
b
x
y
FIGURE 29–37
Simply supported beam with concentrated load.
Estimate: It seems reasonable that the maximum deflection would be at the centre, at x 4 m
from the left end of the beam. But it is equally reasonable that the maximum deflection would
be at the concentrated load, at x 6. So our best guess might be that the maximum deflection
is somewhere between 4 and 6 m.
Solution:
1. We want to find the distance x from the left end at which the deflection is a maximum.
2. The equation is given in the problem statement.
3. We take the derivative using the product rule, noting that every quantity but x or y is a
constant.
dy
Pb
[x (2x) (L2 x2 b2) (1)]
dx 6LEI
4. We set this derivative equal to zero and solve for x.
2x2 L2 x2 b2
3x2 L2 b2
x
兹
2
2
L b
3
We drop the negative value, since x cannot be negative in this problem. Now substituting
L 8.00 m and b 2.00 m gives us
x
兹
64.0 4.00
3
4.47 m
Thus the maximum deflection occurs between the load and the midpoint of the beam, as
expected.
5. It is clear from the physical problem that our point is a maximum, so no test is needed. ◆◆◆
The equation linking the variables in Example 19 had only two variables, x and y. In the
following example, our equation has three variables, one of which must be eliminated before
we take the derivative. We eliminate the third variable by means of a second equation.
◆◆◆
y
x
Example 20: Assume that the strength of a rectangular beam varies directly as its width
and the square of its depth. Find the dimensions of the strongest beam that can be cut from a FIGURE 29–38 Beam cut
round log 30.0 cm in diameter (Fig. 29–38).
from a log.
872
Chapter 29
◆
More Applications of the Derivative
Estimate: A square beam cut from a 30-cm log would have a width of about 20 cm. But we are
informed, and know from experience, that depth contributes more to the strength than width,
so we expect a width less than 20 cm. But from experience we would be suspicious of a very
narrow beam, say, 5 cm or less. So we expect a width between 5 and 20 cm. The depth has to be
greater than 20 cm but cannot exceed 30 cm. These numbers bracket our answer.
Solution:
1. We want to maximize the strength S by varying the width x.
2. The strength S is S kxy2, where k is a constant of proportionality. Note that we have hree
variables, S, x, and y. We must eliminate x or y. By the Pythagorean theorem,
x2 y2 30.02
so
y2 900 x2
Substituting, we obtain
S kx(900 x2)
3. The derivative is
dS
dx
kx(2x) k(900 x2)
3kx2 900k
4. Setting the derivative equal to zero and solving for x gives us 3x2 900, so
x 17.3 cm
We discard the negative value, of course. The depth is
y 兹 900 300 24.5 cm
5. But have we found the dimensions of the beam with maximum strength, or minimum
strength? Let us use the second-derivative test to tell us. The second derivative is
d2 S
6kx
dx2
When x 17.3,
d2 S
104k
dx2
Since k is positive, the second derivative is negative, which tells us that we have found a
maximum.
S
k=
3
k=
2
k=1
0
10
20
17.3
FIGURE 29–39
x(cm)
Graphical Solution or Check: As before, we can solve the problem or check the analytical solution graphically by graphing the function and its derivative. But here the equations
contain an unknown constant k!
Figure 29–39 shows the function graphed with assumed values of k 1, 2, and 3.
Notice that changing k does not change the horizontal location of the maximum point.
Thus we can do a graphical solution or check with any value of k, with k 1 being the
◆◆◆
simplest choice.
Sometimes a graphical solution is the only practical one, as in the following example.
◆◆◆ Example 21: Find the dimensions x and y and the minimum cost for a 10.0-cu. ft.-capacity,
open-top box with square bottom (Fig. 29–40). The sides are aluminum at $1.28/sq. ft., and the
bottom is copper at $2.13/sq. ft. An aluminum-to-aluminum joint costs $3.95/ft., and a copperto-aluminum joint costs $2.08/ft.
Section 29–4
◆
873
Optimization
Solution: The costs are as follows:
an aluminum side
the copper bottom
one alum.-to-alum. joint
one alum.-to-copper joint
$1.28xy
$2.13x2
$3.95y
$2.08x
the total cost C is then
C 4(1.28xy) 2.13x2 4(3.95y) 4(2.08x)
x
x
We can eliminate y from this equation by noting that the volume, 10.0 cu. ft., is equal to x 2 y, so
y 10.0/x2
x
Substituting gives
y
10.0
10.0
C 4(1.28x) 2.13x2 4(3.95) 4(2.08x)
2
x
x2
This simplifies to
C 2.13x2 8.32x 51.2x1 158x2
Taking the derivative,
FIGURE 29–40
C 4.26x 8.32 51.2x
2
316x
3
For an analytical solution, we would set C to zero and attempt to solve for x. For a numerical solution, we would use Newton’s method, the midpoint method, or a similar technique for
finding the root. We will do a graphical solution. As in the preceding examples, we graph the
function and its derivative as shown in Fig. 29–41 and find a minimum point at x 2.83 ft.
150
C
100
78.42
dC
dx
50
0
1
2
3
4
5
x (ft.)
FIGURE 29–41
Substituting back to get y,
10.0
10.0
y 1.25 ft.
x2
(2.83)2
We can get the total cost C by substituting back, or we can read it off the graph. Either way we
◆◆◆
get C $78.42.
Exercise 4
◆
Optimization
Number Problems
1. What number added to half the square of its reciprocal gives the smallest sum?
2. Separate the number 10 into two parts such that their product will be a maximum.
x
874
Chapter 29
◆
More Applications of the Derivative
3. Separate the number 20 into two parts such that the product of one part and the square of
the other part is a maximum.
4. Separate the number 5 into two parts such that the square of one part times the cube of the
other part will be a maximum.
Neighbor
432 m2
Minimum Perimeter
5. A rectangular garden (Fig. 29–42) laid out along your neighbor’s lot contains 432 m2. It is
to be fenced on all sides. If the neighbor pays for half the shared fence, what should be the
dimensions of the garden so that your cost is a minimum?
6. It is required to enclose a rectangular field by a fence (Fig. 29–43) and then divide it into
two lots by a fence parallel to the short sides. If the area of the field is 2.50 ha, find the
lengths of the sides so that the total length of fence will be a minimum.
7. A rectangular pasture 162 m2 in area is built so that a long, straight wall serves as one side
of it. If the length of the fence along the remaining three sides is the least possible, find the
dimensions of the pasture.
FIGURE 29–42 Rectangular
garden.
2.50 ha
Total area
Maximum Volume of Containers
8. Find the volume of the largest open-top box that can be made from a rectangular sheet of
metal 6.00 cm by 16.0 cm (Fig. 29–44) by cutting a square from each corner and turning
up the sides.
9. Find the height and base diameter of a cylindrical, topless tin cup of maximum volume if
its area (sides and bottom) is 100 cm2.
10. The slant height of a certain cone is 50.0 cm. What cone height will make the volume a
maximum?
FIGURE 29–43 Rectangular
field.
16.0 cm
Maximum Area of Plane Figures
11. Find the maximum area of a rectangle with a perimeter of 20.0 cm.
12. A window composed of a rectangle surmounted by an equilateral triangle is 10.0 m in
perimeter (Fig. 29–45). Find the dimensions that will make its total area a maximum.
6.00 cm
FIGURE 29–44 Sheet metal
for box.
13. Two corners of a rectangle are on the x axis between x 0 and 10 (Fig. 29–46). The
other two corners are on the lines whose equations are y 2x and 3x y 30. For
what value of y will the area of the rectangle be a maximum?
Maximum Cross-Sectional Area
14. A trough is to be made of a long rectangular piece of metal by bending up two edges so as
to give a rectangular cross section. If the width of the original piece is 14.0 cm, how deep
should the trough be made in order that its cross-sectional area be a maximum?
15. A gutter is to be made of a strip of metal 30.0 cm. wide, with the cross section having the
form shown in Fig. 29–47. What depth x gives a maximum cross-sectional area?
FIGURE 29–45 Window.
Minimum Distance
y=3
y=
16. Ship A is travelling due south at 40.0 km/h, and ship B is travelling due west at the same
speed. Ship A is now 10.0 km from the point at which their paths will eventually cross,
and ship B is now 20.0 km from that point. What is the closest that the two ships will get
to each other?
3x +
2x
y
0
10.0 cm
y
x
10.0 cm
0
x
FIGURE 29–46
FIGURE 29–47
10.0 cm
Section 29–4
◆
875
Optimization
y
y
x2
6
6
y= 2
Q
4
4
Q
2
2
(4, 1)
P(6, 0)
0
0
8x
y2 =
2
x
4
2
4
8
x
FIGURE 29–49
FIGURE 29–48
17. Find the point Q on the curve y x2Ⲑ2 that is nearest the point (4, 1) (Fig. 29–48).
18. Given one branch of the parabola y2 8x and the point P(6, 0) on the x axis (Fig. 29–49),
find the coordinates of point Q so that PQ is a minimum.
Inscribed Plane Figures
19. Find the dimensions of the rectangle of greatest area that can be inscribed in an equilateral
triangle, each of whose sides is 10.0 cm, if one of the sides of the rectangle is on a side of
the triangle (Fig. 29–50). (Hint: Let the independent variable be the height x of the rectangle.)
20. Find the area of the largest rectangle with sides parallel to the coordinate axes which can
be inscribed in the figure bounded by the two parabolas 3y 12 x2 and 6y x2 12
(Fig. 29–51).
10.0 cm
10.0 cm
x
y
x
10.0 cm
3y = 12 − x2
FIGURE 29–50
2
−4
0
−2
6y = x2 − 12
2
4
x
14.0
−2
FIGURE 29–51
20.0
21. Find the dimensions of the largest rectangle that can be inscribed in an ellipse whose major
FIGURE 29–52 Rectangle
axis is 20.0 units and whose minor axis is 14.0 units (Fig. 29–52).
inscribed in ellipse.
Inscribed Volumes
22. Find the dimensions of the rectangular parallelepiped of greatest volume and with a square
base that can be cut from a solid sphere 18.0 cm in diameter (Fig. 29–53).
23. Find the dimensions of the right circular cylinder of greatest volume that can be inscribed
in a sphere with a diameter of 10.0 cm (Fig. 29–54).
18.0 cm
10.0 cm
FIGURE 29–53
Parallelepiped
inscribed in sphere.
FIGURE 29–54
Cylinder inscribed
in sphere.
876
Chapter 29
◆
More Applications of the Derivative
24. Find the height of the cone of minimum volume circumscribed about a sphere of radius
10.0 m (Fig. 29–55).
h
10.0 m
FIGURE 29–55
about sphere.
Cone circumscribed
25. Find the altitude of the cone of maximum volume that can be inscribed in a sphere of
radius 9.00 m.
Most Economical Dimensions of Containers
26. What should be the diameter of a can holding 1 L and requiring the least amount of metal
if the can is open at the top?
27. A silo (Fig. 28–26) has a hemispherical roof, cylindrical sides, and circular floor, all made
of steel. Find the dimensions for a silo having a volume of 755 m3 (including the dome)
that needs the least steel.
Minimum Travel Time
28. A man in a rowboat at P (Fig. 29–56), 6.00 km from shore, desires to reach point Q on
the shore at a straight-line distance of 10.0 km from his present position. If he can walk
4.00 km/h and row 3.00 km/h, at what point L should he land in order to reach Q in the
shortest time?
L
Shore
Q
x
6.00 km
10.0 km
P
FIGURE 29–56 Find the fastest route
from boat P to shore at Q.
Beam Problems
29. The strength S of the beam in Fig. 29–38 is given by S kxy2, where k is a constant.
Find x and y for the strongest rectangular beam that can be cut from a 45.0-cm-diameter
cylindrical log.
30. The stiffness Q of the beam in Fig. 29–38 is given by Q kxy3, where k is a constant. Find
x and y for the stiffest rectangular beam that can be cut from a 45.0-cm-diameter cylindrical log.
Section 29–4
◆
877
Optimization
Light
M
x
Min
h
d
N
31. The intensity E of illumination at a point, from a light source at a distance x from the point,
is given by E kI冒x2, where k is a constant and I is the intensity of the source. A light M
has an intensity three times that of N (Fig. 29–57). At what distance from M is the illumina100 cm
tion a minimum?
FIGURE 29–57
32. The intensity of illumination I from a given light source is given by
k sin I d2
where k is a constant, is the angle at which the rays strike the surface, and d is the distance between the surface and the light (Fig. 29–58). At what height h should a light be
suspended directly over the centre of a circle 3.00 m in diameter so that the illumination at
the circumference will be a maximum?
L
d
Electrical
33. The power delivered to a load by a 30-V source of internal resistance 2.0 is 30i 2.0i 2 W,
where i is the current in amperes. For what current will this source deliver the maximum
3.00 m
power?
34. When 12 cells, each having an EMF of e and an internal resistance r, are connected to a FIGURE 29–58
variable load R, as in Fig. 29–59, the current in R is
3e
i 3r
R
4
Show that the maximum power (i 2R) delivered to the load is a maximum when the load R
is equal to the equivalent internal resistance of the source, 3r/4.
i
R
FIGURE 29–59
35. A certain transformer has an efficiency E when delivering a current i, where
115i 25 i2
E 115i
At what current is the efficiency of the transformer a maximum?
F
Mechanisms
1.00 m
36. If the lever in Fig. 29–60 weighs 168 N/m, find its length so as to make the lifting force F
a minimum.
37. The efficiency E of a screw is given by
5460 N
x x2
E x
FIGURE 29–60
where is the coefficient of friction and x the tangent of the pitch angle of the screw. Find
x for maximum efficiency if 0.45.
878
Chapter 29
◆
More Applications of the Derivative
x
x
$7.42/m
x
r
$4.23/sq ft.
Volume = 28.7m3
2
y
h
$5.87/m 2
$5.85/ft.
x
$3.81/sq ft.
$4.29/m
FIGURE 29–61
Volume = 10 000 cu ft.
FIGURE 29–62
Graphical Solution
The following problems require a graphical solution.
38. Find the dimensions x and y and the minimum cost for a 28.7-m3-capacity box with square
bottom (Fig. 29–61). The material for the sides costs $5.87/m2, and the joints cost $4.29/m.
The box has both a top and a bottom, at $7.42/m2.
39. A cylindrical tank (Fig. 29–62) with capacity 10 000 cu. ft. has ends costing $4.23/sq. ft.
and cylindrical side costing $3.81/sq. ft. The welds cost $5.85/ft. Find the radius, height,
and total cost, for a tank of minimum cost.
Case Study Discussion—Robot Motion
Although the implementation of this concept would be done by a team programmer, you
can come up with the idea based on your knowledge of calculus and physics found in
this chapter. If the robot scans the area of the lift and calculates a path, adding intended
displacement and time, the calculation can be checked for “jerk.” The second derivative
of displacement versus time is acceleration. The third derivative of displacement (the
derivative of acceleration) is jerk, the rate of change of acceleration with respect to time.
A program line or subroutine can be added to check the third derivative of displacement.
If it is not equal to zero, the plan is rejected and a new calculation done.
v
s
Water
jet
FIGURE 29–63 Turbine
blade.
24.0 m
x
Anchor
FIGURE 29–64 Pole braced
by a cable.
◆◆◆
CHAPTER 29 REVIEW PROBLEMS ◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆◆
1. Airplane A is flying south at a speed of 120 m/s. It passes over a bridge 12 min before
another airplane, B, which is flying east at the same height at a speed of 160 m/s. How fast
are the airplanes separating 12 min after B passes over the bridge?
2. Find the approximate change in the function y 3x2 2x 5 when x changes from 5 to
5.01.
3. A person walks toward the base of a 60-m-high tower at the rate of 5.0 km/h. At what rate
does the person approach the top of the tower when 80 m from the base?
4. A point moves along the hyperbola x2 y2 144 with a horizontal velocity vx 15.0 cm/s.
Find the total velocity when the point is at (13.0, 5.00). 41.8 cm/s
5. A conical tank with vertex down has a vertex angle of 60.0
. Water flows from the tank at a
rate of 5.00 cm3/min. At what rate is the inner surface of the tank being exposed when the
water is 6.00 cm deep?
6. Find the instantaneous velocity and acceleration at t 2.0 s for a point moving in a straight
line according to the equation s 4t 2 6t.
7. A turbine blade (Fig. 29–63) is driven by a jet of water having a speed s. The power output
from the turbine is given by P k(s v v 2), where v is the blade speed and k is a constant.
Find the blade speed v for maximum power output.