Download 5.2. Increasing and decreasing functions

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5. Suppose a population of bacteria satisfies that after t days,
P(t) =
6 t2
---------mg
(2/3)
t
.
(a) Find P'(t).
(b) Find the instantaneous rate of change of P when t = 8.
(c) Find the equation of the line tangent to the curve when t = 27
Ans: (a) 8 t(1/3) mg/day
(b) 16 mg/day
(c) P = 24 t - 162 mg
5.2. Increasing and decreasing functions
A function f is increasing on a set A if, for any u and v in A, whenever u<v, then f(u) < f(v).
This means that as you follow the graph moving toward the right, the graph also moves
upward and hence increases.
8
6
4
2
1.5
2
2.5
3
3.5
4
Figure. The function f(x) = x2 - 2 x for 1≤x≤4. It is increasing.
Similarly, a function f is decreasing on a set A if, for any u and v in A, whenever u<v, then
f(u) > f(v). As you follow the graph moving toward the right, the graph moves downward,
or decreases.
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4
3
2
1
0.5
1
1.5
2
Figure. The function f(x) = 4 - x2 is decreasing for 0≤x≤2.
Most functions are increasing some places and decreasing other places:
1
0.5
1
2
3
4
5
-0.5
-1
Figure. This example is increasing for 0<x<0.5, 1.5<x<2.5, and 3.5<x<4.5. It is decreasing
for 0.5<x<1.5, 2.5 < x < 3.5, and 4.5 < 5.
There is a relative maximum at x = a if f(x) ≤ f(a) for all x near a. Thus if there is a
relative maximum at x = a, then x = a is the top of a hill. There is a relative minimum at x
= a if f(x) ≥ f(a) for all x near a. Thus if there is a relative minimum at x = a, then x = a is
the bottom of a valley.
In the first example, there is a relative maximum at x = 0.5, at x = 2.5, and at x = 4.5. There
is a relative minimum at x = 1.5, at x = 3.5.
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4
2
1
2
3
4
5
6
-2
Figure. This example is increasing for 0<x<1.25 and 3.25 < x < 5. It is decreasing for 1.25
< x < 3.25 and 5<x. There is a relative maximum at x = 1.25 and x = 5. There is a relative
minimum at x = 3.25.
Note that at either a relative maximum or a relative minimum in the picture, f '(x) = 0.
It is important to be able to tell when a function is increasing and decreasing.
Theorem. If f '(x) > 0 for a<x<b then f is increasing for a<x<b.
If f '(x) < 0 for a<x<b then f is decreasing for a<x<b.
Idea. If f '(x) > 0 then the tangent line has positive slope, so it is moving upward to the
right. If f '(x) < 0 then the tangent line has negative slope, so it is moving downward to the
right.
Procedure for finding when f (x) is increasing and when it is decreasing.
(1) Solve f '(x) = 0, and tell when f '(x) has no value.
(2) List these values in order. These numbers are called the critical numbers.
(3) At every point other than those values, f '(x) is either positive (+) or negative (-). Tell
which.
(4) Every interval in which f '(x) > 0 is an interval where f is increasing.
(5) Every interval in which f '(x) < 0 is an interval where f is decreasing.
Example. If f(x) = 3 - 4 x, tell where f is increasing and tell where f is decreasing.
Solution. Note f '(x) = - 4
Hence f is decreasing everywhere, and f is increasing nowhere.
Example. Let f(x) = x2 - 6 x + 5. Tell where f is increasing and where f is decreasing.
Solution. Note f '(x) = 2 x - 6.
The critical numbers are where 2 x- 6 is undefined or 0:
2 x - 6 = 0 means x = 3.
Hence x = 3 is the only critical number. When x < 3, 2x-6 < 0; when x >3, 2x-6 >0.
We thus obtain this diagram:
(-)
3 (+)
Hence f is increasing for x > 3; f is decreasing for x < 3.
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Example. Let f(x) = 2 x3 - 3 x2 - 36 x + 4.
Find where f is increasing, and where f is decreasing.
Solution. f '(x) = 6 x2 - 6 x - 36.
f '(x) always has a value. We solve f ' (x) = 0.
6 x2 - 6 x - 36 = 0
6( x2 - x - 6) = 0
6 (x - 3) (x + 2) = 0.
The roots are 3 and -2.
List them in order
-2
3
Outside these points we figure out the sign:
For x < -2, f ' (x) = (+) (-) (-) = +.
For -2 < x < 3, f '(x) = (+) (-) (+) = -.
For 3<x, f '(x) = (+) (+) (+) = +.
Thus we obtain the chart:
(+) -2
(-)
3
(+)
So f is increasing for x < - 2 and for 3 < x since that is where f '(x) >0.
Also f is decreasing for -2 < x < 3 since that is where f '(x) < 0.
It is common to write intervals using a different notation: (a,b) means a<x<b. In this form
the answer would be
f is increasing on (-∞, -2) and (3, ∞).
f is decreasing on (-2, 3).
Example. Let f(x) =√(16 - x2).
Find where f is increasing, and where f is decreasing.
Solution.
f(x) = (16 - x2)1/2
f '(x) = (1/2) (16 - x2)-(1/2) (-2x) = - x /(16 - x2)(1/2)
f '(x) makes sense only 16 - x2 > 0 hence only for -4 < x < 4.
f ' (x) = 0 when x = 0. Hence the critical numbers are:
-4
0
4
There are no values to the left of -4 or to the right of 4. We find
-4
(+)
0
(-)
4
f is increasing for -4< x < 0.
f is decreasing for 0 < x < 4.
You may use your calculator to graph the function. f(x).
Example. Let f(x) = 3/ (x-1). Tell where f is increasing and where f is decreasing.
Solution.
f '(x) = (x-1) (0) - 3
-3
------------=
----------(x-1)2
(x-1)2
The only critical number is x = 1, since there is no value at x = 1.
The chart is
(-)
1
(-)
Thus f is never increasing. Moreover f is decreasing for x<1 or x > 1.
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In fact, note that there is no point when x = 1.
Problems for 5.2
1. Tell where the function is increasing and where it is decreasing.
(a) f(x) = 2 x - 5
(b) f(x) = x2 - 6 x + 5
(c) f(t) = 3 + 2x - x 2
(d) f(x) = 2 x3 + 3 x2 - 36 x + 5
(e) f(x) = x3 - 3x + 4
Ans: (a) increasing for all x, never decreasing
(b) increasing for x> 3; decreasing for x < 3
(c) increasing for t<1; decreasing for t>1
(d) increasing for x<-3 or x>2; decreasing for -3<x<2
(e) increasing for x<-1 or x>1; decreasing for -1<x<1.
2. A person is being medicated. The concentration of a certain drug t hours after the start
of treatment is given by
C(t) = (t2+1) / (t + 1) mg/l.
(a) Find C'(t).
(b) Find the instantaneous rate of change of the concentration when t = 1.
(c) When t = 1, is the concentration increasing or decreasing?
(d) Find the equation of the line tangent to the curve when t = 1.
Ans: (a) (t2 + 2 t - 1) / (t+1)2
(b) 1/2 mg/l/hour
(c) increasing
(d) C = (1/2)t + (1/2)
3. A person is being medicated. The concentration of a certain drug t hours after the start
of treatment is given by
C(t) = (t2+1) / (t 3 + 1) mg/l.
(a) Find C'(t).
(b) Find the instantaneous rate of change of the concentration when t = 1/2.
(c) Find the instantaneous rate of change of the concentration when t = 1.
(d) When t = 1, is the concentration increasing or decreasing?
(e) Find the equation of the line tangent to the curve when t = 1.
Ans: (a) (- t4 - 3 t2 + 2 t) / (t3 + 1)2
(b) 0.148 mg/l/hour
(c) -0.5 mg/l/hour
(d) decreasing
(e) C = 1.5 - 0.5 t
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5.3. Relative maxima and minima
Recall that the function f has a relative maximum at x = a if f(x) ≤ f(a) for all x near a.
Thus if there is a relative maximum at x = a, then x = a is the top of a hill on the graph.
There is a relative minimum at x = a if f(x) ≥ f(a) for all x near a. Thus if there is a
relative minimum at x = a, then x = a is the bottom of a valley on the graph.
4
2
1
2
3
4
5
6
-2
Figure. In this example there is a relative maximum at x = 1.25 and x = 5. There is a
relative minimum at x = 3.25.
The general shape of the graph is found by locating these relative extrema. For this
purpose, the following theorem is useful.
Theorem. (First Derivative Test) Suppose f '(a) = 0.
(1) There is a relative maximum at x = a provided f '(x) > 0 to the left of a (but
near a) and f '(x) < 0 to the right of a (but near a).
(2) There is a relative minimum at x = a provided f '(x) < 0 to the left of a (but near
a) and f '(x) > 0 to the right of a (but near a).
Again, this is obvious from the picture. If f is increasing to the left of a and decreasing to
the right of a, then there is a relative maximum at x = a. If f is decreasing to the left of a and
increasing to the right of a, then there is a relative minimum at x = a. If f is decreasing to the
left of a and also decreasing to the right of a, then there is neither a relative maximum nor a
relative minimum at x = a. If f is increasing to the left of a and also increasing to the right
of a, then there is neither a relative maximum nor a relative minimum at x = a.
(3) If f '(a) = 0, f '(x) > 0 to the left of a (but near a) and f '(x) > 0 to the right of a
(but near a), then there is neither a relative maximum nor a relative minimum.
(Instead, f is increasing near a.)
(4) If f '(a) = 0, f '(x) < 0 to the left of a (but near a) and f '(x) < 0 to the right of a
(but near a), then there is neither a relative maximum nor a relative minimum.
(Instead, f is decreasing near a.)
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Example. Let f(x) = x2 + 2 x + 5.
(a) Tell where f is increasing.
(b) Tell where f is decreasing.
(c) Locate all relative maxima.
(d) Locate all relative minima.
Solution.
f '(x) = 2x + 2. Hence f ' (x) = 0 when x = -1.
We find the following chart for f ':
(-)
-1
(+)
(a) f is increasing for x > -1.
(b) f is decreasing for x < -1.
(c) there are no relative maxima.
(d) there is a relative minimum at x = -1.
A shorter way to ask you to find all the relative maxima and all the relative minima is to say
"locate and classify all the critical numbers." This means to find (locate) all the critical
numbers and then, for each critical number, tell whether it corresponds to a relative
maximum, a relative minimum, or neither.
Example. Let f(x) = 4 x +5 - x 2.
(a) Tell where f is increasing.
(b) Tell where f is decreasing.
(c) Locate and classify all critical numbers.
Solution:
f '(x) = 4 - 2 x
Hence f '(x) = 0 when 4 - 2 x = 0 or x = 2.
We obtain the chart for f ' as follows:
(+)
2
(-)
(a) f is increasing for x < 2.
(b) f is decreasing for x > 2.
(c)
x = 2 is a relative maximum
Example. Let P(t) = 2 t3 - 3 t2 - 36 t + 4.
Find the locations of all relative maxima and the locations of all relative minima.
Solution.
P'(t) = 6 t2 - 6 t - 36 = 6 (t2 - t - 6) = 6 (t-3) (t + 2).
Solve P'(t) = 0 to find that the critical numbers are 3 and -2.
We obtain the chart
(+)
-2
(-)
3
(+)
To the left of 3, f '(t) < 0 while to the right of 3, f '(t) > 0. By the First Derivative Test there
is a relative minimum at t = 3.
To the left of -2, f '(t) > 0 while to the right of -2, f '(t) < 0. Hence by the First Derivative
Test there is a relative maximum at t = -2.
Write the answer as:
t = -2: relative maximum
t = 3: relative minimum
Example. Let f(x) = 6 x5 + 15 x4 - 80 x3 + 5.
Tell where f is increasing, where f is decreasing. Classify each critical number.
Solution
f '(x) = 30 x4 + 60 x3 - 240 x2.
Solve f '(x) = 0.
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30 x4 + 60 x3 - 240 x2 = 0
30 (x4 + 2 x3 - 8 x2) = 0
30 x2 (x2 + 2 x - 8 ) = 0
30 x2 (x + 4) (x - 2) = 0.
Thus roots are x = 0, x= -4, and x = 2.
Put them in order:
-4
0
2
Find the sign of f '(x):
(+)
-4
(-)
0
(-)
2
(+)
Thus f is increasing for x<-4 and x > 2.
f is decreasing for -4<x<0 and 0<x<2; we just write that f is decreasing for -4 < x < 2.
There is a relative maximum at x = -4. There is a relative minimum at x = 2. Note that f '(0)
= 0 so 0 is a critical number yet there is neither a relative maximum nor a relative minimum
at x = 0. Thus the classification of critical numbers is
-4
relative maximum
0
neither
2
relative minimum
Example. A population of bacteria grows so that after t hours there are
P(t) = 2t3 - 21 t2 + 60 t + 3 grams of bacteria.
(a) Tell when the population is increasing.
(b) Tell when the population is decreasing.
(c) Tell when there is a relative maximum.
(d) Tell when there is a relative minimum.
(e) Tell the initial population.
(f) Tell the population at the relative maximum.
(g) Tell the population at the relative minimum.
Solution. P'(t) = 6 t2 - 42 t + 60.
Hence P'(t) = 0 when 6 t2 - 42 t + 60 = 0
6 (t2 - 7 t + 10) = 0
6 (t - 5) (t - 2) = 0.
Thus P'(t) = 0 when t = 2 or t = 5.
For 0<t<2, the sign of P'(t) is (+) (-) (-) = +, so P is increasing.
For 2<t<5, the sign of P'(t) is (+) (-) (+) = -, so P is decreasing.
For 5<t the sign of P'(t) is (+) (+) (+) = + so P is increasing.
Hence
(a) P is increasing when 0<t<2 and when 5<t.
(b) P is decreasing when 2<t<5.
(c) There is a relative maximum when t = 2.
(d) There is a relative minimum when t = 5.
(e) The initial population is P(0) = 3 grams.
(f) The population at the relative maximum is P(2) = 55 grams.
(g) The population at the relative minimum is P(5) = 28 grams.
What to do if the expression does not factor:
Sometimes it is hard to factor f '(x). In these cases, to find the sign of f '(x), one can test
any point in the interval. We expect that f '(x) can only change sign at a place where it is
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zero (or undefined). Since we will have found all such places on the chart, everywhere else
it cannot change sign.
Example. Let f(x) = -2.2 x3 + 18.2 x2 + 69.1 x + 16.4
(a) Tell where f is increasing.
(b) Tell where f is decreasing.
(c) Locate and classify all critical numbers.
Solution. f '(x) = -6.6 x2 + 36.4 x + 69.1
We want to solve f '(x) = -6.6 x2 + 36.4 x + 69.1 = 0.
We can't factor, so instead we use the quadratic formula to obtain
x = -1.4938 or x = 7.009
The chart for f '(x) is then
-1.4938
7.009
For x < -1.4938 we substitute a value into f ' to find whether f '(x) is positive or negative.
Pick any such value (and avoid the endpoint). I'll choose -2. Then from the calculator
f '(-2) = -6.6 (-2)2 + 36.4 (-2) + 69.1 = -30.1.
Hence for x < -1.4938 we have f '(x) < 0.
For -1.4938 < x < 7.009 we pick any value. I'll choose x = 0. Then
f '(0) = 69.1 > 0.
Hence for -1.4938 < x < 7.009 we have f '(x) > 0.
For x > 7.009 we pick any value. I'll choose x = 8. Then
f '(8) = -6.6 (8)2 + 36.4 (8) + 69.1 = -62.1.
Hence for x > 7.009 we have f '(x) < 0.
The chart for f '(x) is then
(-)
-1.4938
(+)
7.009
(-)
The answer is then
(a) f is increasing for -1.4938 < x < 7.009
(b) f is decreasing for x < -1.4938 or x > 7.009
(c) when x = -1.4938 there is a relative minimum;
when x = 7.009 there is a relative maximum
Problems for 5.3.
1. Let f(x) = 3 + 4 x - x 2.
(a) Tell where f is increasing.
(b) Tell where f is decreasing.
(c) Tell all x where f has a relative maximum.
(d) Tell all x where f has a relative minimum.
(e) Tell the points where f has a relative maximum.
(f) Tell the points where f has a relative minimum.
Ans: (a) x < 2
(b) x > 2
(c) x = 2
(e) (2, 7)
(f) none
2. Let f(x) =2 x2 - 10 x + 5
(a) Tell where f is increasing.
(d) none
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(b) Tell where f is decreasing.
(c) Tell all x where f has a relative maximum.
(d) Tell all x where f has a relative minimum.
(e) Tell the points where f has a relative maximum.
(f) Tell the points where f has a relative minimum.
Ans: (a) x > 2.5
(b) x < 2.5
(c) none
(e) none
(f) (2.5, -7.5)
3. Let f(x) =2 x3 - 15 x2 + 36 x + 4
(a) Tell where f is increasing.
(b) Tell where f is decreasing.
(c) Tell all x where f has a relative maximum.
(d) Tell all x where f has a relative minimum.
(e) Tell the points where f has a relative maximum.
(f) Tell the points where f has a relative minimum.
Ans: (a) x < 2 or x > 3
(b) 2 < x < 3
(d) x = 3
(e) (2, 32)
(f) (3, 31)
(d) x = 2.5
(c) x = 2
4. Let f(x) =2 x3 + 3 x2 - 12 x + 1
(a) Tell where f is increasing.
(b) Tell where f is decreasing.
(c) Locate and classify all critical numbers.
Ans: (a) x < -2 or x > 1
(b) -2 < x < 1
(c) x = -2: relative maximum; x = 1: relative minimum
5 Let f(x) =3 x - x3 + 4
(a) Tell where f is increasing.
(b) Tell where f is decreasing.
(c) Locate and classify all critical numbers.
Ans: (a) -1 < x < 1
(b) x < -1 or x > 1
(c) x = -1: relative minimum; x = 1: relative maximum
6 Let f(x) =x2 / (x-1).
(a) Tell where f is increasing.
(b) Tell where f is decreasing.
(c) Locate and classify all critical numbers.
Ans: (a) x < 0 or x > 2
(b) 0<x<1 or 1<x<2
(c) x = 0: relative maximum; x = 2: relative minimum; x = 1: neither
7 Let f(x) =15 x - x3 + 2
(a) Tell where f is increasing.
(b) Tell where f is decreasing.
(c) Locate and classify all critical numbers.
Ans: (a) -√5 < x < √5
(b) x < -√5 or x > √5
(c) x = -√5: relative minimum; x = √5: relative maximum
8. A mold colony grows so that after t days there are
P(t) = 6 t5 - 45 t4 + 80 t 3 + 4 mm 2 of mold.
(a) Tell when the population is increasing.
(b) Tell when the population is decreasing.
(c) Locate and classify all critical numbers.
(d) Sketch the graph of P(t), showing this information.
Ans: (a) t < 2 or t > 4 (b) 2<t<4
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(c) t = 0: neither relative maximum nor relative minimum; t= 2: relative maximum;
t = 4: relative minimum
9. A population of bacteria grows so that after t hours there are
P(t) = t3 - 6 t2 + 9 t + 12 grams of bacteria.
(a) Tell when the population is increasing.
(b) Tell when the population is decreasing.
(c) Tell when there is a relative maximum.
(d) Tell when there is a relative minimum.
(e) Tell the population at the relative minimum.
(f) Tell the population at the relative maximum.
(g) Tell the initial population.
(h) Sketch the graph of P(t) showing this information.
Ans: (a) 0<t<1 and 3 < t
(b) 1<t<3
(c) t = 1 hour
(d) t = 3 hours
(e) 12 grams (f) 16 grams (g) 12 grams
5.4. Absolute Maxima and Minima
There are many practical situations where you need to find when an expression is a
maximum or a minimum: For example:
(a) When was the population of deer biggest during the period 1940≤t≤2000?
(b) When was the concentration of the drug highest?
(c) What is the maximum possible harvest?
There are two kinds of extrema-- relative maxima or minima, and absolute maxima or
minima. We have already discussed relative maxima and minima.
A function f(x) has an absolute maximum for a≤x≤b at x = c if
a≤c≤b and moreover, f(c) ≥ f(x) whenever a≤x≤b.
A function f(x) has an absolute minimum for a≤x≤b at x = c if
a≤c≤b and moreover, f(c) ≤ f(x) whenever a≤x≤b.
Thus an absolute maximum occurs when the function is biggest for a≤x≤b, and an absolute
minimum occurs when the function is smallest for a≤x≤b.
In the following graph of f(x), we see that for 0≤x≤6, f has an absolute maximum at x = 5
and an absolute minimum at x = 3.125. There are relative maxima at x = 5 and x = 1.25 and
a relative minimum at x = 3.125.
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4
2
1
2
3
4
5
6
-2
As in this example, an absolute maximum can occur at a relative maximum, hence at a
critical number. An absolute minimum can occur at a relative minimum, hence at a critical
number.
For the same graph, where is the absolute maximum for 0≤x≤4? This means we pay
attention only to the part of the graph for 0≤x≤4. So the absolute maximum is at x = 1.25,
which was a relative maximum.
Where is the absolute maximum for 0≤x≤4.5? Here clearly the largest value is at x = 4.5,
which is not a critical number. Instead, x = 4.5 is an endpoint for 0≤x≤4.5 or the interval
[0,4.5].
where is the absolute minimum for 4.5≤x≤6? Clearly it occurs at x = 6, which is an
endpoint of [4.5,6], not a critical number.
Summary: An absolute maximum or minimum of f(x) for a x b occurs either
(i) at a critical number in the interval [a,b]; or
(ii) at an endpoint (x = a or x = b).
An absolute maximum occurs, of course, either at a relative maximum or at an endpoint. An
absolute minimum occurs, of course, either at a relative minimum or at an endpoint.
Procedure for locating an absolute maximum or an absolute minimum for f(x) on
an interval [a,b]:
(1) List all the critical numbers in the interval [a,b].
(2) List the endpoints a and b.
(3) Evaluate f(x) for each listed number.
The largest f(x) in the table is the absolute maximum, and the x where it occurs is
the location of the absolute maximum. The smallest f(x) in the table is the
absolute minimum, and the x where it occurs is the location of the absolute
minimum.
Example. Let f(x) = x2 - 2 x.
(a) Find the absolute extrema for 0≤x≤4 and where they occur.
(b) Find the absolute extrema for 3≤x≤4 and where they occur.
(c) Find the absolute extrema for 0.5≤x≤2 and where they occur.
Solution. (a) f '(x) = 2x - 2
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The critical numbers satisfy 2 x - 2 = 0
2x = 2
x = 1.
Thus the possible places for an absolute extremum are x = 1, 0, or 4. Make a chart:
x
f(x)
1
-1
0
0
4
8
The absolute maximum is 8 occurring at x = 4.
The absolute minimum is -1 occurring at x = 1.
(b) The only critical number is x = 1, as above, but it is not in the interval 3≤x≤4. Hence the
only possible places for an absolute extremum are the endpoints, x = 3 and x = 4. Now the
chart is
x
f(x)
3
3
4
8
The absolute maximum is 8 at x = 4. The absolute minimum is 3 at x = 3.
(c) The critical number 1 lies in the interval. hence the chart now is
x
f(x)
1
-1
0.5
-0.75
2
0
The absolute maximum is 0 at x = 2. The absolute minimum is -1 at x = 1.
Example. Let f(x) = x3 - 6 x2 + 9 x + 5.
(a) Find the absolute maximum of f(x) for 0≤x≤5, and where it occurs.
(b) Find the absolute minimum of f(x) for 0≤x≤5, and where it occurs.
Solution. f '(x) = 3 x2 - 12 x + 9.
= 3(x2 - 4 x + 3) = 3 (x-3) (x-1).
The critical numbers are x = 1, x=3. Both are in [0,5]. Form a chart
x
f(x)
1
9
3
5
0
5
5
25
On [0,5] the absolute maximum is 25 and it occurs when x = 5; the absolute minimum is 5
and it occurs when x = 3 or x = 0.
(c) Find the absolute extrema of f(x) for 0≤x≤2, and tell where they occur.
Now the chart contains the new endpoints (0 and 2) and the critical number 1 but not the
critical number 3, since 3 is not in the relevant interval. The chart is
x
f(x)
1
9
0
5
2
7
On [0,2] the absolute maximum is 9 and it occurs when x = 1; the absolute minimum is 5
and it occurs when x = 0.
(d) Find the absolute extrema of f(x) for 2≤x≤4, and tell where they occur.
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Now the chart contains the new endpoints (2 and 4) and the critical number 3 but not the
critical number 1, since 1 is not in the relevant interval. The chart is
x
f(x)
3
5
2
7
4
9
On [2,4] the absolute maximum is 9 and it occurs when x = 4; the absolute minimum is 5
and it occurs when x = 3.
(d) Find the absolute extrema of f(x) for 4≤x≤5, and tell where they occur.
Now the chart contains the new endpoints (4 and 5) but neither critical number. The chart is
x
f(x)
4
9
5
25
On [4,5] the absolute maximum is 25 and it occurs when x = 5; the absolute minimum is 9
and it occurs when x = 4.
Example. Find the absolute maximum and minimum for f(x) = 4 - x + x2 on [0,2], and tell
where they occur.
Solution. f '(x) = -1 + 2 x.
-1 + 2x = 0 when x = 1/2
The chart is
x
f(x)
1/2
3.75
0
4
2
6
The absolute maximum is 6 (at x = 2) and the absolute minimum is 3.75 (at x = 1/2).
Problems for 5.4.
1. Find the absolute maxima and minima and where they occur:
(a) f(x) = x2 - 6 x + 2 for 0≤x≤5
(b) f(x) = x2 - 6 x + 2 for 2≤x≤6
(c) f(x) = x2 + 4x - 5 for 0≤x≤2
(d) f(x) = x2 + 4x - 5 for -3≤x≤2
Ans: (a) maximum = 2 at x = 0; minimum = - 7 at x = 3
(b) maximum = 2 at x = 6; minimum = -7 at x = 3
(c) maximum = 7 at x = 2; minimum = -5 at x = 0
(d) maximum = 7 at x = 2; minimum = -9 at x = -2
2. Let f(x) = 2 x3 + 3 x2 - 36 x + 5
(a) Find the absolute maximum and where it occurs if 0≤x≤3.
(b) If 0≤x≤3, find the absolute minimum and where it occurs.
(c) If 0≤x≤5, find the absolute maximum and where it occurs.
(d) If 0≤x≤5, find the absolute minimum and where it occurs.
(e) If -4 ≤x≤0, find the absolute maximum and where it occurs.
(f) If -4 ≤x≤0, find the absolute minimum and where it occurs.
(g) If -4≤x≤5, find the absolute maximum and where it occurs.
(h) If -4≤x≤5, find the absolute minimum and where it occurs.
150
Ans: (a) max = 5 at x = 0
(c) max = 150 at x = 5
(e) max = 86 at x = -3
(g) max = 150 at x = 5
(b) min = -39 at x = 2
(d) min = -39 at x = 2
(f) min = 5 at x = 0
(h) min = -39 at x = 2
3. Find the absolute maxima and minima and where they occur:
(e) f(x) = 4 x3 - 5 x2 + 7 for 0≤x≤2
Ans:
(e) maximum = 19 at x = 2; minimum = 5.8426 at x = 5/6 =0.8333
5.5. Applied maximum and minimum problems
Here are some problems that show how maxima and minima can be applied to different
situations:
Example. A mold colony grows and shrinks so that, for 1 ≤t≤ 12, after t days the area of
the colony is
A(t) = 12t - t2 + 2 mm 2.
(a) What is the maximal area of the colony?
(b) When is the colony largest?
(c) What is the minimal area of the colony?
(d) When is the area smallest?
Solution. Here we seek the absolute maximum and minimum area for 1≤t≤12.
A'(t) = 12-2t = 0 when
12 = 2t
t = 6.
The possibilities are therefore
t
A(t)
6
38
1
13
12
2
(a) The maximal area is 38 mm2.
(b) The colony is largest after t= 6 days.
(c) The smallest area is 2 mm2.
(d) The smallest area occurs after t = 12 days.
Example. A bacterial population grows and shrinks so that after t hours, for the relevant
time period, the biomass of the population is
P(t) = -2 t2 + 44 t + 5 mg
(a) What is the maximal biomass of the population?
(b) When is the population largest?
Solution. Here they don't given an interval. Clearly we need t≥0, but otherwise there is no
constraint.
P'(t) = -4t + 44
P'(t) = 0 when t = 11. Hence
the only critical number is 11, and only positive values are meaningful.
151
We find
(+)
11
(-)
so there is a relative maximum at t = 11. Clearly this is the only maximum, since the
function increases for t<11 and decreases for t> 11. So 11 must also be the location of the
absolute maximum.
(a) The maximum population will be P(11) = 247 mg.
(b) The population is largest when t = 11 hours.
Sometimes the formula to be maximized and minimized is not given, but instead you must
find it. The general procedure is as follows:
(1) Find the variables that describe the situation.
(2) Find a formula for the quantity to be maximized or minimized.
(3) Express the quantity to be maximized or minimized in terms of just one variable. To do
this, use other given relationships or geometry.
(4) Use calculus to minimize or maximize the relevant quantity.
Example. An experimenter must fence in a rectangular field with fences on three sides and
the fourth side along an already-existing brick wall that runs the length of the property. The
area must be 3200 square feet. Find the dimensions of the field that uses the least amount
of fence.
Solution. We want to minimize the total length of fence. Let x be the length of the fence
that is parallel to the brick wall. Let y be the length of the fence perpendicular to the brick
wall. Then the total length of fence is
L = x + 2y
This has two variables. We need to find an expression with only one variable.
Since the area is A = xy = 3200, it follows that y = 3200/x.
Hence L = x + 2(3200/x) = x + 6400/x
The problem is then: Find x that will minimize L. Clearly we want an absolute minimum.
Dx[L] = Dx[ x + 6400 x-1] = 1 - 6400 x -2
so the derivative is 0 when 1 = 6400 x-2
Multiply by x2 to get
x2 = 6400
x = ±80
but only positive values are relevant.
The derivative does not make sense when x = 0.
The chart of relevant x is
0
(-)
80
(+)
Clearly the minimum occurs when x = 80.
Since y = 3200/x, we have y = 3200/80 = 40. Thus the field is 40 feet by 80 feet, with the
80 feet along the brick wall.
Example. A drug is being developed for the market. Because of economies of scale, the
cost of producing a bottle depends on the number x of bottles manufactured. Assume that
the cost of producing x bottles is 40x + 500 dollars.
The company can sell x bottles provided the selling price of one bottle is 70- 0.05 x.
(a) How many bottles should the company produce in order to maximize its profit?
(b) What is the maximum profit?
(c) What is the selling price of one bottle?
152
Solution. The profit is the money received minus the costs. If x bottles are produced, the
total cost is 40x + 500, while the income is x(70-0.05 x).
Hence the profit is
P(x) = x (70 - 0.05 x) - (40 x + 500) dollars
P(x) = 70 x - 0.05 x 2 - 40 x - 500
P(x) = 30 x - 0.05 x 2 - 500
To maximize the profit, we find P'(x) = 30 - 0.1 x = 0
30 = 0.1 x
300 = x.
Note that the chart for P ' is
(+)
300
(-)
so the maximum occurs when x = 300.
(a) The manufacturer should produce 300 bottles.
(b) The maximum profit is P(300) = $4000
(c) The selling price of one bottle is 70 - 0.05(300) = $55
Example. A rectangular box with an open top and square base is being made from sheet
metal. You have 768 square inches of metal. Find the dimensions of the box that has the
maximum volume.
Solution. Let x be the length of the square base, and y be the height of the box, both
measured in inches. Then the volume is
V = x2 y
Since the base is a square with side x, its area is x2. Each of the 4 sides is a rectangle with
base x and height y, hence area xy. Thus the total area is x2 + 4 xy. Clearly we want to
use all the sheet metal for the largest box. Hence
x2 + 4xy = 768.
We want to maximize V, so we need to express it using only one variable. From the
formula for area we find
y=
768 - x2
--------4x
Hence V =
x2 y =
x2 (768 - x2)
----------------4x
=
x(768 - x2)
------------4
or V = (1/4) (768 x - x3).
Now V ' = (1/4) (768 - 3 x2)
The maximum will occur when V ' = 0
(1/4) (768 - 3 x2) = 0
768 = 3 x2
768 = 3 x2
x2 = 256
x = √256 = 16
Hence y = (768 - 162)/ (4(16)) = 8.
Answer: The box should have base 16 inches and height 8 inches.
153
Problems for 5.5.
1. An experimenter must fence in a rectangular field with fences on three sides and the
fourth side along an already-existing wall. The area must be 1800 square feet. Find the
dimensions of the field that uses the least amount of fence.
Ans: 30 feet perpendicular to the wall; 60 feet parallel to the wall.
2. An experimenter must fence in a rectangular field. The fences on the north and south
cost $10 per foot while the fences on the east and west cost $15 per foot. The area must be
600 square feet. Find the dimensions of the field that requires the cheapest fence.
Ans: 30 feet on north and south; 20 feet on east and west
3. An experimenter must fence in a rectangular field. The fence on the south costs $10 per
foot while the fences on the other three sides cost $15 per foot. The area must be 3000
square feet. Find the dimensions of the field that requires the cheapest fence.
Ans: 60 feet on north and south; 50 feet on east and west
4. An experimenter must fence in a rectangular field. He has 900 feet of fencing. Find the
dimensions of the largest field possible.
Ans: square 225 feet on a side
5. An experimenter must fence in a rectangular field. No fence is on the north side
because it is on a straight river. The other three sides require fencing. If she has 900 feet of
fencing, find the dimensions of the largest field possible.
Ans: 450 feet on north and south; 225 feet on east and west
6. A rectangular box with an open top and square base is being made from sheet metal. You
have 432 square inches of metal. Find the dimensions of the box that has the maximum
volume.
Ans: The base is 12 inches on a side, while the height is 6 inches.
7. A rectangular box with a closed top and square base is being made from sheet metal.
You have 1350 square inches of metal. Find the dimensions of the box that has the
maximum volume.
Ans: The box is a cube with sides of length 15 inches.
8. A mold colony grows and shrinks so that after t days the area of the colony is
A(t) = 10 t - t2 + 2 mm 2.
(a) What is the maximal area of the colony?
(b) When is the colony largest?
Ans: (a) 27 mm2
(b) t = 5 days
9. A mold colony grows and shrinks so that after t days the area of the colony is
A(t) = 24.8 t - 2 t2 + 1.4 mm 2 for 0≤t≤8.
(a) What is the maximal area of the colony during this time?
(b) When is the colony largest?
(c) What is the minimal area of the colony during this time?
(d) When is the area minimized during this time?
Ans: (a) 78.28 mm2
(b) t = 6.2 days
(c) 1.4 mm2 (d) t = 0 days
154
10. A bacterial population grows and shrinks so that after t hours the biomass of the
population is
P(t) = -2.2 t2 + 155.32 t + 1.65 mg for 20≤t≤60.
(a) What is the maximal biomass of the population?
(b) When is the population largest?
(c) When is the population smallest during this time?
(d) What is the smallest population during this time?
Ans: (a) 2743 mg
(b) 35.3 hours
(c) t = 60 hours
(d) 1401 mg
11. A patient is being medicated. For 0≤t≤7, the drug concentration in her blood t hours
after the beginning of treatment is
D(t) = - 5 t2 + 34 t + 15 mg/liter.
(a) When is the drug concentration largest during this time?
(b) What is the maximum drug concentration?
(c) When is the drug concentration smallest during this time?
(d) What is the minimum drug concentration during this time?
Ans: (a) t = 3.4 hours
(b) 72.8 mg/liter
(c) t = 7 hours
(d) 8 mg/liter
12. A bacterial population grows and shrinks so that after t hours the biomass of the
population is
P(t) = -2 t3 + 15 t 2 + 300 t + 4 mg.
(a) What is the maximal biomass of the population?
(b) When is the population largest?
Ans: (a) 2504 mg
(b) 10 hours
13. A patient is being medicated. Over the relevant time interval, the drug concentration in
her blood t hours after the beginning of treatment is
D(t) = - 2 t3 + 3.9 t 2 + 25.2 t + 9 mg/liter.
(a) When is the drug concentration largest?
(b) What is the maximum drug concentration?
Ans: (a) 2.8 hours after the start
(b) 66.232 mg/liter
14. A chemical reaction A + B → AB
occurs at the rate R(x) = 3 (4 - x) (6 - x) for 0 ≤ x ≤ 4
where x is the concentration of the product AB in centimoles/liter and R is the reaction rate
in centimoles/ second.
(a) Find the concentration x that will maximize the reaction rate.
(b) Find the maximum reaction rate
Ans: (a) x = 0 centimoles/liter
(b) 72 centimoles/second
15. The yield of corn is a function
Y = 10 F / (2 + F2) for F≥0
where F is the amount of fertilizer in the soil in tons/acre
and Y is measured in tons/acre
(a) Find the level of fertilizer that will maximize the yield.
(b) Find the maximum yield.
Ans: (a) 1.414 tons/acre
(b) 3.536 tons/acre
155
5.6. Harvesting natural populations
Some biologists end up in environmental science. For example, they might get jobs
managing fisheries or wild populations that are being hunted. What might such
management entail?
Example. Suppose that the breeding fish population at the start of a year is P.
At the start of the year one year later, the population is F(P). For example, maybe you know
from studies that
F(P) = P + 0.4 P (1 - 0.0001 P)
when the population is left to a state of nature.
If the population at the start of the 2004 is P and the population is left alone in nature, then
one year later at the start of 2005 the population is F(P).
Suppose that the population in year 2000 is 3,000 fish (on January 1). The population in
year 2001 (on January 1) is then
F(3000)
= 3000 + 0.4 (3000) (1 - 0.0001 (3000))
= 3840 fish.
The population in year 2002 (on January 1) is then
F(3840)
= 3840 + 0.4 (3840) (1 - 0.0001 (3840))
= 4786 fish
The equilibrium population (in the state of nature) is the population P so F(P) = P. This
is the population that will remain the same year after year.
Example. Find the equilibrium population for the fish population above.
Solution. Solve F(P) = P.
P + 0.4 P (1 - 0.0001 P) = P
0.4 P (1 - 0.0001 P) = 0
P = 0 or 1 - 0.0001 P = 0
P = 0 or 0.0001 P = 1
P = 0 or P = 1/0.0001 = 10000 fish
Answer: There are two equilibria: 0 and 10,000 fish.
In fact, people want to "harvest" some fish (catch them to eat).
Example. Suppose that the population in year 2000 is 3,000 fish (on January 1). Suppose
that we harvest 1700 fish on January 1, 2001. How many fish will there be on January 1,
2002?
Solution. At the start of January 1, 2001 there are F(3000) = 3840 fish. We harvest 1700
of them, leaving 3840-1700 = 2140 fish alive to breed. Hence the population on January 1,
2002 is F(2140)
= 2140 + 0.4 (2140) (1 - 0.0001 (2140))
= 2812 fish
Actually, people usually want a big harvest, so they want to "maximize" the harvest.
Example. Suppose that the population in year 2000 is 3,000 fish (on January 1). Suppose
that we harvest the maximal possible harvest of fish on January 1, 2001.
(a) What is the maximal possible harvest of fish on January 1, 2001?
156
(b) If you make the maximal harvest on January 1, 2001, how many fish will there be on
January 1, 2002?
Solution. (a) On January 1, before any fish are harvested, there are
F(3000) = 3000 + 0.4 (3000) (1 - 0.0001 (3000))
= 3840 fish.
The maximal possible harvest is to take them all. So the maximal harvest is 3840 fish.
(b) Now the breeding population on January 1, 2001 is 0 fish. The following year there are
F(0) = 0 + 0.4 (0) (1 - 0.0001 (0)) = 0 fish.
So getting a maximal harvest one year drives the population to extinction, so there are no
future harvests.
A harvest H is sustainable provided there is a population P so that the population one year
later, after harvesting, is again P. This value of P is called the equilibrium population for
a harvest of H.
Suppose the population is P at the start of the year. One year later the population, before
harvest, is F(P). After harvest the breeding population is F(P)-H.
Since P is sustainable, we must have
F(P) - H = P
or H = F(P) - P.
Here P is the equilibrium population for the harvest H.
Example. Show that a harvest of 500 fish is sustainable, and find the equilibrium
population under a harvest of 500 fish.
Solution. If the harvest H is sustainable, we must have
H = F(P) - P.
500 = F(P) - P
500 = P + 0.4 P (1 - 0.0001 P) - P
500 = 0.4 P (1 - 0.0001 P)
500 = 0.4 P - 0.00004 P 2
0.00004 P2 - 0.4 P + 500 = 0
P = 8536 fish or P = 1464 fish.
Example. Show that a harvest of 8000 fish is not sustainable.
Solution. If the harvest H is sustainable, we must have
H = F(P) - P.
8000 = F(P) - P
8000 = P + 0.4 P (1 - 0.0001 P) - P
8000 = 0.4 P (1 - 0.0001 P)
8000 = 0.4 P - 0.00004 P 2
0.00004 P2 - 0.4 P + 8000 = 0
P = 5000 ± 13228 i fish.
This is meaningless. The harvest is not sustainable.
The biggest harvest that is sustainable is called the maximal sustainable harvest. It is the
biggest harvest that can be obtained year after year after year.
The sustainable harvest H when the equilibrium population is P is given by
H = F(P) - P
Hence the maximum sustainable harvest is found by maximizing H.
157
Example. Find the maximal sustainable harvest and the equilibrium population under that
harvest where F(P) = P + 0.4 P (1 - 0.0001 P)
and P is the number of fish in the population.
(a) Find the maximal sustainable harvest.
(b) Find the equilibrium population for the maximal sustainable harvest.
(c) What percent of the equilibrium population is harvested for a maximal sustainable
harvest?
.
Solution. H = F(P) - P
H=
P + 0.4 P (1 - 0.0001 P) - P
H = 0.4 P (1 - 0.0001 P)
H = 0.4 P - 0.00004 P2
H' = 0.4 - 0.00008 P
Solve H' = 0:
0.4 - 0.00008 P = 0
0.4 = 0.00008 P
P = 0.4 / 0.00008 = 5000 fish.
Note that the chart for H' is
(+)
5000 (-)
so there is a maximum when P = 5000.
The harvest H when P = 5000 is
H = 0.4 (5000) - 0.00004 (5000) 2
H = 1000
(a) The maximal sustainable harvest is 1000 fish.
(b) The equilibrium population for a maximal sustainable harvest is P = 5000 fish.
(c) The harvest fraction is 1000/5000 = 0.2 = 20%.
Example. A population of deer in a certain forest satisfies that, when there are P deer one
year, then the following year the population is
1.3 P - 0.0003 P 2
deer. The local bureau of forestry wants to manage the deer population for the benefit of
hunters.
(a) Find the maximum sustainable harvest.
(b) Find the equilibrium population of deer when there is a maximal sustainable harvest.
Solution. Here F(P) = 1.3 P - 0.0003 P 2
A sustainable harvest H would satisfy
H = F(P) - P = 1.3 P - 0.0003 P2 - P
H = 0.3 P - 0.0003 P2
The harvest is maximized when H ' = 0.
Hence 0.3 - 2 (0.0003 P) = 0
0.3 - 0.0006 P = 0
- 0.0006 P = -0.3
P = 500
The harvest is then H = 0.3 (500) - 0.0003 (500)2
= 75.
Answers: (a) The maximum sustainable harvest is 75 deer.
(b) The maximum sustainable harvest equilibrium is 500 deer.
158
Problems for 5.6.
1. Suppose that a population of fish contains P fish at the end of the year. It is known that
the population at the end of the next year, under natural conditions, is F(P), which satisfies
the equation
F(P)= P + 0.35 P (1 - .000025 P)
(a) Suppose the population at the end of one year is 30000 fish. What is the
population at the end of the next year?
(b) Suppose the population at the end of one year is 30000 fish. What is the
maximal possible harvest at the end of the next year?
(c) Suppose the population at the end of one year is 30,000 fish. Suppose the
maximal possible harvest is made at the end of the next year. What will be the maximal
possible harvest the year after that?
(d) What is the positive equilibrium population under natural conditions?
(e) Find the equilibrium population when there is a maximum sustainable harvest.
(f) What is the maximum sustainable harvest?
(g) If we harvest a constant fraction h of the sustainable equilibrium population,
what value of h leads to the maximum sustainable harvest?
Answers:
(a) 32625 fish
(b) 32625 fish
(c) 0 fish
(d) 40,000 fish
(e) 20,000 fish
(f) 3500 fish
(g) .175 or 17.5%
2. Suppose that a population of fish contains P tons of fish at the end of the year. It is
known that at the end of the next year, under natural conditions, the population will consist
of G(P) tons of fish, where
G(P) = P + 0.62 P (1 - .0003 P)
(a) Suppose the population at the end of one year is 7000 tons of fish. What is the
population at the end of the next year?
(b) What is the positive equilibrium population under natural conditions?
(c) Find the equilibrium population when there is a maximum sustainable harvest.
(d) What is the maximum sustainable harvest?
Answers:
(a) 2226 tons of fish
(b) 3333 tons of fish
(c) 1667 tons of fish
(d) 517 tons of fish
3. Suppose that a population of fish contains P tons of fish at the end of the year. It is
known that at the end of the next year, under natural conditions, the population will consist
of F(P) tons of fish, where
F(P) = 1.27 P - 0.0000027 P2
(a) Suppose the population at the end of one year is 7000 tons of fish. What is the
population at the end of the next year?
(b) What is the positive equilibrium population?
(c) Find the equilibrium population when there is a maximum sustainable harvest.
(d) What is the maximum sustainable harvest?
(e) What fraction of the equilibrium for a sustainable harvest is fished each year?
Answers:
(a) 8757.7 tons
(b) 100,000 tons of fish
(c) 50,000 tons of fish
(d) 6750 tons of fish
(e) 0.135
4. Suppose that a population of fish contains P tons of fish at the end of the year. It is
known that the the following year the natural population will consist of F(P) tons of fish,
where
F(P) = P + 0.000075 P2 (1 - .00001 P)
159
(a) Suppose the population at the end of one year is 30000 fish. What is the
population at the end of the next year?
(b) Suppose the population at the end of one year is 30000 fish. What is the
maximal possible harvest at the end of the next year?
(c) Find the equilibrium population in the natural state.
(d) Find the equilibrium population when there is a maximum sustainable harvest.
(e) What is the maximum sustainable harvest?
(f) Explain how the maximum sustainable harvest can be larger than the equilibrium
population for a maximum sustainable harvest.
Answers:
(a) 77,250 tons of fish
(b) 77,250 tons of fish
(c) 100,000 tons
(d) 66,667 tons
(e) 111,111 tons
5. Suppose that a population of deer in a certain region contains D deer at the beginning of
the year. It is known that the the following year the natural population will consist of F(D)
deer, where
F(D) = D + 0.75 D (1 - .00002 D)
Deer hunters want the population to be managed so that, each year, they can hunt as many
deer as possible. Assuming that each license allows exactly one deer to be killed, how many
hunting licenses should be issued each year?
Ans: 9375 licenses
5.7. Differentiating the exponential and logarithm
functions
We wish to find and use derivatives for functions of the form f(x) = ax, where a is a
constant. By far the most convenient such function for this purpose is the exponential
function with base the special number e.
Definition. The number e, which is approximately 2.7182818284590..., is the number
such that
lim
eh - 1
h→0 ------ = 1
h
The number e is called Euler's number, after the great mathematician Leonard Euler
(1707-1783).
The major reason for the use of e is the following theorem, which says that ex is its own
derivative:
Theorem. Dx [ex ] = ex .
Proof. For any function f(x), recall that
f '(x) =
lim
f(x+h) - f(x)
h→0 ---------------h
Hence