APPLIED MAXIMUM AND MINIMUM PROBLEMS 0 121
I Let R be the point where the boat lands, and let O be the center of the circle. Since AOPR is isosceles,
PR = 2cos0. The arc length RQ = 26. Hence, the time T= PR/1.5 + RQ/3 = | cosfl + §0. So, D
-!sin0+ \. Setting D e T=0, we find sin0= |, Q=Tt/6. Since T is a continuous function on the
closed interval [0, Tr/2], we can use the tabular method. List the critical number ir/6 and the endpoints 0 and
7T/2, and compute the corresponding values of T. The smallest of these values is the absolute minimum.
Clearly, 7r/3<3, and it is easy to check that 7r/3<(6V5 + IT) 19. (Assume the contrary and obtain the
false consequence that ir > 3V5.) Thus, the absolute minimum is attained when 0 — Tr/2. That means that
the man walks all the way.
16.15
16.14 Find the answer to Problem 16.13 when, instead of rowing, the man can paddle a canoe at 4 miles per hour.
I Using the same notation as in Problem 16.13, we find T=\ cos0 + §0, D e T= — \ sin 6 + §. Setting
D g T=0, We obtain sin 0=5, which is impossible. Hence, we use the tabular method for just the
endpoints 0 and itII. Then, since \ < ir/3, the absolute minimum is \, attained when 0 = 0. Hence, in
this case, the man paddles all the way.
A wire 16 feet long has to be formed into a rectangle. What dimensions should the rectangle have to maximize
the area?
I Let x and y be the dimensions. Then 16 = 2x + 2y, 8 = x + y. Thus, 0 s x < 8. The area A = xy =
x(8 — x) = Sx — x
2 , so D X A = 8 - 2x, D
2
X A = —2. Hence, the only critical number is x = 4. We can use
the tabular method. Then the maximum value 16 is attained when x = 4. When x = 4, y = 4. Thus, the
rectangle is a square.
16.16
Find the height h and radius r of a cylinder of greatest volume that can be cut within a sphere of radius b.
I The axis of the cylinder must lie on a diameter of the sphere. From Fig. 16-6, b
2 = r
2 + (h/2)
2 , so
the volume of the cylinder V= -rrr
2 h = Tr(b
2 - H
2 /4)h = ir(b
2 h - A
3 /4). Then D h V= ir(b
2 - 3h
2 /4) and
D
2
h V= -(3ir/2)h, so the critical number is h = 2b/V3. Since the second derivative is negative, there is a
relative maximum at h = 2ft/V3, which, by virtue of the uniqueness of the critical number, is an absolute
maximum. When h = 2b/V3, r = 6 VI •
Fig. 16-6
16.17
Among all pairs of nonnegative numbers that add up to 5, find the pair that maximizes the product of the square of
the first number and the cube of the second number.
I Let x and y be the numbers. Then x + y = 5. We wish to maximize P = x
2 y
3 = (5 - y)
2
.y
3
. Clearly,
0 < y < 5. D,P = (5 - y)2(3y2) + y3[2(5 - >>)(-!)] = (5 - y)y2[3(5 -y)- 2y] = (5 - y)y2(15 - Sy). Hence,
the critical numbers are 0, 3, and 5. By the tabular method, the absolute maximum for P is 108, corresponding to
y = 3. When y = 3, x = 2.
e
T
7T/6
(6V3+77)/9
0
4
77/2
7T/3
0
T
0
i
7T/2
7T/3
Jt
A
4
16
0
0
8
0
I Let R be the point where the boat lands, and let O be the center of the circle. Since AOPR is isosceles,
PR = 2cos0. The arc length RQ = 26. Hence, the time T= PR/1.5 + RQ/3 = | cosfl + §0. So, D
-!sin0+ \. Setting D e T=0, we find sin0= |, Q=Tt/6. Since T is a continuous function on the
closed interval [0, Tr/2], we can use the tabular method. List the critical number ir/6 and the endpoints 0 and
7T/2, and compute the corresponding values of T. The smallest of these values is the absolute minimum.
Clearly, 7r/3<3, and it is easy to check that 7r/3<(6V5 + IT) 19. (Assume the contrary and obtain the
false consequence that ir > 3V5.) Thus, the absolute minimum is attained when 0 — Tr/2. That means that
the man walks all the way.
16.15
16.14 Find the answer to Problem 16.13 when, instead of rowing, the man can paddle a canoe at 4 miles per hour.
I Using the same notation as in Problem 16.13, we find T=\ cos0 + §0, D e T= — \ sin 6 + §. Setting
D g T=0, We obtain sin 0=5, which is impossible. Hence, we use the tabular method for just the
endpoints 0 and itII. Then, since \ < ir/3, the absolute minimum is \, attained when 0 = 0. Hence, in
this case, the man paddles all the way.
A wire 16 feet long has to be formed into a rectangle. What dimensions should the rectangle have to maximize
the area?
I Let x and y be the dimensions. Then 16 = 2x + 2y, 8 = x + y. Thus, 0 s x < 8. The area A = xy =
x(8 — x) = Sx — x
2 , so D X A = 8 - 2x, D
2
X A = —2. Hence, the only critical number is x = 4. We can use
the tabular method. Then the maximum value 16 is attained when x = 4. When x = 4, y = 4. Thus, the
rectangle is a square.
16.16
Find the height h and radius r of a cylinder of greatest volume that can be cut within a sphere of radius b.
I The axis of the cylinder must lie on a diameter of the sphere. From Fig. 16-6, b
2 = r
2 + (h/2)
2 , so
the volume of the cylinder V= -rrr
2 h = Tr(b
2 - H
2 /4)h = ir(b
2 h - A
3 /4). Then D h V= ir(b
2 - 3h
2 /4) and
D
2
h V= -(3ir/2)h, so the critical number is h = 2b/V3. Since the second derivative is negative, there is a
relative maximum at h = 2ft/V3, which, by virtue of the uniqueness of the critical number, is an absolute
maximum. When h = 2b/V3, r = 6 VI •
Fig. 16-6
16.17
Among all pairs of nonnegative numbers that add up to 5, find the pair that maximizes the product of the square of
the first number and the cube of the second number.
I Let x and y be the numbers. Then x + y = 5. We wish to maximize P = x
2 y
3 = (5 - y)
2
.y
3
. Clearly,
0 < y < 5. D,P = (5 - y)2(3y2) + y3[2(5 - >>)(-!)] = (5 - y)y2[3(5 -y)- 2y] = (5 - y)y2(15 - Sy). Hence,
the critical numbers are 0, 3, and 5. By the tabular method, the absolute maximum for P is 108, corresponding to
y = 3. When y = 3, x = 2.
e
T
7T/6
(6V3+77)/9
0
4
77/2
7T/3
0
T
0
i
7T/2
7T/3
Jt
A
4
16
0
0
8
0