Chapter 3: Differentials and Integrals in R

1 Derivatives and Marginals (R)

2 Maxima and Minima (C)

3 Derivatives of Implicit Functions

4 Indefinite Integrals and Applications

5 Definite Integrals and Applications

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3.2 Maxima and Minima

Discussion
Suppose that the weekly sales revenue of a film company is modelled
by
50t
R(t) = 2
t + 196
where R is in millions of dollars and t is the number of weeks. The
company wants to know when the revenue is increasing or decreasing
and when it is maximized.

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3.2 Maxima and Minima
Definition 1
The point (x1 , f (x1 )) is said a relative maximum of f if there exists an
interval I containing x1 such that
f (x) ≤ f (x1 ), ∀ x ∈ I.
Similarly, (x2 , f (x2 )) is said a relative minimum of f if
f (x) ≥ f (x2 ), ∀ x ∈ I.
If I is the domain of f or the interested interval, we obtain the absolute
maximum and absolute minimum .

y = f (x)

x2 x1 x

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Definition 2
Let f : (a, b) → R. We say that f is increasing (decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) ≥ f (x2 ) (resp, f (x1 ) ≤ f (x2 )).

f is called strictly increasing (strictly decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) > f (x2 ) (resp, f (x1 ) < f (x2 )).

Theorem 1
Let f be differentiable on (a, b). If
f 0 (x) ≥ 0 ∀ x ∈ (a, b) then f is increasing on (a, b);
f 0 (x) ≤ 0 ∀ x ∈ (a, b) then f is decreasing on (a, b);
f 0 (x) > 0 ∀ x ∈ (a, b) then f is strictly increasing on (a, b);
f 0 (x) < 0 ∀ x ∈ (a, b) then f is strictly decreasing on (a, b).
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3.2 Maxima and Minima
Definition 2
Let f : (a, b) → R. We say that f is increasing (decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) ≥ f (x2 ) (resp, f (x1 ) ≤ f (x2 )).

f is called strictly increasing (strictly decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) > f (x2 ) (resp, f (x1 ) < f (x2 )).

Theorem 1
Let f be differentiable on (a, b). If
f 0 (x) ≥ 0 ∀ x ∈ (a, b) then f is increasing on (a, b);
f 0 (x) ≤ 0 ∀ x ∈ (a, b) then f is decreasing on (a, b);
f 0 (x) > 0 ∀ x ∈ (a, b) then f is strictly increasing on (a, b);
f 0 (x) < 0 ∀ x ∈ (a, b) then f is strictly decreasing on (a, b).
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3.2 Maxima and Minima
Definition 2
Let f : (a, b) → R. We say that f is increasing (decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) ≥ f (x2 ) (resp, f (x1 ) ≤ f (x2 )).

f is called strictly increasing (strictly decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) > f (x2 ) (resp, f (x1 ) < f (x2 )).

Theorem 1
Let f be differentiable on (a, b). If
f 0 (x) ≥ 0 ∀ x ∈ (a, b) then f is increasing on (a, b);
f 0 (x) ≤ 0 ∀ x ∈ (a, b) then f is decreasing on (a, b);
f 0 (x) > 0 ∀ x ∈ (a, b) then f is strictly increasing on (a, b);
f 0 (x) < 0 ∀ x ∈ (a, b) then f is strictly decreasing on (a, b).
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Definition 2
Let f : (a, b) → R. We say that f is increasing (decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) ≥ f (x2 ) (resp, f (x1 ) ≤ f (x2 )).

f is called strictly increasing (strictly decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) > f (x2 ) (resp, f (x1 ) < f (x2 )).

Theorem 1
Let f be differentiable on (a, b). If
f 0 (x) ≥ 0 ∀ x ∈ (a, b) then f is increasing on (a, b);
f 0 (x) ≤ 0 ∀ x ∈ (a, b) then f is decreasing on (a, b);
f 0 (x) > 0 ∀ x ∈ (a, b) then f is strictly increasing on (a, b);
f 0 (x) < 0 ∀ x ∈ (a, b) then f is strictly decreasing on (a, b).
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Definition 2
Let f : (a, b) → R. We say that f is increasing (decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) ≥ f (x2 ) (resp, f (x1 ) ≤ f (x2 )).

f is called strictly increasing (strictly decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) > f (x2 ) (resp, f (x1 ) < f (x2 )).

Theorem 1
Let f be differentiable on (a, b). If
f 0 (x) ≥ 0 ∀ x ∈ (a, b) then f is increasing on (a, b);
f 0 (x) ≤ 0 ∀ x ∈ (a, b) then f is decreasing on (a, b);
f 0 (x) > 0 ∀ x ∈ (a, b) then f is strictly increasing on (a, b);
f 0 (x) < 0 ∀ x ∈ (a, b) then f is strictly decreasing on (a, b).
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3.2 Maxima and Minima
Definition 2
Let f : (a, b) → R. We say that f is increasing (decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) ≥ f (x2 ) (resp, f (x1 ) ≤ f (x2 )).

f is called strictly increasing (strictly decreasing) if

∀x1 , x2 ∈ (a, b), if x1 > x2 then f (x1 ) > f (x2 ) (resp, f (x1 ) < f (x2 )).

Theorem 1
Let f be differentiable on (a, b). If
f 0 (x) ≥ 0 ∀ x ∈ (a, b) then f is increasing on (a, b);
f 0 (x) ≤ 0 ∀ x ∈ (a, b) then f is decreasing on (a, b);
f 0 (x) > 0 ∀ x ∈ (a, b) then f is strictly increasing on (a, b);
f 0 (x) < 0 ∀ x ∈ (a, b) then f is strictly decreasing on (a, b).
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3.2 Maxima and Minima

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Definition 3
A relative maximum or a relative minimum is called a relative
extremum.

A point x0 ∈ D is called a critical value of f if f 0 (x0 ) = 0.

Theorem 2 (Fermat)
Suppose f is differentiable at x0 which is a relative extremum of f , then
f 0 (x0 ) = 0.

Theorem 3
Let f ∈ C 2 (a, b) and x0 ∈ (a, b). Suppose that f 0 (x0 ) = 0. If
f 00 (x0 ) < 0 then (x0 , f (x0 )) is a relative maximum of f ;
f 00 (x0 ) > 0 then (x0 , f (x0 )) is a relative minimum of f .

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3.2 Maxima and Minima

Definition 3
A relative maximum or a relative minimum is called a relative
extremum.

A point x0 ∈ D is called a critical value of f if f 0 (x0 ) = 0.

Theorem 2 (Fermat)
Suppose f is differentiable at x0 which is a relative extremum of f , then
f 0 (x0 ) = 0.

Theorem 3
Let f ∈ C 2 (a, b) and x0 ∈ (a, b). Suppose that f 0 (x0 ) = 0. If
f 00 (x0 ) < 0 then (x0 , f (x0 )) is a relative maximum of f ;
f 00 (x0 ) > 0 then (x0 , f (x0 )) is a relative minimum of f .

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3.2 Maxima and Minima

Definition 3
A relative maximum or a relative minimum is called a relative
extremum.

A point x0 ∈ D is called a critical value of f if f 0 (x0 ) = 0.

Theorem 2 (Fermat)
Suppose f is differentiable at x0 which is a relative extremum of f , then
f 0 (x0 ) = 0.

Theorem 3
Let f ∈ C 2 (a, b) and x0 ∈ (a, b). Suppose that f 0 (x0 ) = 0. If
f 00 (x0 ) < 0 then (x0 , f (x0 )) is a relative maximum of f ;
f 00 (x0 ) > 0 then (x0 , f (x0 )) is a relative minimum of f .

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3.2 Maxima and Minima

Definition 3
A relative maximum or a relative minimum is called a relative
extremum.

A point x0 ∈ D is called a critical value of f if f 0 (x0 ) = 0.

Theorem 2 (Fermat)
Suppose f is differentiable at x0 which is a relative extremum of f , then
f 0 (x0 ) = 0.

Theorem 3
Let f ∈ C 2 (a, b) and x0 ∈ (a, b). Suppose that f 0 (x0 ) = 0. If
f 00 (x0 ) < 0 then (x0 , f (x0 )) is a relative maximum of f ;
f 00 (x0 ) > 0 then (x0 , f (x0 )) is a relative minimum of f .

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3.2 Maxima and Minima

Definition 3
A relative maximum or a relative minimum is called a relative
extremum.

A point x0 ∈ D is called a critical value of f if f 0 (x0 ) = 0.

Theorem 2 (Fermat)
Suppose f is differentiable at x0 which is a relative extremum of f , then
f 0 (x0 ) = 0.

Theorem 3
Let f ∈ C 2 (a, b) and x0 ∈ (a, b). Suppose that f 0 (x0 ) = 0. If
f 00 (x0 ) < 0 then (x0 , f (x0 )) is a relative maximum of f ;
f 00 (x0 ) > 0 then (x0 , f (x0 )) is a relative minimum of f .

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3.2 Maxima and Minima

Example 1
Find the relative maxima and minima of f where

f (x) = 2x 3 − 12x 2 + 6.

Exercise 1
Find the relative maxima and minima of f where
a) f (x) = x 3 − 3x 2 + 6x + 1 b) f (x) = 2x 5 + 5x 4 + 11.

c) f (x) = ex − x + 5 d) f (x) = ln x − 12 x 2 , x > 0.

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3.2 Maxima and Minima

Example 1
Find the relative maxima and minima of f where

f (x) = 2x 3 − 12x 2 + 6.

Exercise 1
Find the relative maxima and minima of f where
a) f (x) = x 3 − 3x 2 + 6x + 1 b) f (x) = 2x 5 + 5x 4 + 11.

c) f (x) = ex − x + 5 d) f (x) = ln x − 12 x 2 , x > 0.

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3.2 Maxima and Minima

Solution of Example 1
We have: f 0 (x) = 6x 2 − 24x and f 00 (x) = 12x − 24. The equation
f 0 (x) = 0 has 2 roots: x1 = 0, x2 = 4.
∗ f 00 (0) = −24 < 0 ⇒ (0, 6) is a relative maximum.
∗ f 00 (4) = 24 > 0 ⇒ (4, −58) is a relative minimum.

Example 2
Find the relative maxima, relative minima, and horizontal points of
inflection of h(x) = 14 x 4 − 23 x 3 − 2x 2 + 8x + 4.

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3.2 Maxima and Minima

Solution of Example 1
We have: f 0 (x) = 6x 2 − 24x and f 00 (x) = 12x − 24. The equation
f 0 (x) = 0 has 2 roots: x1 = 0, x2 = 4.
∗ f 00 (0) = −24 < 0 ⇒ (0, 6) is a relative maximum.
∗ f 00 (4) = 24 > 0 ⇒ (4, −58) is a relative minimum.

Example 2
Find the relative maxima, relative minima, and horizontal points of
inflection of h(x) = 14 x 4 − 23 x 3 − 2x 2 + 8x + 4.

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3.2 Maxima and Minima
Solution of Example 2
Note that h0 (x) = x 3 − 2x 2 − 4x + 8 = (x 2 − 4)(x − 2) = (x + 2)(x − 2)2 .
Thus there are two critical points (−2, − 32 32
3 ) and (2, 3 ).
Using test values x = −3, x = 0 and x = 3, we can find the sign of
h0 (x).
Thus (−2, − 32 32
3 ) is a relative minima and (2, 3 ) is a horizontal point of
inflection.

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3.2 Maxima and Minima

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3.2 Maxima and Minima
Question
How to find the maximum value and minimum value of f on [a, b]?

Answer
Step 1: Solve the equation f 0 (x) = 0 to obtain roots x1 , x2 , . . . , xk .
Step 2: Compute f (x1 ), f (x2 ), . . . , f (xk ) and f (a), f (b) to find the
maximum value and minimum value of f .

Example 3
The total revenue in dollars for a firm is given by

R(x) = 6000x − 15x 2 − x 3

where x is the number of units sold per day. If only 60 units can be sold
per day, find the number of units that must be sold to maximize
revenue. Find the maximum revenue.
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3.2 Maxima and Minima
Question
How to find the maximum value and minimum value of f on [a, b]?

Answer
Step 1: Solve the equation f 0 (x) = 0 to obtain roots x1 , x2 , . . . , xk .
Step 2: Compute f (x1 ), f (x2 ), . . . , f (xk ) and f (a), f (b) to find the
maximum value and minimum value of f .

Example 3
The total revenue in dollars for a firm is given by

R(x) = 6000x − 15x 2 − x 3

where x is the number of units sold per day. If only 60 units can be sold
per day, find the number of units that must be sold to maximize
revenue. Find the maximum revenue.
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3.2 Maxima and Minima
Question
How to find the maximum value and minimum value of f on [a, b]?

Answer
Step 1: Solve the equation f 0 (x) = 0 to obtain roots x1 , x2 , . . . , xk .
Step 2: Compute f (x1 ), f (x2 ), . . . , f (xk ) and f (a), f (b) to find the
maximum value and minimum value of f .

Example 3
The total revenue in dollars for a firm is given by

R(x) = 6000x − 15x 2 − x 3

where x is the number of units sold per day. If only 60 units can be sold
per day, find the number of units that must be sold to maximize
revenue. Find the maximum revenue.
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Solution
We want to find the maximum value of R(x) on [0, 60]. Note that

R 0 (x) = 6000 − 30x − 3x 2 = 3(40 − x)(x + 50).

We have R 0 (x) = 0 ⇔ x = 40 or x = −50. Since x ≥ 0, we have

R(0) = 0, R(40) = 152, 000 and R(60) = 90, 000

we conclude that the maximum revenue is $152, 000 when x = 40

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3.2 Maxima and Minima

Example 4
Suppose that the weekly sales revenue of a film company is modelled
by
50t
R(t) = 2
t + 196
where R is in millions of dollars and t is the number of weeks.
Determine when the weekly revenue is maximized and find the
maximum revenue.

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3.2 Maxima and Minima

Hint
* Solution 1: Find R 0 and solve R 0 (t) = 0.

* Solution 2: Note that (t − 14)2 ≥ 0 ⇔ t 2 + 196 ≥ 28t. Thus

50t 50t
R(t) = ≤ ≈ 1.79
t2 + 196 28t
Thus the maximum revenue is approximately 1.79 millions USD after
t = 14 weeks.

Example 5
A firm in a competitive market must sell its product for $300 per unit.
The cost per unit (per day) is 90 + x, where x represents the number
of units sold per day. Find the maximum profit.

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3.2 Maxima and Minima

Hint
* Solution 1: Find R 0 and solve R 0 (t) = 0.

* Solution 2: Note that (t − 14)2 ≥ 0 ⇔ t 2 + 196 ≥ 28t. Thus

50t 50t
R(t) = ≤ ≈ 1.79
t2 + 196 28t
Thus the maximum revenue is approximately 1.79 millions USD after
t = 14 weeks.

Example 5
A firm in a competitive market must sell its product for $300 per unit.
The cost per unit (per day) is 90 + x, where x represents the number
of units sold per day. Find the maximum profit.

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Solution
The profit is

P(x) = R(x) − C(x) = 300x − (90 + x)x = 210x − x 2 .

Thus
P 0 (x) = 210 − 2x and P 0 (x) = 0 ⇔ x = 105.
Since P 00 (x) = −2 < 0, the maximum profit is P(105) = 11, 025 USD
when x = 105 units are sold per day.

Exercises
Page 652: 15–19, Page 662: 3–5, 8–10, 15, 18–19, 23, 27–29 . . .

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3.2 Maxima and Minima

Solution
The profit is

P(x) = R(x) − C(x) = 300x − (90 + x)x = 210x − x 2 .

Thus
P 0 (x) = 210 − 2x and P 0 (x) = 0 ⇔ x = 105.
Since P 00 (x) = −2 < 0, the maximum profit is P(105) = 11, 025 USD
when x = 105 units are sold per day.

Exercises
Page 652: 15–19, Page 662: 3–5, 8–10, 15, 18–19, 23, 27–29 . . .

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