Calculus: This section is built on the formula of an average gradient.
𝑚=
𝑦2 −𝑦1
𝑥2 −𝑥1
=
Δ𝑦
Δ𝑥
=
𝑑𝑦
𝑑𝑥
Calculus is a study of change.
This section includes:






theory, i.e. first principles of finding the derivative of a function
limits
applying the rules of differentiation
sketching a cubic function, using calculus
determine the equation of a tangent
application in finance using Calculus
Average gradient/ derivative:
The average gradient formula
𝑚=
𝑦2 −𝑦1
𝑥2 −𝑥1
=
𝑓(𝑥+ℎ)−𝑓(𝑥)
(𝑥+ℎ)−𝑥
𝑓 ′ (𝑥) = lim
𝑓(𝑥+ℎ)−𝑓(𝑥)
𝑓 ′ (𝑥) = lim
𝑓(𝑥+ℎ)−𝑓(𝑥)
ℎ→0
ℎ→0
(𝑥+ℎ)−𝑥
becomes the first principle formula
when ℎ → 0 and (𝑥 + ℎ) − 𝑥 = ℎ
ℎ
We use 𝑓 (𝑥 + ℎ) 𝑓𝑜𝑟 𝑦2 𝑎𝑛𝑑 𝑓(𝑥) 𝑓𝑜𝑟 𝑦1 .
The denominator has been simplified to become ℎ.
The difference in meaning between finding average gradient and the
gradient at a point is that the distance between the two 𝑥 values tends to
become zero when we talk of the derivative. When we find the average
gradient, it may be two points far away from each other.
Practice Example
from First Principles.
Determine the Derivative, 𝑓 ′ (𝑥) if 𝑓(𝑥) = 𝑥 2 + 9,
𝑓(𝑥) = 𝑥 2 + 9
Solution:
𝑓(𝑥) = (𝑥 + ℎ)2 + 9
𝑓(𝑥) = 𝑥 2 + 2𝑥ℎ + ℎ2 + 9
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ→0
ℎ
𝑓 ′ (𝑥) = lim
𝑓
′ (𝑥)
𝑓
𝑥 2 + 2𝑥ℎ + ℎ2 + 9 − (𝑥 2 + 9)
= lim
ℎ→0
ℎ
′ (𝑥)
𝑥 2 + 2𝑥ℎ + ℎ2 + 9 − 𝑥 2 − 9
= lim
ℎ→0
ℎ
𝑓
′ (𝑥)
2𝑥ℎ + ℎ2
= lim
ℎ→0
ℎ
ℎ(2𝑥 + ℎ)
ℎ→0
ℎ
𝑓 ′ (𝑥) = 𝑓 ′ (𝑥) = lim
𝑓 ′ (𝑥) = lim 2𝑥 + ℎ
ℎ→0
= 2𝑥
Calculations dealing with Limits:
lim
𝑝−2
𝑝→−2 𝑝+5
= lim
−2−2
𝑝→−2 −2+5
=
−4
3
In the above example you may insert the value of 𝑝, because there is no
denominator that may become 0, and hence turn the expression into an
undefined status.
lim
𝑝→0
𝑝
𝑝2 −7𝑝+1
lim 16
𝑝→2
=
0
= 16
(0)2 −7(0)+1
(there is no ‘𝑝’ for substitution)
=0
lim
𝑥 2 +3𝑥
lim
𝑥→−3 𝑥+3
= lim
𝑥 2 +𝑥−6
𝑥→−3
𝑥(𝑥+3)
𝑥+3
= lim
𝑥→−3 𝑥+3
(𝑥+3)(𝑥−2)
𝑥→−3
= −3
𝑥+3
= −5
(By substituting ‘ − 3’ into 𝑥, from the beginning, in the examples above,
the denominator will be zero. You have to prevent that by factorizing and
making the denominator = 1, by canceling of the denominator with a
factor in the numerator)
lim
(2+ℎ)2 −22
lim
ℎ
ℎ→0
= lim
= lim
= lim 1
ℎ
𝑥→−2
4ℎ+ℎ2
=1
ℎ
ℎ→0
= lim
𝑥→−2 𝑥+2
4+4ℎ+ℎ2 −4
ℎ→0
𝑥+2
ℎ(4+ℎ)
ℎ
ℎ→0
=4
lim
𝑥 2 −3𝑥
𝑥→0
= lim
𝑥→0
lim
4𝑥 2 −1
𝑥→∞ 7𝑥+2𝑥 2
𝑥
𝑥(𝑥−3)
𝑥
= lim(𝑥 − 3)
𝑥→0
= −3
4𝑥2 1
−
𝑥2 𝑥2
2
𝑥→∞ 7𝑥2 +2𝑥2
𝑥
𝑥
= lim
= lim
4−0
𝑥→∞ 0+2
=2
1 𝑛
lim ( )
9
𝑛→∞
lim (𝑥 4 )
𝑥→∞
=∞
=0
lim (12𝑛 )
𝑛→∞
=∞
lim (1 − 3𝑥 )
𝑥→−∞
=1−
1
3𝑥
=1−0
=1
Limits: ‘When dealing with limits we only care about where we are
going, not whether we get there.’ Classroom mathematics, page 147.
Differentiation:
How to preparation your expression to find the derivative, using rules.
1.
Have separate terms.
E.g. 𝑓(𝑥) = (𝑥 − 2)(𝑥 + 1) = 𝑥 2 − 𝑥 − 2 in separate terms.
𝑓(𝑥) =
𝑓(𝑥) =
2.
𝑥 2 +2𝑥
𝑥
𝑥2
=
𝑥 2 −4𝑥−4
𝑥
=
2𝑥 2 −4𝑥
+
2𝑥
𝑥
=𝑥+2
(𝑥−2)(𝑥+2)
2𝑥(𝑥−2)
=
𝑥
2𝑥
+
2
2𝑥
1
= + 𝑥 −1
2
Have co-efficient in front of variable.
4𝑥
E.g.:
𝑥 1
= 𝑥
4 4
3.
If we differentiate with respect to 𝑥, then no x is allowed in a ‘root’.
1
𝑓(𝑥) = √𝑥 = 𝑥 2
1
3
𝑓(𝑥) = √𝑥 == 𝑥 3
5
4
𝑓(𝑥) = √𝑥 5 == 𝑥 4
4.
If we differentiate with respect to 𝑥, then no 𝑥 is allowed in the
denominator.
1
𝑓(𝑥) = = 𝑥 −1
𝑥
𝑓(𝑥) =
𝑥 2 −𝑥
8𝑥 5
1
1
8
8
= 𝑥 −3 − 𝑥 4
NOTE
Take note of the following:
1.
2.
𝑚
𝑛
√𝑎 𝑛 = 𝑎 𝑚
1
𝑎𝑚
= 𝑎−𝑚
The Derivative of a constant is always ZERO.
𝐼𝑓 𝑓(𝑥) = 𝑘, 𝑎𝑛𝑑 𝑘 𝑖𝑠 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡, 𝑡ℎ𝑒𝑛 𝑓 ′ (𝑥) = 0
Sometimes, the expression becomes much more complicated and
we need other rules as well.
1.
Product rule
If 𝑦 = (2𝑥 − 3)(𝑥 2 + 1),
𝑙𝑒𝑡 𝑢 = (2𝑥 − 3) 𝑎𝑛𝑑 𝑣 = (𝑥 2 + 1)
𝑑𝑦
𝑑𝑥
= 𝑢′ . 𝑣 + 𝑣 ′ . 𝑢
𝑢′ = 2 𝑎𝑛𝑑 𝑣 ′ = 2𝑥
∴
𝑑𝑦
= 2(𝑥 2 + 1) + 2𝑥(2𝑥 − 3)
𝑑𝑥
= 2𝑥 2 + 2 + 4𝑥 2 − 6𝑥
= 6𝑥 2 − 6𝑥 + 2
Let us check the answer, by doing the differentiation in a different way
𝑦 = (2𝑥 − 3)(𝑥 2 + 1) = 2𝑥 3 − 3𝑥 2 + 2𝑥 − 3
𝑑𝑦
= 6𝑥 2 − 6𝑥 + 2
𝑑𝑥
2.
Quotient rule:
(2𝑥 − 2𝑥 2 )
𝑦=
(𝑥 2 − 1)
𝑙𝑒𝑡 𝑢 = (2𝑥 − 2𝑥 2 ) 𝑎𝑛𝑑 𝑣 = (𝑥 2 − 1)
𝑑𝑦 𝑢′ . 𝑣 − 𝑣 ′ . 𝑢
=
𝑑𝑥
𝑣2
𝑢′ = 2 − 4𝑥 𝑎𝑛𝑑 𝑣 ′ = 2𝑥
𝑑𝑦 (2 − 4𝑥)(𝑥 2 − 1) − 2𝑥(2𝑥 − 2𝑥 2 )
=
(𝑥 2 − 1)2
𝑑𝑥
𝑑𝑦 −2𝑥 2 + 4𝑥 − 2
=
(𝑥 2 − 1)2
𝑑𝑥
𝑑𝑦 −2(𝑥 − 1)2
=
(𝑥 2 − 1)2
𝑑𝑥
𝑑𝑦
−2(𝑥 − 1)2
=
𝑑𝑥 (𝑥 − 1)2 (𝑥 + 1)2
𝑑𝑦
−2
=
𝑑𝑥 (𝑥 + 1)2
3.
𝑑
𝑑𝑥
Power rule:
(𝑢𝑛 ) = 𝑛𝑢𝑛−1
𝑑𝑢
𝑑𝑥
3.1
If 𝑦 = (𝑥 3 − 1)7 , 𝑓𝑖𝑛𝑑 𝑦 ′
Solution: Since y is a power of a function of x, the power rule applies.
Letting 𝑢(𝑥) = 𝑥 3 − 1 𝑎𝑛𝑑 𝑛 = 7, 𝑤𝑒 ℎ𝑎𝑣𝑒
𝑦 ′ = 𝑛[𝑢(𝑥)]𝑛−1 𝑢′ (𝑥)
= 7(𝑥 3 − 1)7−1
𝑑
𝑑𝑥
(𝑥 3 − 1)
= 7(𝑥 3 − 1)6 (3𝑥 2 )
= 21𝑥 2 (𝑥 3 − 1)6
3.2
3
If 𝑦 = √(4𝑥 2 + 3𝑥 − 2)2 , 𝑓𝑖𝑛𝑑
𝑑𝑦
𝑑𝑥
𝑤ℎ𝑒𝑛 𝑥 = −2
2
Solution: since 𝑦 = (4𝑥 2 + 3𝑥 − 2)3 , 𝑤𝑒 𝑢𝑠𝑒 𝑡ℎ𝑒 𝑝𝑜𝑤𝑒𝑟 𝑟𝑢𝑙𝑒 𝑤𝑖𝑡ℎ
2
𝑢 = 4𝑥 2 + 3𝑥 − 2, 𝑎𝑛𝑑 𝑛 = .
3
𝑑𝑦
We have that
𝑑𝑥
2
2
= (4𝑥 2 + 3𝑥 − 2)3−1
3
𝑑
𝑑𝑥
(4𝑥 2 + 3𝑥 − 2)
−1
2
= (4𝑥 2 + 3𝑥 − 2) 3 (8𝑥 + 3)
3
=
Thus
=
𝑑𝑦
2(−13)
3
3 √4𝑥 2 +3𝑥−2
𝑤ℎ𝑒𝑛 𝑥 = −2
𝑑𝑥
3 √8
2(8𝑥+3)
3
=
−13
3
3.3
If
𝑦=
1
𝑥 2 −2
, 𝑓𝑖𝑛𝑑
𝑑𝑦
𝑑𝑥
Solution: although the quotient rule can be used here, a more efficient
approach is to treat the right hand side as the power (𝑥 2 − 2)−1 and
use the power rule. Let 𝑢 = 𝑥 2 − 2
Let 𝑢 = 𝑥 2 − 2, 𝑎𝑛𝑑 𝑡ℎ𝑒𝑛 𝑦 = 𝑢−1
And
𝑑𝑦
= (−1)(𝑥 2 − 2)−1−1
𝑑𝑥
𝑑
𝑑𝑥
= (−1)(𝑥 2 − 2)−2 (2𝑥)
2𝑥
= − (𝑥 2
−2)2
3.5
2𝑠+5 4
𝑑𝑧
𝑠 2 +1
𝑑𝑠
If 𝑧 = (
) , 𝑓𝑖𝑛𝑑
(𝑥 2 − 2)
Solution: Since 𝑧 is a power of a function, we first use the power rule:
𝑑𝑧
𝑑𝑠
2𝑠+5 4−1 𝑑
= 4(
𝑠 2 +1
)
2𝑠+5
(
)
𝑑𝑠 𝑠 2 +1
Now use the quotient rule:
𝑑𝑧
2𝑠 + 5 3 (𝑠 2 + 1)(2) − (2𝑠 + 5)(2𝑠)
= 4( 2
) (
)
(𝑠 2 + 1)2
𝑑𝑠
𝑠 +1
Simplifying, we have
𝑑𝑧
2𝑠 + 5 3 −2𝑠 2 − 10𝑠 + 2
= 4( 2
) (
)
(𝑠 2 + 1)2
𝑑𝑠
𝑠 +1
8(𝑠 2 + 5𝑠 − 1)(2𝑠 + 5)3
=−
(𝑠 2 + 1)5
4.
Chain rule
If 𝑦 is a differentiable function of u and u is a differentiable function of x,
then y is a differentiable function of x and
dy
dx
=
dy du
.
du dx
.
I will explain the chain rule by considering rate of change.
Suppose 𝑦 = 8𝑢 + 5 𝑎𝑛𝑑 𝑢 = 2𝑥 − 3
Let 𝑥 change by one unit. How does u change? To answer this question,
we differentiate and find
is a change in 𝑦 of
𝑑𝑦
𝑑𝑢
𝑑𝑢
𝑑𝑥
= 2. But for each one unit change in 𝑢, there
= 8. Therefore, what is the change in 𝑦 if 𝑥
changes by one unit; that is, what is
𝑑𝑦 𝑑𝑢
.
𝑑𝑢 𝑑𝑥
𝑑𝑦
𝑑𝑥
? The answer is 8.2, which is
.
Thus,
𝑑𝑦
𝑑𝑥
=
𝑑𝑦 𝑑𝑢
.
𝑑𝑢 𝑑𝑥
4.1
We will now use the chain rule to redo the problem at the beginning of
this section. If 𝑦 = 𝑢2 𝑎𝑛𝑑 𝑢 = 2𝑥 + 1,
Then
𝑑𝑦
=
𝑑𝑥
𝑑𝑦 𝑑𝑢
.
𝑑𝑢 𝑑𝑥
=
𝑑
𝑑𝑢
(𝑢2 ).
𝑑
𝑑𝑥
Replacing 𝑢 𝑏𝑦 2𝑥 + 1 gives
(2𝑥 + 1) = (2𝑢)2 = 4𝑢
𝑑𝑦
𝑑𝑥
= 4(2𝑥 + 1) = 8𝑥 + 4,
Which agrees with our previous result?
4.2
If 𝑦 = 2𝑢2 − 3𝑢 − 2 𝑎𝑛𝑑 𝑢 = 𝑥 2 + 4, 𝑓𝑖𝑛𝑑
𝑑𝑦
𝑑𝑥
Solution: By the chain rule,
𝑑𝑦
𝑑𝑥
=
𝑑𝑦 𝑑𝑢
.
𝑑𝑢 𝑑𝑥
=
𝑑
𝑑𝑢
(2𝑢2 − 3𝑢 − 2).
𝑑
𝑑𝑥
(𝑥 2 + 4) = (4𝑢 − 3)(2𝑥)
We can write our answer in terms of x alone by replacing u by 𝑥 2 + 4
𝑑𝑦
𝑑𝑥
= [4(𝑥 2 + 4) − 3](2𝑥) = 8𝑥 3 + 26𝑥
4.3
𝐼𝑓 𝑦 = √𝑤 𝑎𝑛𝑑 𝑤 = 7 − 𝑡 3 , 𝑓𝑖𝑛𝑑
𝑑𝑦
𝑑𝑡
.
Solution: here y is a function of w and w is a function of t, so we can
view y as a function of t. By the chain rule,
𝑑𝑦
𝑑𝑡
=
𝑑𝑦 𝑑𝑤
.
𝑑𝑤 𝑑𝑡
=
𝑑
𝑑
(√𝑤). 𝑑𝑡 (7 − 𝑡 3 )
𝑑𝑤
−1
1
= ( 𝑤 2 ) (−3𝑡 2 ) =
2
=−
3𝑡 2
=−
2√𝑤
1
2 √𝑤
(−3𝑡 2 )
3𝑡 2
2√7−𝑡 3
4.4
𝐼𝑓 𝑦 = 4𝑢3 + 10𝑢2 − 3𝑢 − 7 𝑎𝑛𝑑 𝑢 =
𝑓𝑖𝑛𝑑
𝑑𝑦
𝑑𝑥
4
3𝑥−5
,
𝑤ℎ𝑒𝑛 𝑥 = 1.
Solution: By the chain rule,
𝑑𝑦
𝑑𝑥
=
𝑑𝑦 𝑑𝑢
.
𝑑𝑢 𝑑𝑥
=
𝑑
𝑑𝑢
(4𝑢3 + 10𝑢2 − 3𝑢 − 7) .
𝑑
4
(
)
𝑑𝑥 3𝑥−5
=
(12𝑢2
+ 20𝑢 − 3).
(3𝑥−5)
𝑑
𝑑
(4)−4 (3𝑥−5)
𝑑𝑥
𝑑𝑥
(3𝑥−5)2
−12
= (12𝑢2 + 20𝑢 − 3). (3𝑥−5)2
Even though
𝑑𝑦
𝑑𝑥
is expressed in terms of 𝑥’𝑠 and 𝑢’𝑠, we can evaluate it
when 𝑥 = 1 if we determine the corresponding value of 𝑢. when 𝑥 = 1,
𝑢=
4
(3(1)−5)
Thus,
𝑑𝑦
𝑑𝑥
= −2
, 𝑤ℎ𝑒𝑛 𝑥 = 1
−12
= [12(−2)2 + 20(−2) − 3]. [3(1)−5]2 = −15
Cubic Graph
Given: 𝑓(𝑥) = (𝑥 − 2)(−𝑥 2 + 2𝑥 + 15) = −𝑥 3 + 4𝑥 2 + 11𝑥 − 30
SHAPE: the shape of a cubic graph is determined by the sign of 𝑥 3 .
If 𝑥 3 has a positive coefficient, then the graph starts with a positive
gradient, turns and becomes negative, turns again and becomes positive
again.
If
𝑥 3 has a negative coefficient, then the graph starts with a
negative gradient, turns and becomes positive, turns again and
becomes negative again.
Turning Point(s) also stationary points.
When you look at the graph, it has two turning points.
𝑦 = 𝑥 3 Has a maximum point first and then a minimum point.
𝑦 = −𝑥 3 Has a minimum and then a maximum turning point ( when you take the 𝑥
values of the turning points from left to right.)
Turning points are where the curve changes from being negative to positive.
Between the negative and positive gradients, the gradient becomes zero.
(In the switch-over from positive to negative or negative to positive)
You know already that the gradient is describe by
We then work out
𝑑𝑦
𝑑𝑥
𝑜𝑟 𝑓 ′ (𝑥 )
𝑓 ′ (𝑥) = −3𝑥 2 + 8𝑥 + 11
Put the gradient = 0, e.g. 𝑓 ′ (𝑥) = −3𝑥 2 + 8𝑥 + 11 = 0
Find the 2 x-values (critical values) which are the solutions to the equation, which will
be the x coordinates of the maximum/minimum.
Substitute the x-values, one by one, back into the equation of the graph to find the y
value on the curve.
−3𝑥 2 + 8𝑥 + 11 = 0
3𝑥 2 − 8𝑥 − 11 = 0
(3𝑥 − 11)(𝑥 + 1) = 0
(3𝑥 − 11) = 0
𝑜𝑟
(𝑥 + 1) = 0
11
𝑜𝑟
𝑥 = −1
3
2
11 3
11
11
𝑦 = − ( ) + 4 ( ) + 11 ( ) − 30
3
3
3
𝑥=
8
𝑦 = 189
𝑦 = −(−1)3 + 4(−1)2 + 11(−1) − 30
𝑦 = −36
We now have (-1; -36) as a minimum turning point and
2
(3 ; 1889 ) as a maximum turning point.
3
2
3
(3 ; 1889 )
(-1;-36)
Intercepts
Y-intercept: we find the y-intercept, only when 𝑥 = 0
𝑓(𝑥) = −(0)3 + 4(0)2 + 11(0) − 30
In this example 𝑦 = − 30
𝑋-intercept: we find the 𝑥-intercept, only when 𝑦 = 0
𝑦 = −𝑥 3 + 4𝑥 2 + 11𝑥 − 30 = (𝑥 − 2)(−𝑥 2 + 2𝑥 + 15) = 0
when 𝑦 = 0, we deal with values on the x-axis. To find 𝑥 −values that will
result into a y-value which is zero, we have to factorize the equation into
linear factors.
(𝑥 − 2)(−𝑥 2 + 2𝑥 + 15) = −(𝑥 − 2)(𝑥 2 − 2𝑥 − 15) = (𝑥 − 2)(𝑥 2 − 2𝑥 − 15) = 0
−(𝑥 − 2)(𝑥 2 − 2𝑥 − 15) = 0
(𝑥 − 2)(𝑥 2 − 2𝑥 − 15) = 0
(𝑥 − 2)(𝑥 2 − 2𝑥 − 15) = (𝑥 − 2)(𝑥 − 5)(𝑥 + 3) = 0
𝑊ℎ𝑒𝑛 𝑥 = 2,
𝑤𝑒 𝑐𝑎𝑙𝑙 ′2′ 𝑎𝑛 𝑥 − 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛,
𝑡ℎ𝑒𝑛 (𝑥 − 2)𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟. 𝑇ℎ𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 𝑖𝑠 𝑖𝑚𝑝𝑜𝑟𝑡𝑎𝑛𝑡.
The same with
𝑥 = −3
𝑜𝑟
𝑥=5
Finally, let us look at the final product, let us put in the x- and y- axes!!!
(323; 1889 )
(-3;0)
(2;0)
(5;0)
(0;-30)
(-1;-36)
Make sure that you can work out the gradient at any point on the graph.
Make sure that you can work out the average gradient between two
points on the graph.
Be able to find the equation of a tangent to any point on the curve.
Understand what is meant by 𝑓(𝑥) > 0 𝑜𝑟 𝑓(𝑥) < 0
Understand what is meant by 𝑓’(𝑥) > 0 𝑜𝑟 𝑓’(𝑥) < 0
Relationships in economics, that can be describe in Mathematical
equations.
• Identities
Example: price x units = revenue
Profit = revenue – cost
Total cost = average cost per unit x units produced
• Relationships
Example: Systems in the economy depend on one another, e.g. capital
available, labor and input costs, and profits which will again provide for
capital to pay for labor and input costs.
Rate of change of price with respect to Quantity
Let 𝑝 = 100 − 𝑞 2 be the demand function for a manufacturer’s product.
Find the rate of change of price 𝑝 per unit with respect to quantity 𝑞.
How fast is the price changing with respect to 𝑞, 𝑤ℎ𝑒𝑛 𝑞 = 5? assume
that 𝑝 is in Rands?
Solution: The rate of change of 𝑝 with respect to 𝑞 is
𝑑𝑝
𝑑𝑞
=
𝑑
𝑑𝑞
(100 − 𝑞 2 ) = −2𝑞
Thus when 𝑞 = 5 ,
𝑑𝑝
𝑑𝑞
= −2(5) = 10
This means that when five units are demanded, an increase of one extra
unit demanded corresponds to a decrease of approximately R10 in the
price per unit that consumers are willing to pay.
The price of an item is often described as the demand or supply
function: 𝑝 = ____________. (lower key)
The price that the consumer is willing to pay for an item is the demand
function.
The price that a company is willing to sell its product for, is called the
supply function.
Revenue
Example: Sam has 100 T shirts in his shop. The price per T-shirt is R85.
He sells 60 T-shirts last month. What was his income (revenue) for last
month?
𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑝𝑟𝑖𝑐𝑒 × 𝑢𝑛𝑖𝑡𝑠 𝑠𝑜𝑙𝑑
Please note that the price is also called the demand function. When we
use 𝑝 for price in a formula, we use the lower case 𝑝. Price often
depends on the demand for a specific product. Example: during a
drought, food prices go up, because there is a shortage of fresh
produce.
Marginal revenue
Suppose a manufacturer sells a product at R2
per unit. If 𝑞 units are sold, the total revenue is given by
𝑟 = 2𝑞
The marginal revenue function is
𝑑𝑟
𝑑𝑞
=
𝑑
𝑑𝑞
(2𝑞) = 2 which is a constant.
Thus, the marginal revenue is 2 regardless of the number of units sold.
This is what we would expect, because the manufacturer receives R2 for
each unit sold.
𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑢𝑛𝑖𝑡𝑠 𝑥 𝑝𝑟𝑖𝑐𝑒
Suppose “𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑓(𝑞)” is the total revenue function for a
manufacturer. The equation 𝑇𝑅 = 𝑓(𝑞) states that the Rand value
received for selling q units of the product is the revenue. The demand
function for a manufacturer’s product is also described in terms of 𝑞, ∴
𝑝 = 𝑓(𝑞). (another function described in terms of 𝑞, but 𝑝 in the lower
key)
𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 × 𝑝𝑟𝑖𝑐𝑒
𝑇𝑜𝑡𝑎𝑙 𝑟𝑒𝑣𝑒𝑛𝑢𝑒 = 𝑞. 𝑓(𝑞)
Marginal Revenue If the demand equation for a manufacturer’s product
is 𝑝 =
1000
𝑞+5
where 𝑝 is in Rands, find the marginal revenue function and
evaluate it when 𝑞 = 45.
Solution: First we find the revenue function. The revenue 𝑟 received
for selling 𝑞 units when the price per unit is 𝑝 is given by 𝑟𝑒𝑣𝑒𝑛𝑢𝑒 =
(𝑝𝑟𝑖𝑐𝑒)(𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦), that is 𝑟 = 𝑝𝑞
𝑟=(
1000
1000𝑞
)𝑞 =
𝑞+5
𝑞+5
we will express 𝑟 in terms of 𝑞 only
Thus, the marginal-revenue function is given by
𝑑𝑟
𝑑𝑞
=
1000(𝑞+5)−1(1000𝑞)
(𝑞+5)2
5000
= (𝑞+5)2
And, when 𝑞 = 45, marginal revenue
𝑑𝑟
𝑑𝑞
=
5000
(45+5)2
=
5000
2500
=2
This means that selling one additional unit beyond 45 results in
approximately R2 more in revenue.
Maximizing Revenue
product
is 𝑝 =
80−𝑞
4
The demand equation for a manufacturer’s
0 ≤ 𝑞 ≤ 80 (𝑞 is the number of units and 𝑝 is the
price per unit.)
1.
At what value of 𝑞 will there be maximum revenue?
2.
What is the maximum revenue?
Solution:
𝑅𝑒𝑣𝑒𝑛𝑢𝑒 = (𝑝𝑟𝑖𝑐𝑒)(𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦)
𝑟=
𝑟(𝑞) =
To maximize:
80 − 𝑞
.𝑞
4
80𝑞−𝑞 2
4
1
= 20𝑞 − 𝑞 2
4
𝑑𝑟
𝑑𝑞
=0
1
20 − 𝑞 = 0
2
𝑞 = 40
1
Thus 40 is the only critical value. The second derivative 𝑟 ′′ (𝑞) = − ,
2
which is negative. Therefore, at 𝑞 = 40, the revenue will be at a
maximum.
(80)(40) − (40)2
∴ 𝑟(40) =
= 400
4
The Maximum revenue is R400
Average cost:
If 𝑐 is the total cost of producing 𝑞 units of a product,
𝑐
then the average cost per unit, 𝑐,
̅ is 𝑐̅ = .
𝑞
If the total cost of 20 units is R100, then the average cost per unit is
𝑐̅ =
100
20
= 𝑅5.
By multiplying both sides of 𝑐̅ =
𝑐
𝑞
by 𝑞, we have 𝑐̅𝑞 = 𝑐
𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 = 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 ×
𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡.
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡:
𝑇𝐶 = 𝐶(𝑥)
𝑎𝑛𝑑 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑐𝑜𝑠𝑡, 𝐴𝐶 = 𝐶̅ (𝑥) =
𝑇𝐶 C(𝑥)
=
𝑞
𝑥
𝑞 𝑖𝑠 𝑡ℎ𝑒 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 ; 𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠
Say it costs R 16 to produce a T-shirt. What is the Total cost to produce
10 T-shirts?
Or, the total cost the manufacture 10 chairs is R2500. What is the
average cost to manufacture a chair?
Marginal cost. A manufacturer’ total cost function, 𝑐 = 𝑓(𝑞) gives the
total cost 𝑐 of producing and marketing 𝑞 units of a product. The rate of
change of 𝑐 with respect to 𝑞 is called the Marginal cost.
Thus,
𝑀𝑎𝑟𝑔𝑖𝑛𝑎𝑙 𝑐𝑜𝑠𝑡 =
𝑑𝑐
𝑑𝑞
For example, suppose 𝑐 = 𝑓(𝑞) = 0.1𝑞 2 + 3, where 𝑐 is in Rands and 𝑞
is in kilograms. The marginal cost when 4 kg are produced is
Evaluated when 𝑞 = 4: When 𝑞 = 4,
𝑑𝑐
𝑑𝑞
𝑑𝑐
𝑑𝑞
,
= 0.2𝑞 = 0.2(4) = 0.8
This means that if production is increased by 1 kg from 4 kg to 5 kg, the
change in cost is approximately R0.80. That is, the additional kilogram
costs about R0.80. In general, we interpret marginal cost as the
approximate cost of one additional unit of output. After all, the difference
𝑓(𝑞 + 1) − 𝑓(𝑞) can be seen as a difference quotient
𝑓(𝑞+1)−𝑓(𝑞)
1
(the
case where ℎ = 1)
The derivative is often easier to compute than the exact value. (In the
case at hand, the actual cost of producing one more kilogram beyond 4
kg is
𝑓(5) − 𝑓(4) = 5.5 − 4.6 = 𝑅0.90
Minimizing average cost
by 𝐶 = 𝑐(𝑞) =
𝑞2
4
A manufacturer’s total cost function is given
+ 3𝑞 + 400 (𝑐 is the total cost of producing 𝑞 units)
1.
At what level of output will average cost per unit be a minimum?
2.
What is this minimum?
Solution: The quantity to be minimized is the average cost 𝑐.
̅
The average cost function is:
𝑐
𝑞2
+3𝑞+400
4
𝑞
𝑞
𝑐̅(𝑞) = =
𝑞
400
4
𝑞
= +3+
𝑑𝑐̅
To minimize 𝑐̅, we differentiate:
𝑑𝑞
1
400
4
𝑞2
= −
𝑑𝑐̅
To get the critical values, we solve
𝑑𝑞
1
400
4
𝑞2
= −
=0
𝑞2 − 1600 = 0
(𝑞 − 40)(𝑞 + 40) = 0
𝑞 = 40, 𝑠𝑖𝑛𝑐𝑒 𝑞 > 0
(Here q must be positive.)
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑐𝑜𝑠𝑡 × 𝑢𝑛𝑖𝑡𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑.
𝐶̅ (𝑥) =
𝐶(𝑥)
𝑥
… … … … … … … … . .1
Let us minimize the average cost per unit produced. In order to
minimize, we need: 𝐶̅ ′ (𝑥) = 0
′
′
𝑢 𝑣−𝑣 𝑢
Quotient rule: 𝐶̅ ′ (𝑥) =
2
𝑣
Let 𝑢 = 𝑐(𝑥), 𝑡ℎ𝑒𝑛 𝑢′ = 𝑐 ′ (𝑥) Let
𝑣 = 𝑥, 𝑡ℎ𝑒𝑛 𝑣 ′ = 1
∴ 𝐶̅ ′ (𝑥) =
𝑥𝐶 ′ (𝑥) − 1. 𝐶(𝑥)
𝑥2
The minimum occurs at 𝐶̅ ′ (𝑥) = 0
𝑥𝐶 ′ (𝑥)−𝐶(𝑥)
𝑥2
=0
For this equation to be equal to zero, the numerator needs to equal zero.
∴ 𝑥𝐶 ′ (𝑥) − 𝐶(𝑥) = 0
𝐶 ′ (𝑥) =
𝐶(𝑥)
𝑥
𝐶(𝑥)
= 𝐶̅ (𝑥) … … … … … … 𝑓𝑟𝑜𝑚 1
𝑥
∴ 𝑓𝑖𝑛𝑎𝑙𝑙𝑦, 𝐶̅ (𝑥) = 𝐶 ′ (𝑥), both equal
𝐶(𝑥)
𝑥
.
Profit (𝑃 in the upper case)
𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑟𝑒𝑣𝑒𝑛𝑢𝑒 − 𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡.
In a question, it is more likely that you will be given the demand function
and an average cost.
In order to work out profit, you need to calculate total revenue and total
cost.
Maximum Profit.
The maximum profit happens at the stationary points of the profit
function.
∴ 𝑃𝑟𝑜𝑓𝑖𝑡’ = 0
𝑟𝑒𝑣𝑒𝑛𝑢𝑒’ − 𝑐𝑜𝑠𝑡’ = 0
If 𝑟𝑒𝑣𝑒𝑛𝑢𝑒 ′′ − 𝑐𝑜𝑠𝑡 ′′ < 0, 𝑡ℎ𝑒𝑛 𝑃𝑟𝑜𝑓𝑖𝑡 will be maximized
𝑟𝑒𝑣𝑒𝑛𝑢𝑒’ < 𝑐𝑜𝑠𝑡’
This example involves maximizing profit when the demand and average
cost functions are known. This discussion leads to the economic
principle that when profit is a maximum, marginal revenue is equal to
marginal cost.
Suppose that the demand equation for a monopolist’s product is
𝑝 = 400 − 2𝑞 and the average cost function: 𝑐̅ = 0.2𝑞 + 4 +
400
𝑞
,
𝑞 Is number of units and both 𝑝 and 𝑐̅ are expressed in Rands per unit.
a.
Determine the level of output at which profit is maximized.
b.
Determine the price at which maximum profit occurs
c.
Determine the maximum profit
d.
If, as a regular device, the government imposes a tax of R22 per
unit on the monopolist, what is the new price for profit
maximization?
Solution:
We know that
𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑟𝑒𝑣𝑒𝑛𝑢𝑒 – 𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡
Since total revenue 𝑟 and total cost c are given by
𝑟 = 𝑝𝑞 = 400𝑞 − 2𝑞 2
𝑐 = 𝑞𝑐̅ = 0.2𝑞 2 + 4𝑞 + 400
The profit is
𝑃 =𝑟−𝑐
𝑃 = 400𝑞 − 2𝑞 2 − (0.2𝑞2 + 4𝑞 + 400)
So that
𝑃(𝑞) = 396𝑞 − 2.2𝑞 2 − 400 𝑓𝑜𝑟 𝑞 > 0
a.
To maximize profit, we put
𝑑𝑃
=0
𝑑𝑞
𝑑𝑃
= 396 − 4.4𝑞 = 0
𝑑𝑞
𝑞 = 90
Now
𝑑2𝑃
𝑑𝑞 2
= −4.4 is always negative, so it is negative at the critical value
𝑞 = 90. By the second derivative test, then there is a relative maximum
there. Since 𝑞 = 90 is the only critical value on (0, ∞) we must have an
absolute maximum there.
b.
The price at which maximum profit is obtained by setting 𝑞 = 90 in
the demand equation:
𝑝 = 400 − 2(90) = 220
c.
The maximum profit is obtained by evaluating
𝑃(90) = 396(90) − 2.2(90)2 − 400
= 17,420
11.
𝑀𝑅 = 𝑀𝐶, when profit is at a maximum.
𝑃𝑟𝑜𝑓𝑖𝑡’ = 0
𝑟𝑒𝑣𝑒𝑛𝑢𝑒’ − 𝑐𝑜𝑠𝑡’ = 0
∴ 𝑟𝑒𝑣𝑒𝑛𝑢𝑒’ = 𝑐𝑜𝑠𝑡’
If we assume that the maximum profit will occur at a critical point such
that 𝑃′ (𝑥) = 0 we can then say the following, P’(x) = R’(x) − C’(x) = 0 ⟹
R’(x) = C’(x) We then will know that this will be a maximum we also
where to know that the profit was always concave down or,
P’(x) = R′′ (x) − 𝐶′′(x) < 0 ⟹ R′′ (x) < 𝐶′′(x)So, if we know that
R′′ (x) < 𝐶′′(x) then we will maximize the profit if R′ (x) < 𝐶′(x) or if the
marginal cost equals the marginal revenue.
Marginal revenue (compare at different units produced.)
The weekly cost to produce 𝑥 knifes is given by
𝐶(𝑥) = 75000 + 100𝑥 − 0.03𝑥 2 + 0.000004𝑥 3
0 ≤ 𝑥 ≤ 10000
and the demand function for the knifes is given by,
𝑝(𝑥) = 200 − 0.005𝑥
0 ≤ 𝑥 ≤ 10000
Determine the marginal cost, marginal revenue and marginal profit when
2500 knifes are sold and when 7500 knifes are sold. Assume that the
company sells exactly what they produce.
Solution
Okay, the first thing we need to do is get all the various functions that
we’ll need. Here are the revenue and profit functions.
𝑅(𝑥) = 𝑥(200 − 0.005𝑥) = 200𝑥 − 0.005𝑥 2
𝑃(𝑥) = 200𝑥 − 0.005𝑥 2 − (75000 + 100𝑥 − 0.03𝑥 2 + 0.000004𝑥 3 )
= −75000 + 100𝑥 + 0.025𝑥 2 − 0.000004𝑥 3
Now, all the marginal functions are,
𝐶 ′ (𝑥) = 100 − 0.06𝑥 + 0.000012𝑥 2
𝑅′ (𝑥) = 200 − 0.01𝑥
𝑃′ (𝑥) = 100 + 0.05𝑥 − 0.000012𝑥 2
The marginal functions when 2500 knifes are sold are,
𝐶 ′ (2500) = 25
𝑅′ (2500) = 175
𝑃′ (2500) = 150
The marginal functions when 7500 are sold are,
𝐶 ′ (7500) = 325
𝑅′ (7500) = 125
𝑃′ (7500) = −22
So, upon producing and selling the 2501st knife it will cost the company
approximately R25 to produce the widget and they will see and added
R175 in revenue and R150 in profit.
On the other hand when they produce and sell the 7501st knife it will
cost an additional R325 and they will receive an extra R125 in revenue,
but lose R200 in profit.
Profit maximizing
This example involves maximizing profit when the demand and average
cost functions are known. This discussion leads to the economic
principle that when profit is maximum, marginal revenue is equal to
marginal cost.
Suppose that the demand equation for a monopolist’s product is 𝑝 =
400 − 2𝑞 and the average cost function:
𝑐̅ = 0.2𝑞 + 4 +
400
𝑞
,
𝑞 is number of units and both 𝑝 and 𝑐̅ are expressed in Rands per unit.
a.
Determine the level of output at which profit is maximized.
b.
Determine the price at which maximum profit occurs.
c.
Determine the maximum profit.
Solution:
𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑟𝑒𝑣𝑒𝑛𝑢𝑒 – 𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡
Since total revenue 𝑟 and total cost c are given by
𝑟 = 𝑝𝑞 = 400𝑞 − 2𝑞 2
𝑐 = 𝑞𝑐̅ = 0.2𝑞 2 + 4𝑞 + 400
𝑃𝑟𝑜𝑓𝑖𝑡 = 𝑡𝑜𝑡𝑎𝑙 𝑟𝑒𝑣𝑒𝑛𝑢𝑒 – 𝑡𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡
𝑃(𝑞) = 400𝑞 − 2𝑞 2 − (0.2𝑞 2 + 4𝑞 + 400) = 396𝑞 − 2.2𝑞2 − 400 𝑓𝑜𝑟 𝑞
>0
a.
To maximize profit, we say
𝑑𝑃
𝑑𝑞
=0
𝑑𝑃
= 396 − 4.4𝑞 = 0
𝑑𝑞
𝑞 = 90
Now
𝑑2𝑃
𝑑𝑞 2
= −4.4 is always negative, so it is negative at the critical
value 𝑞 = 90. By the second derivative test, then there is a relative
maximum there. Since 𝑞 = 90 is the only critical value on (0, ∞) we must
have an absolute maximum there.
b.
The price at which maximum profit is obtained by putting 𝑞 = 90 in
the demand equation: 𝑝 = 400 − 2(90) = 220
c.
The maximum profit is obtained by evaluating
𝑃(90) = 396(90) − 2.2(90)2 − 400 = 𝑅17,420
Maximizing profit.
A complex has 250 apartments to rent. If they
rent x apartments then their monthly profit, is given by,
𝑃(𝑥) = −8𝑥 2 + 3200𝑥 − 80000
How many apartments should they rent in order to maximize their profit?
Solution
All that we’re really being asked to do here is to maximize
the profit subject to the constraint that x must be in the range. 0 ≤ 𝑥 ≤
250
First, we’ll need the derivative and the critical point(s) that fall in the
range 0 ≤ 𝑥 ≤ 250
𝑃′ (𝑥) = −16𝑥 + 3200 ⟹ −16𝑥 + 3200 = 0 ⟹ 𝑥 = 200
Since the profit function is continuous and we have an interval with finite
bounds we can find the maximum value by simply plugging in the only
critical point that we have (which is nicely in the range of acceptable
answers and the end points of the range.)
𝑃(0) = −80000
𝑃(200) = 240000
𝑃(250) = 220000
So, it looks like they will generate the most profit if they only rent out 200
of the apartments instead of all 250 of them.