Honors 2201-4
Homework 13 Solutions
Fall 2003
(1) Domain: x > 0. As x → 0, n(x) → ∞. As x → ∞, n(x) → 1.
27
x2
27
27
D(x) = x 1 + 2 = x +
x
x
27
D 0 (x) = 1 − 2
x
54
D 00 (x) = 3
x
√
There is one critical point in the domain, x = 27. There is one endpoint, x = 0. The
second derivative is always positive on the domain√x > 0, so D(x) is always concave up on
its domain. Therefore, the one critical point x = 27 is a global minimum. The minimum
average upward distance is x ≈ 5.2m and the minimum average number of drops is 2.
n(x) = 1 +
(2) Domain: t ≥ 0. At time t = 0, there is no drug in the bloodstream (q(0) = 20(1−1) = 0).
In the long run, the quantity of drug in the bloodstream approaches zero (limt→∞ q(t) =
20(0 − 0) = 0).
q(t) = 20 e−t − e−2t
q 0 (t) = 20 −e−t + 2e−2t = 20e−t −1 + 2e−t
Solve q 0 (t) = 0.
1 = 2e−t
−t = ln
1
⇒ t = ln 2
2
For 0 ≤ t < ln 2, q 0 (t) > 0. For ln 2 < t, q 0 (t) < 0. Therefore, t = ln 2 is both a
global and local
maximum
point. The maximum amount of drug in the bloodstream is
1
1
q(ln 2) = 20 2 − 4 = 5mg.
(3) Let f (x) be the equation of a line tangent to y = e−x at the point x = t, y = e−t . Then
f (x) has the form f (x) = −e−t (x − t) + e−t = e−t (1 + t − x). The height of the triangle
is determined by the y-intercept of the line f (x), or f (0) = e−t (1 + t). The base of the
1
triangle is determined by the x-intercept of the line f (x), or 0 = e−t (1 + t − x) ⇒ x = 1 + t.
Therefore, the total area of the triangle as a function of the point t is given by A(t) with
domain t ≥ 0.
1
1
A(t) = (1 + t)e−t (1 + t) = e−t (1 + t)2
2
2
1
1
1
1
A0 (t) = − e−t (1 + t)2 + e−t 2(1 + t) = e−t (1 + t)(−1 − t + 2) = e−t (1 + t)(1 − t)
2
2
2
2
The only critical point on the domain t ≥ 0 is t = 1. For 0 ≤ t < 1, A0 (t) > 0. For 1 < t,
A0 (t) < 0. Therefore, t = 1 is both a global and local maximum point. The area of the
largest triangle is therefore, A(1) = 4/(2e) = 2/e ≈ 0.74.
(4) Let x represent the length of the side of the cut out square. Then, the length of the
pizza box sides are 47 − 2x and 45 − 1.5x. The height of the box is x. We find the domain
by noticing that 2x ≤ 47 and 3x ≤ 90. Together these constraints give us a domain of
0 ≤ x ≤ 23.5. Therefore, the volume of the box, V (x), is
V (x) = x(47 − 2x)(45 − 1.5x) = (47x − 2x2 )(45 − 1.5x)
V 0 (x) = (47 − 4x)(45 − 1.5x) + (47x − 2x2 )(−1.5) = 9x2 − 321x + 2115
x=
321 ±
q
3212 − 4(9)(2115)
1
(321 ± 164.02)
18
18
x ≈ 26.95, 8.72
00
V (x) = 18x − 321 ⇒ V 00 (8.72) < 0
≈
The only critical point within the domain 0 ≤ x ≤ 23.5 is x = 8.72. From the second
derivative we know that this critical point is a local maximum. We compare the value of
V (x) at the endpoints to the value at the critical point. V (0) = 0, V (23.5) = 0, V (8.72) =
8230cm3 . The box with maximum volume has dimensions 8.72 x 29.56 x 31.92cm.
(5) R(y) is nonnegative if y ≤ a < b or if a < b ≤ y. The domain of R(y) for the chemical
reaction is 0 ≤ y ≤ a.
R(y) = k(a − y)(b − y)
R (y) = k(−1)(b − y) + k(a − y)(−1) = k(2y − a − b)
R00 (y) = 2k > 0
0
The only critical point, y = (a + b)/2, is not in the domain 0 ≤ y ≤ a < b. Also, we see that
if y < a and y < b, then 2y < a + b, so the first derivative is always negative. If the function
2
R(y) is decreasing on its domain, the maximum must occur at the left most endpoint, or
y = 0 in this case. Therefore, the reaction is fastest when y = 0. It proceeds at a rate of
kab.
(6) Let x represent the number of trees planted per square kilometer, and P (x) represent
the overall productivity. Then, if 0 ≤ x ≤ 200, P (x) = 400x. If x > 200, P (x) =
x(400 − (x − 200)) = x(600 − x). Notice that at the joining point x = 200, the value
of the functions matches up. Therefore, the function P (x) is continuous. Looking first at
the case 0 ≤ x ≤ 200, we see that P 0 (x) = 400 > 0 so productivity is increasing at a constant
rate. The maximum productivity over the interval 0 ≤ x ≤ 200 must occur at the right most
endpoint, or x = 200. For the case 200 < x, we have P 0 (x) = 600 − 2x. The second
derivative is P 00 (x) = −2 < 0 so this part of the productivity function is always concave
down. There is one critical point at x = 300. Since the second derivative is negative it must
be a local maximum. We also see that when 200 < x < 300, P 0(x) > 0, so the productivity
is increasing. When 300 < x, P 0 (x) < 0, so the productivity is decreasing. On the interval
200 < x, x = 300 is a global maximum. Due to the fact that the overall productivity
function P (x) is continuous, we see that P (x) is actually increasing for 0 ≤ x < 300,
making x = 300 a global maximum for the entire domain. Another way to see this is to
compare the productivity for x = 200 and x = 300. P (200) = 400(200) = 80000 and
P (300) = 300(300) = 90000. The farmer should plant 300 trees per square kilometer to get
the maximum yield of 90, 000kg/km2.
3