Math 211 Business Calculus Applications of Derivatives

– p. 1/34
Math 211 Business Calculus
Applications of Derivatives
Professor Richard Blecksmith
richard@math.niu.edu
Dept. of Mathematical Sciences
Northern Illinois University
http://math.niu.edu/∼richard/Math211

– p. 2/34
Asymptotes
A nonvertical line with equation y = mx+b is called
an asymptote of the graph of y = f(x)

– p. 2/34
Asymptotes
A nonvertical line with equation y = mx+b is called
an asymptote of the graph of y = f(x) if the
differencef(x)−(mx+b) tends to 0

– p. 2/34
Asymptotes
A nonvertical line with equation y = mx+b is called
an asymptote of the graph of y = f(x) if the
differencef(x)−(mx+b) tends to 0 asxtakes on
arbitrarily large positive values

– p. 2/34
Asymptotes
A nonvertical line with equation y = mx+b is called
an asymptote of the graph of y = f(x) if the
differencef(x)−(mx+b) tends to 0 asxtakes on
arbitrarily large positive values or arbitrarily large
negative values.

– p. 2/34
Asymptotes
A nonvertical line with equation y = mx+b is called
an asymptote of the graph of y = f(x) if the
differencef(x)−(mx+b) tends to 0 asxtakes on
arbitrarily large positive values or arbitrarily large
negative values.
Ifm = 0 theny = b is called a horizontal asymptote.

– p. 2/34
Asymptotes
A nonvertical line with equation y = mx+b is called
an asymptote of the graph of y = f(x) if the
differencef(x)−(mx+b) tends to 0 asxtakes on
arbitrarily large positive values or arbitrarily large
negative values.
Ifm = 0 theny = b is called a horizontal asymptote.
Otherwisey = mx+b is called a slant asymptote.

– p. 2/34
Asymptotes
A nonvertical line with equation y = mx+b is called
an asymptote of the graph of y = f(x) if the
differencef(x)−(mx+b) tends to 0 asxtakes on
arbitrarily large positive values or arbitrarily large
negative values.
Ifm = 0 theny = b is called a horizontal asymptote.
Otherwisey = mx+b is called a slant asymptote.
A vertical linex = a is called a vertical asymptote

– p. 2/34
Asymptotes
A nonvertical line with equation y = mx+b is called
an asymptote of the graph of y = f(x) if the
differencef(x)−(mx+b) tends to 0 asxtakes on
arbitrarily large positive values or arbitrarily large
negative values.
Ifm = 0 theny = b is called a horizontal asymptote.
Otherwisey = mx+b is called a slant asymptote.
A vertical linex = a is called a vertical asymptote if
|f(x)| takes arbitrarily large values

– p. 2/34
Asymptotes
A nonvertical line with equation y = mx+b is called
an asymptote of the graph of y = f(x) if the
differencef(x)−(mx+b) tends to 0 asxtakes on
arbitrarily large positive values or arbitrarily large
negative values.
Ifm = 0 theny = b is called a horizontal asymptote.
Otherwisey = mx+b is called a slant asymptote.
A vertical linex = a is called a vertical asymptote if
|f(x)| takes arbitrarily large values asx → a from the
right or the left.

– p. 3/34
Example
Lety = f(x) = x+ 1 x

Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2 – p. 3/34

Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2
f ′′ (x) = 2x −3 = 2 x 3 – p. 3/34

Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2
f ′′ (x) = 2x −3 = 2 x 3
Since lim
x→0
+ f(x) – p. 3/34

– p. 3/34
Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2
f ′′ (x) = 2x −3 = 2 x 3
Since lim
x→0
+
f(x) = lim
x→0 +x+ 1 x

Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2
f ′′ (x) = 2x −3 = 2 x 3
Since lim
x→0
+
f(x) = lim
x→0 +x+ 1 x = ∞ – p. 3/34

– p. 3/34
Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2
f ′′ (x) = 2x −3 = 2 x 3
Since lim f(x) = lim
1
x→0
+
x→0 +x+ x = ∞
x = 0 (they-axis) is a vertical asymptote.

– p. 3/34
Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2
f ′′ (x) = 2x −3 = 2 x 3
Since lim f(x) = lim
1
x→0
+
x→0 +x+ x = ∞
x = 0 (they-axis) is a vertical asymptote.
Since lim
x→∞
f(x)−x

– p. 3/34
Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2
f ′′ (x) = 2x −3 = 2 x 3
Since lim f(x) = lim
1
x→0
+
x→0 +x+ x = ∞
x = 0 (they-axis) is a vertical asymptote.
1
Since lim f(x)−x = lim
x→∞ x→∞ x

Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2
f ′′ (x) = 2x −3 = 2 x 3
Since lim f(x) = lim
1
x→0
+
x→0 +x+ x = ∞
x = 0 (they-axis) is a vertical asymptote.
1
Since lim f(x)−x = lim
x→∞ x→∞ x = 0 – p. 3/34

– p. 3/34
Example
Lety = f(x) = x+ 1 x
f ′ (x) = 1− 1 x 2 = 1−x −2
f ′′ (x) = 2x −3 = 2 x 3
Since lim f(x) = lim
1
x→0
+
x→0 +x+ x = ∞
x = 0 (they-axis) is a vertical asymptote.
Since lim f(x)−x = lim
x→∞ x = 0
the liney = x is a slant asymptote.
x→∞
1

Graph of x+1/x
– p. 4/34

– p. 4/34
Graph of x+1/x
♣
(1,2)

Graph of x+1/x
1
lim
x→0 +
x = – p. 4/34
♣
(1,2)

Graph of x+1/x
1
lim
x→0 +
x = ∞ – p. 4/34
♣
(1,2)

Graph of x+1/x
1
lim
x→0 +
x = ∞ – p. 4/34
♣
(1,2)

– p. 4/34
Graph of x+1/x
1
lim
x→0 + x = ∞
y-axis
vertical asymptote
♣
(1,2)

Graph of x+1/x
1
lim
x→0 + x = ∞
y-axis
vertical asymptote
(1,2)
♣
– p. 4/34

– p. 4/34
Graph of x+1/x
1
lim
x→0 + x = ∞
y-axis
vertical asymptote
(1,2)
♣
y = x
slant asymptote

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the vertical asymptotes of f(x)

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the vertical asymptotes of f(x)
Solution: Factor the denominator:

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the vertical asymptotes of f(x)
Solution: Factor the denominator:
x 3 −2x 2 −3x =

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the vertical asymptotes of f(x)
Solution: Factor the denominator:
x 3 −2x 2 −3x = x(x 2 −2x−3) =

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the vertical asymptotes of f(x)
Solution: Factor the denominator:
x 3 −2x 2 −3x = x(x 2 −2x−3) = x(x−3)(x+1)

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the vertical asymptotes of f(x)
Solution: Factor the denominator:
x 3 −2x 2 −3x = x(x 2 −2x−3) = x(x−3)(x+1)
So vertical asymptotes are:
x = 0,

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the vertical asymptotes of f(x)
Solution: Factor the denominator:
x 3 −2x 2 −3x = x(x 2 −2x−3) = x(x−3)(x+1)
So vertical asymptotes are:
x = 0,x = 3,

– p. 5/34
Vertical Asymptotes
To find vertical asymptotes
look for values ofxwhich make the denominator zero
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the vertical asymptotes of f(x)
Solution: Factor the denominator:
x 3 −2x 2 −3x = x(x 2 −2x−3) = x(x−3)(x+1)
So vertical asymptotes are:
x = 0,x = 3, x = −1

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)
Solution: Compute the limit

Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)
Solution: Compute the limit
lim
x→∞ f(x) – p. 6/34

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)
Solution: Compute the limit
5x 3 +4x−8
lim f(x) = lim
x→∞ x→∞x 3 −2x 2 −3x

Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)
Solution: Compute the limit
5x 3 +4x−8 1/x 3
lim f(x) = lim
x→∞ x→∞x 3 −2x 2 −3x 1/x 3 – p. 6/34

Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)
Solution: Compute the limit
5x 3 +4x−8 1/x 3
lim f(x) = lim
x→∞ x→∞x 3 −2x 2 −3x 1/x 3
= lim
x→∞
5+4/x 2 −8/x 3
1−2/x−3/x 2 – p. 6/34

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)
Solution: Compute the limit
5x 3 +4x−8 1/x 3
lim f(x) = lim
x→∞ x→∞x 3 −2x 2 −3x 1/x 3
5+4/x 2 −8/x 3
= lim
x→∞ 1−2/x−3/x = 5+0−0
2 1−0−0

Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)
Solution: Compute the limit
5x 3 +4x−8 1/x 3
lim f(x) = lim
x→∞ x→∞x 3 −2x 2 −3x 1/x 3
= lim
x→∞
5+4/x 2 −8/x 3
1−2/x−3/x 2 = 5+0−0
1−0−0 = 5 – p. 6/34

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)
Solution: Compute the limit
5x 3 +4x−8 1/x 3
lim f(x) = lim
x→∞ x→∞x 3 −2x 2 −3x 1/x 3
5+4/x 2 −8/x 3
= lim
x→∞ 1−2/x−3/x = 5+0−0
2 1−0−0 = 5
So the horizontal asymptote is:

– p. 6/34
Horizontal Asymptotes
To find horizontal asymptotes
compute the limit asx → ∞ (or−∞)
Example: Consider the function
f(x) = 5x3 +4x−8
x 3 −2x 2 −3x
Find the horizontal asymptotes of f(x)
Solution: Compute the limit
5x 3 +4x−8 1/x 3
lim f(x) = lim
x→∞ x→∞x 3 −2x 2 −3x 1/x 3
5+4/x 2 −8/x 3
= lim
x→∞ 1−2/x−3/x = 5+0−0
2 1−0−0 = 5
So the horizontal asymptote is: y = 5

– p. 7/34
Computing Limits
When finding the limit asx → ∞ of fractions of
polynomials, the rule is

Computing Limits
When finding the limit asx → ∞ of fractions of
polynomials, the rule is
Multiply the fraction by 1/xm
1/x m, where – p. 7/34

– p. 7/34
Computing Limits
When finding the limit asx → ∞ of fractions of
polynomials, the rule is
Multiply the fraction by 1/xm
1/xm, where
m is the degree of the polynomial in the denominator.

Computing Limits
When finding the limit asx → ∞ of fractions of
polynomials, the rule is
Multiply the fraction by 1/xm
1/xm, where
m is the degree of the polynomial in the denominator.
and use the fact that for any positive integer k,
lim
x→∞
1
x k = 0 – p. 7/34

– p. 7/34
Computing Limits
When finding the limit asx → ∞ of fractions of
polynomials, the rule is
Multiply the fraction by 1/xm
1/xm, where
m is the degree of the polynomial in the denominator.
and use the fact that for any positive integer k,
lim
x→∞
1
x k = 0
The same idea applies when computing limits where
x → −∞.

– p. 8/34
Example 1.
Compute the limit

– p. 8/34
Example 1.
Compute the limit
x 2 +8
= lim
x→∞ x 3 −4x

Example 1.
Compute the limit
x 2 +8 1/x 3
= lim
x→∞ x 3 −4x
1/x 3 – p. 8/34

Example 1.
Compute the limit
x 2 +8 1/x 3
= lim
x→∞ x 3 −4x 1/x 3
= lim
x→∞
1/x−8/x 3
1−4/x 2 – p. 8/34

– p. 8/34
Example 1.
Compute the limit
x 2 +8
= lim
x→∞ x 3 −4x
1/x 3
1/x 3
= lim
x→∞
1/x−8/x 3
1−4/x 2
= 0−0
1−0

– p. 8/34
Example 1.
Compute the limit
x 2 +8
= lim
x→∞ x 3 −4x
1/x 3
1/x 3
= lim
x→∞
1/x−8/x 3
1−4/x 2
= 0−0
1−0
= 0

– p. 9/34
Second example
This limit may end up being infinity

– p. 9/34
Second example
This limit may end up being infinity
Example: Compute the limit

– p. 9/34
Second example
This limit may end up being infinity
Example: Compute the limit
2x 4 +x
= lim
x→∞ x 2 +5

Second example
This limit may end up being infinity
Example: Compute the limit
2x 4 +x 1/x 2
= lim
x→∞ x 2 +5
1/x 2 – p. 9/34

Second example
This limit may end up being infinity
Example: Compute the limit
2x 4 +x 1/x 2
= lim
x→∞ x 2 +5 1/x 2
= lim
x→∞
2x 2 +1/x
1+5/x 2 – p. 9/34

– p. 9/34
Second example
This limit may end up being infinity
Example: Compute the limit
2x 4 +x 1/x 2
= lim
x→∞ x 2 +5 1/x 2
2x 2 +1/x
= lim
x→∞ 1+5/x 2
= ∞+0
1+0

– p. 9/34
Second example
This limit may end up being infinity
Example: Compute the limit
2x 4 +x 1/x 2
= lim
x→∞ x 2 +5 1/x 2
= lim
x→∞
2x 2 +1/x
1+5/x 2
= ∞+0
1+0
= ∞

– p. 10/34
Graphs with asymptotes
Sketch the graph ofy = f(x) = x−2
x+1

– p. 10/34
Graphs with asymptotes
Sketch the graph ofy = f(x) = x−2
x+1
We need the derivatives

Graphs with asymptotes
Sketch the graph ofy = f(x) = x−2
x+1
We need the derivatives
By the quotient rule
f ′ (x) = (1)(x+1)−(x−2)(1)
(x+1) 2 – p. 10/34

Graphs with asymptotes
Sketch the graph ofy = f(x) = x−2
x+1
We need the derivatives
By the quotient rule
f ′ (x) = (1)(x+1)−(x−2)(1)
(x+1) 2
= x+1−x+2
(x+1) 2 – p. 10/34

Graphs with asymptotes
Sketch the graph ofy = f(x) = x−2
x+1
We need the derivatives
By the quotient rule
f ′ (x) = (1)(x+1)−(x−2)(1)
(x+1) 2
= x+1−x+2
(x+1) 2
3
=
(x+1) 2 – p. 10/34

Graphs with asymptotes
Sketch the graph ofy = f(x) = x−2
x+1
We need the derivatives
By the quotient rule
f ′ (x) = (1)(x+1)−(x−2)(1)
(x+1) 2
= x+1−x+2
(x+1) 2
=
3
(x+1) 2
By the power rule
f ′′ (x) = −6(x+1) −3 – p. 10/34

– p. 11/34
Graphs continued
Since the first derivative f ′ (x) =
positive,
3
is always
(x+1)
2

– p. 11/34
Graphs continued
Since the first derivative f ′ 3
(x) = is always
(x+1)
2
positive, the function is always increasing.

Graphs continued
Since the first derivative f ′ 3
(x) = is always
(x+1)
2
positive, the function is always increasing.
Since the second derivative f ′′ (x) = − 6
(x+1) is 3 – p. 11/34

Graphs continued
Since the first derivative f ′ 3
(x) = is always
(x+1)
2
positive, the function is always increasing.
Since the second derivative f ′′ (x) = − 6
(x+1) is
{ 3 negative ifx > −1
positive ifx < −1 , – p. 11/34

Graphs continued
Since the first derivative f ′ 3
(x) = is always
(x+1)
2
positive, the function is always increasing.
Since the second derivative f ′′ (x) = − 6
(x+1) is
{ 3 negative ifx > −1
positive ifx < −1 ,
{ concave down if x > −1
the graph is
concave up if x < −1 , – p. 11/34

– p. 11/34
Graphs continued
Since the first derivative f ′ 3
(x) = is always
(x+1)
2
positive, the function is always increasing.
Since the second derivative f ′′ (x) = − 6
(x+1) is
{ 3 negative ifx > −1
positive ifx < −1 ,
{ concave down if x > −1
the graph is
concave up if x < −1 ,
There are no critical points and

– p. 11/34
Graphs continued
Since the first derivative f ′ 3
(x) = is always
(x+1)
2
positive, the function is always increasing.
Since the second derivative f ′′ (x) = − 6
(x+1) is
{ 3 negative ifx > −1
positive ifx < −1 ,
{ concave down if x > −1
the graph is
concave up if x < −1 ,
There are no critical points and no inflection points.

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1

Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
Note that
lim
x→−1 +f(x) – p. 12/34

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1

Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1 = −3
0 + – p. 12/34

Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1 = −3
0 = −∞ + – p. 12/34

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1 = −3
0 = −∞ +
To find the horizontal asymptote(s), compute the limit

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1 = −3
0 = −∞ +
To find the horizontal asymptote(s), compute the limit
x−2
= lim
x→∞ x+1

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1 = −3
0 = −∞ +
To find the horizontal asymptote(s), compute the limit
x−2 1/x
= lim
x→∞ x+1 1/x

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1 = −3
0 = −∞ +
To find the horizontal asymptote(s), compute the limit
x−2 1/x
= lim
x→∞ x+1 1/x = lim 1−2/x
x→∞ 1+1/x

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1 = −3
0 = −∞ +
To find the horizontal asymptote(s), compute the limit
x−2 1/x
= lim
x→∞ x+1 1/x = lim 1−2/x
x→∞ 1+1/x = 1−0
1+0

Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1 = −3
0 = −∞ +
To find the horizontal asymptote(s), compute the limit
x−2 1/x
= lim
x→∞ x+1 1/x = lim 1−2/x
x→∞ 1+1/x = 1−0
1+0 = 1 – p. 12/34

– p. 12/34
Find the Asymptotes
For y = f(x) = x−2
x+1
Vertical asymptotes occur when the denominator
x+1 = 0
or whenx = −1
x−2
Note that lim
x→−1 +f(x)
= lim
x→−1 + x+1 = −3
0 = −∞ +
To find the horizontal asymptote(s), compute the limit
= lim
x→∞
x−2
1/x
1/x = lim
x→∞
x+1
Vertical asymptote: x = −1
Horizontal asymptote: y = 1
1−2/x
1+1/x = 1−0
1+0 = 1

Graph of (x−2)/(x+1)
– p. 13/34

Graph of (x−2)/(x+1)
– p. 13/34

– p. 13/34
Graph of (x−2)/(x+1)
x = −1

– p. 13/34
Graph of (x−2)/(x+1)
x = −1
vertical asymptote

– p. 13/34
Graph of (x−2)/(x+1)
x = −1
vertical asymptote
y = 1

– p. 13/34
Graph of (x−2)/(x+1)
x = −1
vertical asymptote
y = 1
horizontal asymptote

– p. 13/34
Graph of (x−2)/(x+1)
x = −1
vertical asymptote
y = 1
horizontal asymptote

– p. 13/34
Graph of (x−2)/(x+1)
x = −1
vertical asymptote
y = 1
horizontal asymptote

– p. 14/34
Business Applications
Three Inportant functions:
• Revenue R(x)

– p. 14/34
Business Applications
Three Inportant functions:
• Revenue R(x)
• Profit P(x)

– p. 14/34
Business Applications
Three Inportant functions:
• Revenue R(x)
• Profit P(x)
• Cost C(x)

– p. 14/34
Business Applications
Three Inportant functions:
• Revenue R(x)
• Profit P(x)
• Cost C(x)
Their derivatives are called

– p. 14/34
Business Applications
Three Inportant functions:
• Revenue R(x)
• Profit P(x)
• Cost C(x)
Their derivatives are called
• Marginal Revenue =R ′ (x)

– p. 14/34
Business Applications
Three Inportant functions:
• Revenue R(x)
• Profit P(x)
• Cost C(x)
Their derivatives are called
• Marginal Revenue =R ′ (x)
• Marginal Profit =P ′ (x)

– p. 14/34
Business Applications
Three Inportant functions:
• Revenue R(x)
• Profit P(x)
• Cost C(x)
Their derivatives are called
• Marginal Revenue =R ′ (x)
• Marginal Profit =P ′ (x)
• Marginal Cost =C ′ (x)

– p. 15/34
Business Applications
The goal of business management is to

– p. 15/34
Business Applications
The goal of business management is to
• Maximize Profit

– p. 15/34
Business Applications
The goal of business management is to
• Maximize Profit
• Maximize Revenue

– p. 15/34
Business Applications
The goal of business management is to
• Maximize Profit
• Maximize Revenue
• Minimize Cost

– p. 15/34
Business Applications
The goal of business management is to
• Maximize Profit
• Maximize Revenue
• Minimize Cost
Since maximums and minimums occur when the
derivative is zero, our goal is to find values for which

– p. 15/34
Business Applications
The goal of business management is to
• Maximize Profit
• Maximize Revenue
• Minimize Cost
Since maximums and minimums occur when the
derivative is zero, our goal is to find values for which
marginal revenue, marginal profit, marginal cost = 0.

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by
R = −0.05p 2 +0.98p+18

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by
R = −0.05p 2 +0.98p+18
What price maximizes revenue

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by
R = −0.05p 2 +0.98p+18
What price maximizes revenue
Solution:

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by
R = −0.05p 2 +0.98p+18
What price maximizes revenue
Solution:
R ′ (p) = −0.1p+0.98 = 0

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by
R = −0.05p 2 +0.98p+18
What price maximizes revenue
Solution:
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ 0.1p = 0.98

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by
R = −0.05p 2 +0.98p+18
What price maximizes revenue
Solution:
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ 0.1p = 0.98
⇐⇒ p = 9.8

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by
R = −0.05p 2 +0.98p+18
What price maximizes revenue
Solution:
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ 0.1p = 0.98
⇐⇒ p = 9.8
Question: How do we know thatp = 9.8 gives a
maximum (and not a minimum)

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by
R = −0.05p 2 +0.98p+18
What price maximizes revenue
Solution:
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ 0.1p = 0.98
⇐⇒ p = 9.8
Question: How do we know thatp = 9.8 gives a
maximum (and not a minimum)
We are not usually interested in minimizing revenue

– p. 16/34
Maximize Revenue
If the price for a product is p, the revenue (in
thousands of dollars) is approximated by
R = −0.05p 2 +0.98p+18
What price maximizes revenue
Solution:
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ 0.1p = 0.98
⇐⇒ p = 9.8
Question: How do we know thatp = 9.8 gives a
maximum (and not a minimum)
We are not usually interested in minimizing revenue
unless you are hoping to get government bailout
money.

– p. 17/34
Max versus Min
R = −0.05p 2 +0.98p+18

– p. 17/34
Max versus Min
R = −0.05p 2 +0.98p+18
R ′ (p) = −0.1p+0.98 = 0

– p. 17/34
Max versus Min
R = −0.05p 2 +0.98p+18
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ p = 9.8

– p. 17/34
Max versus Min
R = −0.05p 2 +0.98p+18
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ p = 9.8
To tell whether this critical point is a max or a min,

– p. 17/34
Max versus Min
R = −0.05p 2 +0.98p+18
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ p = 9.8
To tell whether this critical point is a max or a min,
Use the second derivative test

– p. 17/34
Max versus Min
R = −0.05p 2 +0.98p+18
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ p = 9.8
To tell whether this critical point is a max or a min,
Use the second derivative test
R ′′ (p) = −0.1

– p. 17/34
Max versus Min
R = −0.05p 2 +0.98p+18
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ p = 9.8
To tell whether this critical point is a max or a min,
Use the second derivative test
R ′′ (p) = −0.1
And a negative second derivative implies

– p. 17/34
Max versus Min
R = −0.05p 2 +0.98p+18
R ′ (p) = −0.1p+0.98 = 0
⇐⇒ p = 9.8
To tell whether this critical point is a max or a min,
Use the second derivative test
R ′′ (p) = −0.1
And a negative second derivative implies
p = 9.8 is a Maximum

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500
What value of x maximizes revenue

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500
What value of x maximizes revenue
Solution:

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500
What value of x maximizes revenue
Solution:
R ′ (x) = 3x 2 −42x+120

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500
What value of x maximizes revenue
Solution:
R ′ (x) = 3x 2 −42x+120
= 3(x 2 −14x+40)

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500
What value of x maximizes revenue
Solution:
R ′ (x) = 3x 2 −42x+120
= 3(x 2 −14x+40)
= 3(x−4)(x−10)

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500
What value of x maximizes revenue
Solution:
R ′ (x) = 3x 2 −42x+120
= 3(x 2 −14x+40)
= 3(x−4)(x−10)
The critical points are

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500
What value of x maximizes revenue
Solution:
R ′ (x) = 3x 2 −42x+120
= 3(x 2 −14x+40)
= 3(x−4)(x−10)
The critical points are
x = 4 and

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500
What value of x maximizes revenue
Solution:
R ′ (x) = 3x 2 −42x+120
= 3(x 2 −14x+40)
= 3(x−4)(x−10)
The critical points are
x = 4 andx = 10

– p. 18/34
Maximize Revenue
The revenue for sellingxthousand units of a product
can be approximated by
R = x 3 −21x 2 +120x+500
What value of x maximizes revenue
Solution:
R ′ (x) = 3x 2 −42x+120
= 3(x 2 −14x+40)
= 3(x−4)(x−10)
The critical points are
x = 4 andx = 10
Question: How do we know whetherx = 4 and
x = 10 gives a maximum (and not a minimum)

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x)

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl
4

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl
4 708

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl
4 708 24−42 = −18

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl
4 708 24−42 = −18 Max

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl
4 708 24−42 = −18 Max
10

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl
4 708 24−42 = −18 Max
10 600

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl
4 708 24−42 = −18 Max
10 600 60−42 = 18

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl
4 708 24−42 = −18 Max
10 600 60−42 = 18 Min

– p. 19/34
Max versus Min
R = x 3 −21x 2 +120x+500
R ′ (x) = 3x 2 −42x+120
Critical points: 4 and 10
R ′′ (x) = 6x−42
Use the second derivative test
x R(x) R ′′ (x) = 6x−42 Concl
4 708 24−42 = −18 Max
10 600 60−42 = 18 Min
Revenue is maximum atx = 4

– p. 20/34
Maximize Profit
As a general rule

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)
By calculus we know

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)
By calculus we know
P ′ (x) = R ′ (x)−C ′ (x)

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)
By calculus we know
P ′ (x) = R ′ (x)−C ′ (x)
and

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)
By calculus we know
P ′ (x) = R ′ (x)−C ′ (x)
and
P ′′ (x) = R ′′ (x)−C ′′ (x)

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)
By calculus we know
P ′ (x) = R ′ (x)−C ′ (x)
and
P ′′ (x) = R ′′ (x)−C ′′ (x)
ThusP ′ (x) = 0 if and only if

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)
By calculus we know
P ′ (x) = R ′ (x)−C ′ (x)
and
P ′′ (x) = R ′′ (x)−C ′′ (x)
ThusP ′ (x) = 0 if and only if
R ′ (x) = C ′ (x)

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)
By calculus we know
P ′ (x) = R ′ (x)−C ′ (x)
and
P ′′ (x) = R ′′ (x)−C ′′ (x)
ThusP ′ (x) = 0 if and only if
R ′ (x) = C ′ (x)
This critical point is a maximum

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)
By calculus we know
P ′ (x) = R ′ (x)−C ′ (x)
and
P ′′ (x) = R ′′ (x)−C ′′ (x)
ThusP ′ (x) = 0 if and only if
R ′ (x) = C ′ (x)
This critical point is a maximum
if and only ifP ′′ (x) < 0

– p. 20/34
Maximize Profit
As a general rule
P(x) = R(x)−C(x)
By calculus we know
P ′ (x) = R ′ (x)−C ′ (x)
and
P ′′ (x) = R ′′ (x)−C ′′ (x)
ThusP ′ (x) = 0 if and only if
R ′ (x) = C ′ (x)
This critical point is a maximum
if and only ifP ′′ (x) < 0
if and only if
R ′′ (x) < C ′′ (x)

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20
What level of sales maximizes profit

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20
What level of sales maximizes profit
Solution:

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20
What level of sales maximizes profit
Solution: R ′ (x) = −0.1x+2

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20
What level of sales maximizes profit
Solution: R ′ (x) = −0.1x+2 and C ′ (x) = 1.5

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20
What level of sales maximizes profit
Solution: R ′ (x) = −0.1x+2 and C ′ (x) = 1.5
SettingR ′ (x) = C ′ (x) gives

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20
What level of sales maximizes profit
Solution: R ′ (x) = −0.1x+2 and C ′ (x) = 1.5
SettingR ′ (x) = C ′ (x) gives
−0.1x+2 = 1.5

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20
What level of sales maximizes profit
Solution: R ′ (x) = −0.1x+2 and C ′ (x) = 1.5
SettingR ′ (x) = C ′ (x) gives
−0.1x+2 = 1.5 ⇐⇒ −0.1x = −.5

– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20
What level of sales maximizes profit
Solution: R ′ (x) = −0.1x+2 and C ′ (x) = 1.5
SettingR ′ (x) = C ′ (x) gives
−0.1x+2 = 1.5 ⇐⇒ −0.1x = −.5 ⇐⇒ x = 5

What level of sales maximizes profit
Solution: R ′ (x) = −0.1x+2 and C ′ (x) = 1.5
SettingR ′ (x) = C ′ (x) gives
−0.1x+2 = 1.5 ⇐⇒ −0.1x = −.5 ⇐⇒ x = 5
This is a maximum becauseR ′′ (x) = −.1 is less than
C ′′ (x) = 0
– p. 21/34
Maximize Profit Example
The revenue for sellingxthousand units of a product
is approximated by
R(x) = −0.05x 2 +2x+60
and the cost for producingxthousand units is
approximated by
C(x) = 1.5x+20

– p. 22/34
Ticket Sales Problem
• A movie multiplex sells tickets for $8.

– p. 22/34
Ticket Sales Problem
• A movie multiplex sells tickets for $8.
• On Friday evenings, it averages 720 tickets sold.

– p. 22/34
Ticket Sales Problem
• A movie multiplex sells tickets for $8.
• On Friday evenings, it averages 720 tickets sold.
• Each patron spends an average of $2 on
concessions.

– p. 22/34
Ticket Sales Problem
• A movie multiplex sells tickets for $8.
• On Friday evenings, it averages 720 tickets sold.
• Each patron spends an average of $2 on
concessions.
• A survey shows that for each $0.25 drop in ticket
price, 20 more people will buy tickets on Friday
evenings.

– p. 22/34
Ticket Sales Problem
• A movie multiplex sells tickets for $8.
• On Friday evenings, it averages 720 tickets sold.
• Each patron spends an average of $2 on
concessions.
• A survey shows that for each $0.25 drop in ticket
price, 20 more people will buy tickets on Friday
evenings.
• Assuming that these 20 additional customers also
average $2 in concession sales,

– p. 22/34
Ticket Sales Problem
• A movie multiplex sells tickets for $8.
• On Friday evenings, it averages 720 tickets sold.
• Each patron spends an average of $2 on
concessions.
• A survey shows that for each $0.25 drop in ticket
price, 20 more people will buy tickets on Friday
evenings.
• Assuming that these 20 additional customers also
average $2 in concession sales,
• what ticket price will maximize revenue

– p. 23/34
Ticket Sales Setup
Letx=the number of 25 cent reductions in ticket
price.

– p. 23/34
Ticket Sales Setup
Letx=the number of 25 cent reductions in ticket
price.
The price for a ticket isp = 8.00−0.25x

– p. 23/34
Ticket Sales Setup
Letx=the number of 25 cent reductions in ticket
price.
The price for a ticket isp = 8.00−0.25x
The number of customers isN = 720+20x

– p. 23/34
Ticket Sales Setup
Letx=the number of 25 cent reductions in ticket
price.
The price for a ticket isp = 8.00−0.25x
The number of customers isN = 720+20x
The revenue from movie sales is
pN = (8.00−0.25x)(720+20x)

– p. 23/34
Ticket Sales Setup
Letx=the number of 25 cent reductions in ticket
price.
The price for a ticket isp = 8.00−0.25x
The number of customers isN = 720+20x
The revenue from movie sales is
pN = (8.00−0.25x)(720+20x)
The revenue from concessions is
2N = 2(720+20x)

– p. 23/34
Ticket Sales Setup
Letx=the number of 25 cent reductions in ticket
price.
The price for a ticket isp = 8.00−0.25x
The number of customers isN = 720+20x
The revenue from movie sales is
pN = (8.00−0.25x)(720+20x)
The revenue from concessions is
2N = 2(720+20x)
The total revenue is
R(x) = (8.00−0.25x)(720+20x)+2(720+20x)

– p. 23/34
Ticket Sales Setup
Letx=the number of 25 cent reductions in ticket
price.
The price for a ticket isp = 8.00−0.25x
The number of customers isN = 720+20x
The revenue from movie sales is
pN = (8.00−0.25x)(720+20x)
The revenue from concessions is
2N = 2(720+20x)
The total revenue is
R(x) = (8.00−0.25x)(720+20x)+2(720+20x)
= (10−0.25x)(720+20x)

– p. 24/34
Ticket Sales Solution
R(x) = (10−0.25x)(720+20x)

– p. 24/34
Ticket Sales Solution
R(x) = (10−0.25x)(720+20x)
By the product rule,

– p. 24/34
Ticket Sales Solution
R(x) = (10−0.25x)(720+20x)
By the product rule,
R ′ (x) = (−0.25)(720+20x)+(10−0.25x)(20)

Ticket Sales Solution
R(x) = (10−0.25x)(720+20x)
By the product rule,
R ′ (x) = (−0.25)(720+20x)+(10−0.25x)(20)
= − 1 4 ·720− 1 4 ·20x+10·20−(0.25)(20)x – p. 24/34

– p. 24/34
Ticket Sales Solution
R(x) = (10−0.25x)(720+20x)
By the product rule,
R ′ (x) = (−0.25)(720+20x)+(10−0.25x)(20)
= − 1 4 ·720− 1 4 ·20x+10·20−(0.25)(20)x
= −180−5x+200−5x

– p. 24/34
Ticket Sales Solution
R(x) = (10−0.25x)(720+20x)
By the product rule,
R ′ (x) = (−0.25)(720+20x)+(10−0.25x)(20)
= − 1 4 ·720− 1 4 ·20x+10·20−(0.25)(20)x
= −180−5x+200−5x
= 20−10x

– p. 24/34
Ticket Sales Solution
R(x) = (10−0.25x)(720+20x)
By the product rule,
R ′ (x) = (−0.25)(720+20x)+(10−0.25x)(20)
= − 1 4 ·720− 1 4 ·20x+10·20−(0.25)(20)x
= −180−5x+200−5x
= 20−10x
So we see thatR ′ (x) = 20−10x = 0

– p. 24/34
Ticket Sales Solution
R(x) = (10−0.25x)(720+20x)
By the product rule,
R ′ (x) = (−0.25)(720+20x)+(10−0.25x)(20)
= − 1 4 ·720− 1 4 ·20x+10·20−(0.25)(20)x
= −180−5x+200−5x
= 20−10x
So we see thatR ′ (x) = 20−10x = 0
if and only ifx = 2

– p. 24/34
Ticket Sales Solution
R(x) = (10−0.25x)(720+20x)
By the product rule,
R ′ (x) = (−0.25)(720+20x)+(10−0.25x)(20)
= − 1 4 ·720− 1 4 ·20x+10·20−(0.25)(20)x
= −180−5x+200−5x
= 20−10x
So we see thatR ′ (x) = 20−10x = 0
if and only ifx = 2
The movie multiplex should reduce the ticket price by
50 cents to maximize revenue.

– p. 25/34
Dog Kennel Problem
The owner of a kennel has 90 feet of fencing available
to enclose three pens as shown

– p. 25/34
Dog Kennel Problem
The owner of a kennel has 90 feet of fencing available
to enclose three pens as shown

– p. 25/34
Dog Kennel Problem
The owner of a kennel has 90 feet of fencing available
to enclose three pens as shown

– p. 25/34
Dog Kennel Problem
The owner of a kennel has 90 feet of fencing available
to enclose three pens as shown
Pen 1

– p. 25/34
Dog Kennel Problem
The owner of a kennel has 90 feet of fencing available
to enclose three pens as shown
Pen 1 Pen 2

– p. 25/34
Dog Kennel Problem
The owner of a kennel has 90 feet of fencing available
to enclose three pens as shown
Pen 1 Pen 2 Pen 3

– p. 25/34
Dog Kennel Problem
The owner of a kennel has 90 feet of fencing available
to enclose three pens as shown
Pen 1 Pen 2 Pen 3
What dimensions maximize the total area of the three
pens

– p. 26/34
Dog Kennel Setup
Letx=length of the pen andy = width

– p. 26/34
Dog Kennel Setup
Letx=length of the pen andy = width
We need 2x feet of fencing for the top and bottom

– p. 26/34
Dog Kennel Setup
Letx=length of the pen andy = width
We need 2x feet of fencing for the top and bottom
and4y feet of fencing for sides (including the two red
dividers)

– p. 26/34
Dog Kennel Setup
Letx=length of the pen andy = width
We need 2x feet of fencing for the top and bottom
and4y feet of fencing for sides (including the two red
dividers)
Since only 90 feet of fencing is available,

– p. 26/34
Dog Kennel Setup
Letx=length of the pen andy = width
We need 2x feet of fencing for the top and bottom
and4y feet of fencing for sides (including the two red
dividers)
Since only 90 feet of fencing is available,
2x+4y = 90

– p. 26/34
Dog Kennel Setup
Letx=length of the pen andy = width
We need 2x feet of fencing for the top and bottom
and4y feet of fencing for sides (including the two red
dividers)
Since only 90 feet of fencing is available,
2x+4y = 90
The problem is to maximize the area
A = x·y

– p. 27/34
Dog Kennel Solution
Solve

– p. 27/34
Dog Kennel Solution
Solve
forx:
2x+4y = 90

– p. 27/34
Dog Kennel Solution
Solve
forx:
2x = 90−4y
2x+4y = 90

– p. 27/34
Dog Kennel Solution
Solve
forx:
2x = 90−4y
x = 45−2y
2x+4y = 90

– p. 27/34
Dog Kennel Solution
Solve
2x+4y = 90
forx:
2x = 90−4y
x = 45−2y
Subsitutex = 45−2y intoA = xy

Dog Kennel Solution
Solve
2x+4y = 90
forx:
2x = 90−4y
x = 45−2y
Subsitutex = 45−2y intoA = xy
A = xy = (45−2y)y = 45y −2y 2 – p. 27/34

– p. 27/34
Dog Kennel Solution
Solve
2x+4y = 90
forx:
2x = 90−4y
x = 45−2y
Subsitutex = 45−2y intoA = xy
A = xy = (45−2y)y = 45y −2y 2
Differentiate (with respect toy) and set to 0:

Dog Kennel Solution
Solve
2x+4y = 90
forx:
2x = 90−4y
x = 45−2y
Subsitutex = 45−2y intoA = xy
A = xy = (45−2y)y = 45y −2y 2
Differentiate (with respect toy) and set to 0:
dA
dy = 45−4y = 0 – p. 27/34

– p. 27/34
Dog Kennel Solution
Solve
2x+4y = 90
forx:
2x = 90−4y
x = 45−2y
Subsitutex = 45−2y intoA = xy
A = xy = (45−2y)y = 45y −2y 2
Differentiate (with respect toy) and set to 0:
dA
dy = 45−4y = 0
if and only ify = 45/4 = 11.25

– p. 27/34
Dog Kennel Solution
Solve
2x+4y = 90
forx:
2x = 90−4y
x = 45−2y
Subsitutex = 45−2y intoA = xy
A = xy = (45−2y)y = 45y −2y 2
Differentiate (with respect toy) and set to 0:
dA
dy = 45−4y = 0
if and only ify = 45/4 = 11.25
x = 45−2y = 45−22.5 = 22.5

– p. 27/34
Dog Kennel Solution
Solve
2x+4y = 90
forx:
2x = 90−4y
x = 45−2y
Subsitutex = 45−2y intoA = xy
A = xy = (45−2y)y = 45y −2y 2
Differentiate (with respect toy) and set to 0:
dA
dy = 45−4y = 0
if and only ify = 45/4 = 11.25
x = 45−2y = 45−22.5 = 22.5
The dimensions are 11.25 by 22.5

– p. 28/34
Another Fencing Problem
The owner of a long building wishes to attach a
rectangular fence as shown

– p. 28/34
Another Fencing Problem
The owner of a long building wishes to attach a
rectangular fence as shown

– p. 28/34
Another Fencing Problem
The owner of a long building wishes to attach a
rectangular fence as shown
Building

– p. 28/34
Another Fencing Problem
The owner of a long building wishes to attach a
rectangular fence as shown
Building
4$ y y 4$

– p. 28/34
Another Fencing Problem
The owner of a long building wishes to attach a
rectangular fence as shown
Building
4$ y y 4$
x
5$

– p. 28/34
Another Fencing Problem
The owner of a long building wishes to attach a
rectangular fence as shown
Building
4$ y y 4$
The sides that make up the width cost 4$ per foot
The side that makes up the length costs 5$ per foot
x
5$

– p. 28/34
Another Fencing Problem
The owner of a long building wishes to attach a
rectangular fence as shown
Building
4$ y y 4$
The sides that make up the width cost 4$ per foot
The side that makes up the length costs 5$ per foot
The area must be 810 square feet
x
5$

Another Fencing Problem
The owner of a long building wishes to attach a
rectangular fence as shown
Building
4$ y y 4$
The sides that make up the width cost 4$ per foot
The side that makes up the length costs 5$ per foot
The area must be 810 square feet
What dimensions minimize the cost
x
5$
– p. 28/34

– p. 29/34
Fencing Problem Setup
Letx=length of the pen andy = width

– p. 29/34
Fencing Problem Setup
Letx=length of the pen andy = width
The two side pieces cost4y each
and the one length piece costs 5x

– p. 29/34
Fencing Problem Setup
Letx=length of the pen andy = width
The two side pieces cost4y each
and the one length piece costs 5x
The total cost is

– p. 29/34
Fencing Problem Setup
Letx=length of the pen andy = width
The two side pieces cost4y each
and the one length piece costs 5x
The total cost is
C = 2·4y +5x = 8y +5x

– p. 29/34
Fencing Problem Setup
Letx=length of the pen andy = width
The two side pieces cost4y each
and the one length piece costs 5x
The total cost is
The area isA = xy
C = 2·4y +5x = 8y +5x

– p. 29/34
Fencing Problem Setup
Letx=length of the pen andy = width
The two side pieces cost4y each
and the one length piece costs 5x
The total cost is
C = 2·4y +5x = 8y +5x
The area isA = xy
The problem is to minimize the cost function C

– p. 29/34
Fencing Problem Setup
Letx=length of the pen andy = width
The two side pieces cost4y each
and the one length piece costs 5x
The total cost is
C = 2·4y +5x = 8y +5x
The area isA = xy
The problem is to minimize the cost function C
subject to the constraining equation

– p. 29/34
Fencing Problem Setup
Letx=length of the pen andy = width
The two side pieces cost4y each
and the one length piece costs 5x
The total cost is
C = 2·4y +5x = 8y +5x
The area isA = xy
The problem is to minimize the cost function C
subject to the constraining equation
x·y = 810

– p. 30/34
Fence Solution
Solve

– p. 30/34
Fence Solution
Solve
xy = 810 fory:

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x

Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x = 8· 810
x +5x – p. 30/34

Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x = 8· 810
x
+5x =
6480
x +5x – p. 30/34

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x = 8· 810 6480
+5x =
x x +5x
Differentiate (with respect tox) and set to 0:

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x = 8· 810 6480
+5x =
x x +5x
Differentiate (with respect tox) and set to 0:
dC
dx = −6480 +5 = 0
x 2

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x = 8· 810 6480
+5x =
x x +5x
Differentiate (with respect tox) and set to 0:
dC
dx = −6480 +5 = 0
x 2
if and only if 6480 = 5
x 2

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x = 8· 810 6480
+5x =
x x +5x
Differentiate (with respect tox) and set to 0:
dC
dx = −6480 +5 = 0
x 2
if and only if 6480 = 5
x 2
if and only if5x 2 = 6480

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x = 8· 810 6480
+5x =
x x +5x
Differentiate (with respect tox) and set to 0:
dC
dx = −6480 +5 = 0
x 2
if and only if 6480 = 5
x 2
if and only if5x 2 = 6480
if and only ifx 2 = 1296

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x = 8· 810 6480
+5x =
x x +5x
Differentiate (with respect tox) and set to 0:
dC
dx = −6480 +5 = 0
x 2
if and only if 6480 = 5
x 2
if and only if5x 2 = 6480
if and only ifx 2 = 1296
Taking the square rootx = √ 1296 = 36

– p. 30/34
Fence Solution
Solve xy = 810 fory:
y = 810/x
Substitutey = 810/x into the cost equation
C = 8y +5x = 8· 810 6480
+5x =
x x +5x
Differentiate (with respect tox) and set to 0:
dC
dx = −6480 +5 = 0
x 2
if and only if 6480 = 5
x 2
if and only if5x 2 = 6480
if and only ifx 2 = 1296
Taking the square rootx = √ 1296 = 36
andy = 810/x = 810/36 = 22.5

– p. 31/34
Limit Def Revisited
Recall the limit definition of derivative

– p. 31/34
Limit Def Revisited
Recall the limit definition of derivative
f ′ f(x+h)−f(x)
(x) = lim
h→0 h

– p. 31/34
Limit Def Revisited
Recall the limit definition of derivative
f ′ f(x+h)−f(x)
(x) = lim
h→0 h
This means that ifhis small, thenf ′ (x) is a good
approximation of the ratio

– p. 31/34
Limit Def Revisited
Recall the limit definition of derivative
f ′ f(x+h)−f(x)
(x) = lim
h→0 h
This means that ifhis small, thenf ′ (x) is a good
approximation of the ratio
f(x+h)−f(x)
h

– p. 31/34
Limit Def Revisited
Recall the limit definition of derivative
f ′ f(x+h)−f(x)
(x) = lim
h→0 h
This means that ifhis small, thenf ′ (x) is a good
approximation of the ratio
f(x+h)−f(x)
h
In particular, if we leth = 1

– p. 31/34
Limit Def Revisited
Recall the limit definition of derivative
f ′ f(x+h)−f(x)
(x) = lim
h→0 h
This means that ifhis small, thenf ′ (x) is a good
approximation of the ratio
f(x+h)−f(x)
h
In particular, if we leth = 1 which is sort of small

– p. 31/34
Limit Def Revisited
Recall the limit definition of derivative
f ′ f(x+h)−f(x)
(x) = lim
h→0 h
This means that ifhis small, thenf ′ (x) is a good
approximation of the ratio
f(x+h)−f(x)
h
In particular, if we leth = 1

– p. 32/34
Marginal Revenue, Profit, Cost
Applying the statement

– p. 32/34
Marginal Revenue, Profit, Cost
Applying the statement
f ′ (x) approximatesf(x+1)−f(x)

– p. 32/34
Marginal Revenue, Profit, Cost
Applying the statement
f ′ (x) approximatesf(x+1)−f(x)
to the functionsR(x),P(x),C(x) gives us

– p. 32/34
Marginal Revenue, Profit, Cost
Applying the statement
f ′ (x) approximatesf(x+1)−f(x)
to the functionsR(x),P(x),C(x) gives us
• Marginal Revenue R ′ (x) approximates
R(x+1)−R(x)

– p. 32/34
Marginal Revenue, Profit, Cost
Applying the statement
f ′ (x) approximatesf(x+1)−f(x)
to the functionsR(x),P(x),C(x) gives us
• Marginal Revenue R ′ (x) approximates
R(x+1)−R(x)
• Marginal Profit P ′ (x) approximates
P(x+1)−P(x)

– p. 32/34
Marginal Revenue, Profit, Cost
Applying the statement
f ′ (x) approximatesf(x+1)−f(x)
to the functionsR(x),P(x),C(x) gives us
• Marginal Revenue R ′ (x) approximates
R(x+1)−R(x)
• Marginal Profit P ′ (x) approximates
P(x+1)−P(x)
• Marginal Cost C ′ (x) approximates
C(x+1)−C(x)

– p. 32/34
Marginal Revenue, Profit, Cost
Applying the statement
f ′ (x) approximatesf(x+1)−f(x)
to the functionsR(x),P(x),C(x) gives us
• Marginal Revenue R ′ (x) approximates
R(x+1)−R(x)
• Marginal Profit P ′ (x) approximates
P(x+1)−P(x)
• Marginal Cost C ′ (x) approximates
C(x+1)−C(x)
If you have producedxitems, thenC(x+1)−C(x)
is the cost required to produce one more item.

Marginal Revenue, Profit, Cost
Applying the statement
f ′ (x) approximatesf(x+1)−f(x)
to the functionsR(x),P(x),C(x) gives us
• Marginal Revenue R ′ (x) approximates
R(x+1)−R(x)
• Marginal Profit P ′ (x) approximates
P(x+1)−P(x)
• Marginal Cost C ′ (x) approximates
C(x+1)−C(x)
If you have producedxitems, thenC(x+1)−C(x)
is the cost required to produce one more item.
Likewise forP(x+1)−P(x) andR(x+1)−R(x)
– p. 32/34

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.
What is the current weekly profit

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.
What is the current weekly profit
Answer:
P(50) = −0.006(50) 3 −0.2(50) 2 +900(50)−1200

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.
What is the current weekly profit
Answer:
P(50) = −0.006(50) 3 −0.2(50) 2 +900(50)−1200
= −0.006(125,000)−0.2(2500)+900(50)−1200

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.
What is the current weekly profit
Answer:
P(50) = −0.006(50) 3 −0.2(50) 2 +900(50)−1200
= −0.006(125,000)−0.2(2500)+900(50)−1200
= −750−500+45000−1200

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.
What is the current weekly profit
Answer:
P(50) = −0.006(50) 3 −0.2(50) 2 +900(50)−1200
= −0.006(125,000)−0.2(2500)+900(50)−1200
= −750−500+45000−1200 = 42,550
What would the profit be if they sold one more car

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.
What is the current weekly profit
Answer:
P(50) = −0.006(50) 3 −0.2(50) 2 +900(50)−1200
= −0.006(125,000)−0.2(2500)+900(50)−1200
= −750−500+45000−1200 = 42,550
What would the profit be if they sold one more car
Answer:
P(51) = −0.006(51) 3 −0.2(51) 2 +900(51)−1200

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.
What is the current weekly profit
Answer:
P(50) = −0.006(50) 3 −0.2(50) 2 +900(50)−1200
= −0.006(125,000)−0.2(2500)+900(50)−1200
= −750−500+45000−1200 = 42,550
What would the profit be if they sold one more car
Answer:
P(51) = −0.006(51) 3 −0.2(51) 2 +900(51)−1200
= −0.006(132,651)−0.2(2601)+900(51)−1200

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.
What is the current weekly profit
Answer:
P(50) = −0.006(50) 3 −0.2(50) 2 +900(50)−1200
= −0.006(125,000)−0.2(2500)+900(50)−1200
= −750−500+45000−1200 = 42,550
What would the profit be if they sold one more car
Answer:
P(51) = −0.006(51) 3 −0.2(51) 2 +900(51)−1200
= −0.006(132,651)−0.2(2601)+900(51)−1200
= −795.91−520.2+45900−1200

– p. 33/34
Marginal Example
Flying Corn Motors has a weekly profit in dollars of
P(x) = −0.006x 3 −0.2x 2 +900x−1200
and currently 50 cars are sold per week.
What is the current weekly profit
Answer:
P(50) = −0.006(50) 3 −0.2(50) 2 +900(50)−1200
= −0.006(125,000)−0.2(2500)+900(50)−1200
= −750−500+45000−1200 = 42,550
What would the profit be if they sold one more car
Answer:
P(51) = −0.006(51) 3 −0.2(51) 2 +900(51)−1200
= −0.006(132,651)−0.2(2601)+900(51)−1200
= −795.91−520.2+45900−1200 = 43,383.89

– p. 34/34
Marginal Example Continued
The exact increase in profit is

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50)

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89
What is the marginal profit

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89
What is the marginal profit
Differentiate
P(x) = −0.006x 3 −0.2x 2 +900x−1200

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89
What is the marginal profit
Differentiate
P(x) = −0.006x 3 −0.2x 2 +900x−1200
P ′ (x) = −0.018x 2 −0.4x+900

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89
What is the marginal profit
Differentiate
P(x) = −0.006x 3 −0.2x 2 +900x−1200
P ′ (x) = −0.018x 2 −0.4x+900
What is the marginal profit whenx = 50

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89
What is the marginal profit
Differentiate
P(x) = −0.006x 3 −0.2x 2 +900x−1200
P ′ (x) = −0.018x 2 −0.4x+900
What is the marginal profit whenx = 50
Answer:
P ′ (50) = −0.018(50) 2 −0.4(50)+900

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89
What is the marginal profit
Differentiate
P(x) = −0.006x 3 −0.2x 2 +900x−1200
P ′ (x) = −0.018x 2 −0.4x+900
What is the marginal profit whenx = 50
Answer:
P ′ (50) = −0.018(50) 2 −0.4(50)+900
= −0.018(2500)−0.4(50)+900

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89
What is the marginal profit
Differentiate
P(x) = −0.006x 3 −0.2x 2 +900x−1200
P ′ (x) = −0.018x 2 −0.4x+900
What is the marginal profit whenx = 50
Answer:
P ′ (50) = −0.018(50) 2 −0.4(50)+900
= −0.018(2500)−0.4(50)+900 = −45−20+900

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89
What is the marginal profit
Differentiate
P(x) = −0.006x 3 −0.2x 2 +900x−1200
P ′ (x) = −0.018x 2 −0.4x+900
What is the marginal profit whenx = 50
Answer:
P ′ (50) = −0.018(50) 2 −0.4(50)+900
= −0.018(2500)−0.4(50)+900 = −45−20+900
= 835

– p. 34/34
Marginal Example Continued
The exact increase in profit is
P(51)−P(50) = 43,383.89−42,550 = 833.89
What is the marginal profit
Differentiate
P(x) = −0.006x 3 −0.2x 2 +900x−1200
P ′ (x) = −0.018x 2 −0.4x+900
What is the marginal profit whenx = 50
Answer:
P ′ (50) = −0.018(50) 2 −0.4(50)+900
= −0.018(2500)−0.4(50)+900 = −45−20+900
= 835
Compare835 with actual increase in profit833.89