Calculus Of One Real Variable –
By Pheng Kim Ving |
1.2.2 |
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1. Maxima and
Minima |
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Maximum of f on [a, b] is f(xM). |
Definitions 1.1
Let f be a function and S a subset of dom( f ). We say that: |
Remarks 1.1
i. A maximum attained by f is a value of f, and so is a minimum attained by f.
ii. S may be any interval,
not just a closed finite interval; it may even be a union of a finite number of
intervals. The
definition doesn't require any special type of
subset of the domain of f.
iii. f may be continuous
at all points of S; it may be discontinuous at some points of S. The definition
doesn't require any
special type of function.
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2. Attainment
Of Extrema |
An extremum is either a maximum or a
minimum. The maximum and minimum are referred to collectively as extrema.
We say that [a, b] is finite if both a and b are finite
numbers.
If a function f is continuous
on a closed finite interval [a, b], then it
attains both a maximum and a minimum there. |
The proof of this theorem is omitted because
it involves more theory than is covered in an introductory calculus course
and therefore is beyond the scope of this tutorial. However, the theorem itself
is fairly obvious on an intuitive level.
As an example, the function f in Fig. 1.1 is continuous
on the closed finite interval [a, b]. It attains both
a maximum and
a minimum values there.
Remark 2.1
If one or both of the conditions of the theorem are not satisfied, then the conclusion of the theorem need not hold.
In Fig. 2.1, f isn't continuous on the closed interval [a, b], but it still manages to attain both a maximum and a minimum
there. However, in Fig. 2.2, f isn't continuous on the closed interval [a, b], and f doesn't attain a
maximum there,
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Fig. 2.1 f isn't
continuous on [a, b]; |
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Fig. 2.2 f isn't
continuous on [a, b]; |
although it attains a minimum. In Fig. 2.2,
the only possible maximum of f would be the value of f at the “ first ” point to
the left of x1, but there's no
such point, because if c is a point to the
left of x1, then d = (c + x1)/2, which is the middle
between c and x1, is to the left
of x1 and closer to x1 than c is.
In Fig. 2.3, the interval (a, b] on which f is continuous is half-open,
but f still manages to
attain both a maximum and a
minimum there. However, in Fig. 2.4, the interval (a, b] on which f is continuous is
half-open, and f doesn't attain a
maximum there, although it attains a minimum.
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Fig. 2.3 Interval (a, b] on which f is continuous is |
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Fig. 2.4 Interval (a, b] on which f is continuous
is |
Find the area of the largest rectangular
plot that can be fenced on all sides by 200 m of fence.
We're asked to find the largest rectangular
area that can be enclosed by 200 m of fence, ie, given that the perimeter of a
rectangle is fixed at 200 m, find its dimensions so that its area is the
largest.
A = 100x – x2 = –(x2 – 100x) = –((x – 50)2 – 2500) = 2500 – (x – 50)2.
So A attains the maximum of 2500 when x
– 50 = 0 or x = 50. When x = 50 we have y = 100 – 50 = 50.
The largest
rectangle is a square of side 50 m and area 2500 m2.
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Fig. 2.5 Rectangular Plot. |
EOS
Problems & Solutions |
1. Let f(x)
= x. Does f attain a maximum or a minimum, if so,
specify them, on:
a. R?
b. [1, 2]?
c. (1, 2)?
d. [1, 2)?
e. (1, 2]?
a. No, neither a maximum nor a minimum.
b. Yes, a minimum of f(1) = 1 and a maximum of f(2) = 2.
c. No, neither a maximum nor a
minimum.
d. A minimum of f(1) = 1 but no maximum.
e. A maximum of f(2) = 2 but no minimum.
2.
Let:
a. What's dom( f )?
b. Sketch a graph of f.
c. Is f continuous on its
domain?
d. Does f attain a maximum or a minimum on its domain?
a. dom( f ) = [–1, 1].
b.
3. Let x and y be any two non-negative numbers with a sum of 8. Show that their product attains both a
maximum and
a minimum. What is the maximum of the product?
Solution
4.
Prove that there's a point on the line:
that's closest to the origin. Find that point.
Solution
Let P be a point with
coordinates (x, y) on the given line, and d the distance from
the origin to P. So P is closest to the
origin when d attains its minimum value. We have:
5. Let f(x) = anxn + an–1xn–1 + an–2xn–2 + ... + a1x + a0, where n is an even
positive integer and an > 0. Prove that f
attains a minimum over the entire real line.
Solution
We have:
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