5.2 Applied Extremum Problems
5.2.3 Applications to Economics
5.2. Applied Extremum Problems 153
Since V00(3) = −72 < 0 and x1 = 3 is the only critical number of V in (0,9), it follows from the Second Derivative Test (Special Version) that in (0,9),Vattains its maximum atx1=3. Thus the length of the side of the square that must be cut offis 3 cm and the maximum volume isV(3)=432 cm3.
2 4 6 8
100 200 300 400
V=x(18−2x)2
Figure 5.19
154 Chapter 5. Applications of Differentiation
whereqis the number of units and pis the price per unit, and the average cost function is Cav=q2−8q+57+ 2
q 0<q≤90.
At what value ofqwill there be maximum profit? What is the maximum profit?
ExplanationAlthough the average cost function is undefined atq= 0, we may include 0 in the domain of the cost function. The cost function and the revenue function are differentiable on (0,90). However, we do not know whether maximum profit would be attained in (0,90) or at an endpoint. So we use the method for finding absolute extrema for functions on closed and bounded intervals.
Solution The cost functionCis given by
C(q)=q·Cav=q3−8q2+57q+2 (0≤q≤90), and the revenue functionRis given by
R(q)= p·q= 90−q
2 ·q (0≤q≤90).
Therefore the profit functionPis given by P(q) = R(q)−C(q)
=
45q− q22
−(q3−8q2+57q+2)
= −q3+ 15
2q2−12q−2, (0≤q≤90).
DifferentiatingP(q), we get P0(q) = d
dq
−q3+ 15
2q2−12q−2
= −3q2+15q−12 (0<q<90) SolvingP0(q)=0, that is, −3q2+15q−12 = 0 (0<q<90)
−3(q−1)(q−4) = 0 (0<q<90), we get the critical numbers ofP:q1=1 andq2=4.
Comparing the values ofPat the critical numbers as well as that at the endpoints:
q 0 1 4 90
P(q) −2 −15
2 6 −669332
we see that maximum profit is attained atq2 =4 and the maximum profit is 6 (units of money).
RemarkIf we know that maximum profit is not attained at the endpoints, we can simply compare the values of Patq1andq2.
1 2 3 4 5 6
-20 -10 5
P=−q3+152q2−12q−2
Figure 5.20
5.2. Applied Extremum Problems 155
Exercise 5.2
1. For each of the given function f, find its absolute extrema on the given interval.
(a) f(x)=4x3+3x2−18x+1, [0,3]
(b) f(x)=−3x5+5x3+2, [−2,0]
(c) f(x)=1+2x3−3x4, [−1,1]
2. Find two positive real numbers whose sum is 50 and whose product is a maximum.
3. Find two real numbersxandysatisfying 2x+y=15 such thatx2+y2is minimized.
Can you find a geometric meaning for the result?
4. Find the dimensions of the rectangle of area 100 square units that has the least perimeter.
5. A rectangular field is to be enclosed by a fence and divided equally into two parts by a fence parallel to one pair of the sides. If a total of 600 m of fence is to be used, find the dimensions of the field if its area is to be maximized.
6. A book is to contain 36 in2 of printed matter per page, with margins of 1 in along the sides and 112in along the top and bottom. Find the dimensions of the page that will require the minimum amount of paper.
7. Suppose that a ball is thrown straight up into the air and its height aftertseconds is 5+24t−16t2feet.
Determine how long it will take the ball to reach its maximum height and determine the maximum height.
8. It is known from experiments that the height (in meter) of a certain plant aftertmonths is given (approx- imately) by
h(t)= √
t−t, 0≤t≤1.
How long, on the average, will it take a plant to reach its maximum height? What is the maximum height?
9. A company manufactures and sellsxpieces of a certain product per month. The monthly cost (in dollars) is
C(x)=120000+100x and the price-demand equation is
p=300− x 15
where 0≤ x ≤ 4000. Find the maximum profit, the production level that will give the maximum profit, and the price the company should charge for each piece of the product.
156 Chapter 5. Applications of Differentiation
Chapter 6
Integration
6.1 Definite Integrals
In the introduction of Chapter 3, we consider the area of the region under the curvey= x2and above thex-axis forxbetween 0 and 1. To get approximations for the area, we divide [0,1] intonequal subintervals:
[x0,x1], [x1,x2], . . . , [xn−1,xn], wherexi = i
nfor 0≤i≤n; and for eachi=1, . . . ,n, in the subinterval [xi−1,xi], we take the left endpointxi−1 and consider the sum Pn
i=1
f(xi−1)· 1
n, that is, f(x0)·1
n+· · ·+ f(xn−1)·1
n. We have seen that the sum is close to
1
3(which is the required area) ifnis large. Using limit notation, the result can be written as
n→∞lim Xn
i=1
f(xi−1)·1 n = 1
3. (6.1.1)
The above idea can be generalized to any continuous functions fon any closed and bounded interval. Moreover, f need not be non-negative.
Theorem 6.1.1 Let f be a function that is continuous on a closed and bounded interval [a,b]. Then the follow- ing limit exists:
n→∞lim Xn
i=1
f(xi−1)·b−a n , wherexi =a+ i
n(b−a)for 0≤i≤n.
Explanation
• By the construction of thexi’s, we havex0 =a, xn= b, x0 < x1 < · · · < xn, and for every i = 1, . . . ,n, the subinterval [xi−1,xi] has lengthb−a
n andxi−1is the left- endpoint of the subinterval.
• If f is non-negative on [a,b], that is, f(x) ≥ 0 for all x∈[a,b], then lim
n→∞
Pn
i=1f(xi−1)·b−a
n is the area bounded by the graph of f, thex-axis and the vertical lines given byx=aandx=b.
y= f(x)
x0 x1 · · · xn−1 xn Figure 6.1
158 Chapter 6. Integration
Definition Let f be a function that is continuous on a closed and bounded interval [a,b]. The number
n→∞lim Pn
i=1f(xi−1)· b−a
n , where xi = a+ i
n(b−a) for 0 ≤ i ≤ n, is called the definite integralof f from a to band is denoted byRb
a f(x) dx, that is, Z b
a
f(x) dx= lim
n→∞
Xn
i=1
f(xi−1)·b−a
n . (6.1.2)
Example The result given in (6.1.1) can be written as Z 1
0
x2dx= 1
3. (6.1.3)
Remark In Theorem 6.1.1, in each subinterval [xi−1,xi], instead of taking the left endpoint xi−1, we can take the right endpoint xi(see Figure 6.2) and we have
n→∞lim Xn
i=1
f(xi)·b−a
n =
Z b
a
f(x) dx. (6.1.4)
y= f(x)
x0 x1 · · · xn−1 xn Figure 6.2
y= f(x)
x0 t1 x1 t2 · · · tn xn Figure 6.3
In fact, we can take an arbitrarily point (denoted by ti) in [xi−1,xi]: the sum Pn
i=1 f(ti)· b−a
n is close to Rb
a f(x) dxifnis large enough (see Figure 6.3).
More generally, the subintervals [x0,x1], . . . ,[xn−1,xn] need not be of equal lengths. All we need is that the lengths are small enough: ifa= x0 < x1 <· · ·< xn = band∆x1, . . . ,∆xnare small enough, where∆xi is the length of the ith subinterval [xi−1,xi], then for every choice oft1, . . . ,tn withti ∈ [xi−1,xi] for 1≤ i≤ n, the sum (called aRiemann Sum)
Xn
i=1
f(ti)∆xi is close toRb
a f(x) dx. Many authors use this to define definite integral.
Below, we apply (6.1.4) to deduce the result given in (6.1.3). For this, we take f(x)= x2,a= 0,b=1 and xi= i
n for 0≤i≤n.
6.1. Definite Integrals 159
Example By (6.1.4), we have Z 1
0
x2dx = lim
n→∞
Xn
i=1
i n
2
·1 n
= lim
n→∞
1 n3
Xn
i=1
i2
= lim
n→∞
1
n3 · n(n+1)(2n+1)
6 Sum of Squares Formula
= lim
n→∞
2n3
6n3 Leading Term Rule
= 1 3 FAQ Can we defineRb
a f(x) dxif f is not continuous on [a,b]?
Answer In definingRb
a f(x) dx, we need Theorem 6.1.1. The condition “f is continuous on[a,b]” is used to guarantee that lim
n→∞
Pn i=1
f(xi−1)· 1
nexists.
In general, if f is a function defined on [a,b] such that there exists a (unique) real numberIsatisfying (∗) Pn
i=1
f(ti)∆xiis arbitrarily closeIif∆x1, . . . ,∆xnare sufficiently small, where∆xi= xi−xi−1for 1≤i≤n, a=x0< x1<· · ·< xn =bandti ∈[xi−1,xi] for 1≤i≤n,
then the unique numberI is defined to beRb
a f(x) dx.
Remark
• In view of (∗), we may write Z b
a
f(x) dx= lim
k∆k→0
Xn
i=1
f(ti)∆xi,
wherek∆k → 0 means the lengths∆xi’s tend to zero. However, this kind of limit is different from that discussed in Chapter 3.
• The symbolR
was introduced by Leibniz and is called theintegral sign. It is an elongatedS and was chosen because a definite integral is a limit of sums.
• Since the definite integral of a (continuous) function on [a,b] depends on the function f and the interval [a,b] only, it can simply be denoted byRb
a f, omitting the variablexand the notation dx. However, the notationRb
a f(x) dxis preferred. There are two reasons:
(1) the notation dxreminds us of the factors∆xi in the sumsPn
i=1
f(ti)∆xi;
(2) with the variable included in the notation, it is easier to handle thesubstitution methodfor integra- tion (see Chapter 10).
• In the notationRb
a f(x) dx, the variable x is called adummy variable; it can be replaced by any other symbol. For example, usingtas the dummy variable, (6.1.3) can be written asR1
0 t2dt= 1
3. Note that if we usetas the dummy variable, we have to change dxto dtaccordingly.
160 Chapter 6. Integration
Example Use definition to find Z 2
1
xdx.
Solution Applying (6.1.2) to f(x)=x,a=1,b=2 andxi=1+ i
n for 0≤i≤n, we get Z 2
1
xdx = lim
n→∞
Xn
i=1
1+ i−1 n
!
· 1 n
= lim
n→∞
1 n2
Xn
i=1
(n+i−1)
= lim
n→∞
1
n2 ·n(n+2n−1)
2 Sum of A.P.
= lim
n→∞
3n−1 2n
= lim
n→∞
3n
2n Leading Term Rule
= 3 2.
Remark The value of the definite integral is the area of the trapezoidal region
shown in Figure 6.4. Readers can check that the result agrees with that obtained by using formula for area of trapezoid.
1 2
1 2
Figure 6.4
Definite Integral for Constant FunctionsLetcbe a constant and letaandbbe real numbers witha<b. Then we have
Z b
a
cdx=c·(b−a).
Proof Applying (6.1.2) to f(x)=candxi =a+ i
n(b−a) for 0≤i≤n, we get Z b
a
cdx = lim
n→∞
Xn
i=1
c· b−a n
= lim
n→∞c·b−a
n ×n Sum of Constants
= lim
n→∞c·(b−a) Rule (L1) for Limit
= c·(b−a)
Remark Ifc > 0, thenRb
a cdx is the area of the rectangular region shown in Figure 6.5.
a b
c
Figure 6.5
Rules for Definite Integrals Let f andgbe functions that are continuous on a closed and bounded interval [a,b]. Letαbe a constant and letc∈(a,b). Then we have
6.1. Definite Integrals 161
(Int1) Z b
a
hf(x)+g(x)i dx=
Z b
a
f(x) dx+ Z b
a
g(x) dx
Proof Apply definition and Rule (L4) for limits of sequences.
(Int2) Z b
a
αf(x) dx=α Z b
a
f(x) dx
Proof Apply definition and Rule (L5s) for limits of sequences.
RemarkUsing Rules (Int1) and (Int2), we get Z b
a
hf(x)−g(x)i dx=
Z b
a
f(x) dx− Z b
a
g(x) dx.
In fact, Rule (Int1) is valid for sum and difference of finitely many (continuous) functions.
(Int3) Z b
a
f(x) dx= Z c
a
f(x) dx+ Z b
c
f(x) dx.
Explanation The proof for the result is not easy. For the case where f is nonnegative on [a,b], the result can be seen from the geometric
interpretation shown in Figure 6.6.
y= f(x)
a c b
Figure 6.6
Example Z 1
0
5x2dx = 5 Z 1
0
x2dx Rule (Int2)
= 5× 1
3 by (6.1.3)
= 5 3 Example
Z 2
1
(3−x) dx = Z 2
1
3 dx− Z 2
1
xdx Rule (Int1)
= 3×(2−1)−3 2
Definite Integral for Constant &
Example on Page 160
= 3 2 In definingRb
a f(x) dx, we needa<b. For convenience, we introduce the following:
Convention Let f be a function that is continuous on a closed and bounded interval [a,b] and letc ∈ [a,b].
Then we define (1)
Z a
b
f(x) dx=− Z b
a
f(x) dx (2)
Z c
c
f(x) dx=0
162 Chapter 6. Integration
Example Z 2
2
(1+2x−3x2) dx=0 since the function 1+2x−3x2is continuous on [0,3] (for example) and 2∈[0,3].
Example Z 0
1
x2dx = − Z 1
0
x2dx by convention
= −1
3 by (6.1.3)
Terminology In a definite integralRb
a f(x) dx,
• the function f is called theintegrand;
• the numbersaandbare called thelimits of integration;ais thelower limitandbtheupper limit.
Exercise 6.1
1. For each of the following definite integrals, use the results in this section to find its value:
(a) R1
0(1−3x2) dx (b) R1
2 4xdx 2. Use definition to find the definite integralR1
0 x3dx.
Given: 13+23+· · ·+n3= n2(n+1)2
4