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CHAPTER
4
Calculus
LEARNING OBJECTIVES
• By the end of this chapter, you should be able to
✔ solve simple differentiation problems,
✔ solve simple integration problems,
✔ identify the nature of critical points – maximum point,
minimum point and inflection point,
✔ determine the coordinates of the maximum point and
minimum point,
✔ solve basic microeconomics questions which involve
concepts of differentiation and integration.
4.1 Introduction
• Calculus is a dynamic concept.
• It consists of differential calculus and integral
calculus.
• Differential calculus refers to finding the rate of
change of one quantity with respect to another
quantity.
• Integral calculus is the reverse of differential calculus.
4.2 Differentiation
•
Let y = f(x) be a given function. The derivative of y with respect to x, denoted by f' (x) is defined by
•
f' (x) = lim {[f (x + Δx) – f(x)]/ Δx}
Δx → 0
•
•
The method of finding the derivative using the above definition is called the first principle.
•
If the derivative of a function exists at a particular point, then we say that f is differentiable at that
point. The derivative of y with respect to x is denoted in many ways, namely
The derivative is also called the differential coefficient and the process of calculating the
derivative of a function is called differentiation.
dy/dx, f' (x), or y'.
•
The derivative has several meanings depending on its use; one of them being gradient and the
other, instantaneous rate of change.
Example 1
Find the derivative of f(x) = 2x by using the first principle.
•
Solution
f(x) = 2x
f(x + Δx) = 2(x + Δx) = 2x + 2Δx
By the first principle,
•
f' (x) = lim {[f(x + Δx) – f(x)]/ Δx}
Δx→0
= lim {[(2x + 2Δx) – (2x)]/ Δx}
Δx→0
= lim 2
Δx→0
=2
Example 2
Find the derivative of f(x) = 3x2 by using the first principle.
•
Solution
f(x) = 3x2
f(x + Δx) = 3(x + Δx)2
= 3x2 + 6xΔx + 3(Δx)2
By the first principle,
f'(x) = lim {[f(x + Δx) – f(x)]/ Δx}
Δx→0
= lim {[3x2 + 6xΔx + 3(Δx)2 – 3x2]/ Δx}
Δx→0
= lim 6x + 3Δx
Δx→0
= 6x
4.3 Rules of Differentiation
• Rule 1: If y = f(x) = c, then f' (x) = 0
[Constant rule]
Example 3
Find f' (x) if (a) f(x) = 8,
(b) f(x) = –49.
• Solution
(a) f(x) = 8
f' (x) = 0
(b) f(x) = –49
f' (x) = 0
Rule 2:
If f(x) = mx, then f' (x) = m
[Constant – multiple rule]
Example 4
• Find f' (x) if (a) f(x) = 7x
(b) f'(x) = –12x
• Solution
(a) f(x) = 7x
f'(x) = 7
(b) f(x) = –12x
f'(x) = –12
Example 5
• Find f'(x) if (a) f(x) = axn
(b) f(x) = 9x4
• Solution
(a) f(x) = axn
f'(x) = naxn–1
(b) f(x) = 9x4
f'(x) = 36x3
Rule 4:
If h(x) = f(x) + g(x), then h (x) = f' (x) + g' (x)
[Sum rule]
Using another differential notation,
d(f + g) = df + dg
dx
dx dx
Example 6
Find f'(x) if
(a) f(x) = 7 + 8x2 + 3x3
(b) f(x) = x2 + 5√x
•
Solution
(a) f(x) = 7 + 8x2 + 3x3
f'(x) = 16x + 9x2
f(x) = x2 + x
(b) f(x) = x2 + 5√x
f'(x) = 2x + 1/5 x-4/5
Rule 5:
If f(x) = uv and u' (x) and v' (x) both exist, then
f' (x) = uv' (x) + vu' (x) [Product rule]
Example 7
Find f ’ (x) if
(a) f(x) = (2x + 7)(x – 9), (b) f(x) = x2(2x + 8)
• Solution
(a) f(x) = (2x + 7)(x – 9)
f'(x) = (2x + 7)(x – 9)' + (x – 9)(2x + 7)'
= (2x + 7)(1) + (x –9)(2)
= 4x – 11
(b) f(x) = x2(2x + 8)
f'(x) = x2(2x+ 8)' + (2x + 8)(x2)'
= x2(2) + (2x + 8)(2x)
= 6x2 + 16x
Rule 6:
If f(x) = u/v and u' (x) and v' (x) both exist, and v(x) ≠
0, then
f' (x) = [v(x)u' (x) – u(x)v' (x)]/v2
[Quotient rule]
Example 8
• Find f' (x) if f(x) = (x2 + 1)/x.
• Solution
f (x) = (x2 + 1)/ x
f' (x) = [x(x2 + 1)' – (x2 + 1)(x)‘]/x2
= [x(2x) – (x2 + 1)(1)]/x2
= (x2 – 1)/x2
Rule 7:
If f(x) = u[v(x)] and u' [v(x)] and v' (x) both
exist, then
f' (x) = u' [v(x)] v' (x)
[Chain rule]
Another formulation of the chain rule is as follows.
If y = f(t) and t = g(x), then
dy =
dy dt
dx
dt dx
Example 9
Find f'(x) if
(a) f(x) = (x3 – 4x)30
(b) f(x) = 3√(1 + 8x2)
(c) f(x) = (x2 + 2)3(x3 – 1)4
Solution
(a) f(x) = (x3 – 4x)30
f'(x)
= 30(x3 – 4x)30 – 1(x3 – 4x)
= 30(x3 – 4x)29(3x2 – 4)
(b) f(x) = 3√(1 + 8x2)
f(x) = (1 + 8x2)1/3
f'(x)= 1/3 (1 + 8x2)-2/3 (16x)
= (16/3)x(1 + 8x2)-2/3
Solution (continue)
(c) f(x) = (x2 + 2)3 (x3 – 1)4
•
•
•
•
•
•
•
•
•
By the product rule, we get
f'(x) = (x2 + 2)3[(x3 – 1)4]' + (x3 – 1)4[(x2 + 2)3]'
Apply the chain rule to find
[(x3 – 1)4]' = 4(x3 – 1)3(3x2) = 12x2(x3 – 1)3
[(x2 + 2)3]' = 3(x2 + 2)2(2x) = 6x(x2 + 2)2
Hence,
f' (x) = (x2 + 2)3(12x2)(x3 – 1)3 + (x3 – 1)4(6x)(x2 + 2)2
= 6x(x3 – 1)3(x2 + 2)2[2x(x2 + 2) + (x3 – 1)]
= 6x(x3 – 1)3(x2 + 2)2[3x3 + 4x – 1]
Example 10
Find dy/dx if y = 3t2 – 3 and t = 4x + 2.
• Solution
If y = 3t2 – 3 and t = 4x + 2, then
dy/dx = (3t2 – 3)' (4x + 2)‘
= 6t(4)
= 24t
= 24(4x + 2)
4.4 Higher orders of
differentiation
• If a function y = f(x) is differentiable, then its derivative f'
(x) is called the first derivative or the first derived function.
• If we continue differentiating, we get the higher orders
• The second derivative or the second derived function is
d/dx [f' (x)] = f'' (x) and
• The third derivative is d/dx [f'' (x)] = f''' (x).
• The second derivative are also denoted by d2y/dx2 or y'' and the
third derivative by d3y/dx3 or y'''.
Example 11
If f(x) = x4 + x3 + x2 + x + 1, then
• Solution
f'(x) = 4x3 + 3x2 + 2x + 1 First derivative
f''(x) = 12x2 + 6x + 2
f''' (x) = 24x + 6
f'''' (x) = 24
Second derivative
Third derivative
Fourth derivative
4.5 Applications of derivatives
• A derivative can be used to calculate the gradient of a
curve at any point on a curve.
• It can be used to find the equation of the tangent
line and the normal.
Example 12
Find the equation of the tangent line to the curve
f(x) = x2 + 3x + 2 at the point x = 2.
• Solution
f(x) = x2 + 3x + 2
f(2) = 22 + 3(2) + 2 = 12
f'(x) = 2x + 3
f'(2) = 2(2) + 3 = 7
Substituting m = 7 and (2, 12) into y = mx + c, we get
12 = 7(2) + c
c = –2
The equation of the tangent line to the curve is y = 7x – 2.
Example 13
The curve y = ax2 + bx + c passes through (2,3) and the line y = x is a tangent to the curve at the point of origin.
Find the values of a, b and c.
•
Solution
Since the curve passes through the origin, c must be equal to zero. Hence, the curve becomes
y = ax2 + bx.
From y = ax2 + bx,
dy/dx = 2ax + b
Substituting dy/dx = 1 and x = 0 into the derivative function, we get
1 = 2a(0) + b
b=1
Substituting (2, 3) and b = 1 into the equation y = ax2 + bx, we get
3 = a(2)2 + 1(2)
a = 1/4
Thus , a = 1/4 , b = 1, c = 0.
Example 14
Find the equation of the normal to the curve f(x) = x3 + x – 2 at the
point x = 2.
•
Solution
f(x) = x3 + x – 2
f(2) = 23 + 2 – 2 = 8
f'(x) = 3x2 + 1
f'(2) = 3(2)2 + 1 = 13
Gradient of the normal is
m = – 1/13.
Substituting m = – 1/13 and (2, 8) into y = mx + c, we get
8 = (– 1/13) (2) + c
c = 106/13
Therefore, the required equation is y = – x/13 + 106/13 .
Example 15
Given y = (x2 + h)2 and its gradient is 8 at x = 2. Find the value
of h which is a constant.
• Solution
y = (x2 + h)2 = x4 + 2hx2 + h2
dy/dx = 4x3 + 4xh
Therefore, 8 = 4(2)3 + 8h
8h = 8 – 32
h = -24/8
h = -3
4.6 Critical Points
• The point x = c is called a critical point for a
continuous function, f if
(a) f' (c) = 0 or f' (c) fails to exist, and if
(b) f(c) is well defined.
• The critical points comprise the local maximum
point, local minimum point and inflection point.
Illustrations of some of the critical
points are shown below.
•
Maximum point
Minimum point
Inflexion point
Second derivative and first
derivative tests
• The two tests that can be used to examine the nature of the critical points are the
second derivative test and the first derivative test.
• Second derivative test
• We can use the second derivative test to determine the nature of the critical points.
Three steps are required in the second derivative test. The steps are given below.
1. Find f' (x) and f'' (x).
2. Let f' (x) = 0 and solve for critical values, c.
3. Substitute the critical values, c into f'' (x). If f'' (c) is
(a) positive, the point is a local minimum point.
(b) negative, the point is a local maximum point.
(c) zero, the test fails and the first derivative test has to be used.
First derivative test
• The first derivative test can also be used to determine the critical points.
• However, it is used only if the second derivative test fails.
• Three steps are required in the first derivative test.
1. Find f' (x).
2. Let f' (x) = 0 and solve for critical values, c.
3. Select points in the neighbourhood of c, one slightly less and one slightly more
than the critical value, c and substitute into f' (x).
• If f' (x) (a) changes sign from negative to positive, the point is a local minimum point.
(b) changes sign from positive to negative, the point is a local maximum point.
(c) does not change sign, the point is an inflection point.
Example 16
Determine the critical point(s) for the curve y = x2.
•
Solution
Using the second derivative test
f'(x) = 2x and f'' (x) = 2
Let f' (x) = 0
Hence, 2x = 0
x=0
When x = 0,
f(0) = 02 = 0
The point (0, 0) is a critical point.
When x = 0,
f''(0) = 2
Hence, the point (0, 0) is a local minimum point.
The graph of the curve is in page 249.
Example 17
Find the critical point(s) of the curve f(x) = 2x3 – 3x2 – 12x + 2.
•
Solution
Using the second derivative test
f'(x) = 6x2 – 6x – 12 and f''(x) = 12x – 6
Let f'(x) = 0,
Hence, 6x2 – 6x – 12 = 0
6(x + 1)(x – 2) = 0
x = –1 or 2
When x = –1, f(–1) = 2(–1)3 – 3(–1)2 – 12(–1) + 2 = 9
When x = 2, f(2) = 2(2)3 – 3(2)2 – 12(2) + 2 = –18
The points (–1, 9) and (2, –18) are critical points.
When x = –1, f''(–1) = –18.
Hence, the point (–1, 9) is a local maximum point.
When x = 2, f''(2) = 18.
Hence, the point (2, –18) is a local minimum point.
The graph of the curve is given in 250.
Example 18
Determine the critical point(s) for the curve f(x) = x4.
•
Solution
Using the second derivative test
f'(x) = 4x3 and f''(x) = 12x2
Let f'(x) = 0
Hence, 4x3 = 0, x = 0
When x = 0,
f'(0) = 4(0)3 = 0
The point (0, 0) is a critical point.
When x = 0,
f''(0) = 12(0)2 = 0
The second derivative test fails here. We have to use the first derivative test.
Solution (continue)
Using the first derivative test
We choose two numbers, –0.01 and + 0.01, which are
in the neighbourhood of x = 0.
When x = –0.01, f'(–0.01) = 4(–0.01)3 = negative value
When x = +0.01, f'(+0.01) = 4(0.01)3 = positive value
Since the sign changes from negative to positive, the
point (0, 0) is a local minimum point. The graph of
the curve is given in page 251.
Example 19
Determine the critical point(s) for the curve f(x) = x3.
•
Solution
Using the second derivative test
f'(x) = 3x2 and f''(x) = 6x
Let f'(x) = 0
Hence, 3x2 = 0,
When x = 0,
x=0
f(0) = 3(0)3 = 0
The point (0, 0) is a critical point.
When x = 0,
f'' (0) = 0
The second derivative test fails here. We have to use the first derivative test.
Using the first derivative test
We choose two numbers, –0.01 and +0.01, which are in the neighbourhood of x = 0.
When x = –0.01, f'(–0.01) = 3(–0.01)2 = positive value
When x = +0.01, f' (+0.01) = 3(+0.01)2 = positive value
Note that the sign does not change. Thus, the point (0, 0) is an inflection point. The graph of the curve is given in page 252.
4.7 Integration
• The process of finding the function when its
derivative is given is called integration (the opposite
of differentiation). The function so formed is called
the integral (the opposite of derivative).
• Thus, x2 is the integral of 2x as d/dx(x2) is 2x.
• In general, if f and F are functions such that F(x) =
∫f(x)dx for all x in the domain of f, then F(x) is said
to be an integral of f(x).
Example 20
• I f f(x) = 3/2 x2 + x and F(x) = x3/2 + 1/2 x2, then
F(x) is an integral of f(x).
Indefinite integral
• Let F(x) be an integral of the function f(x). Then, we write ∫f(x)dx = F(x) + c
which is read as ‘the integral of f(x), dx is equal to F(x) + c where c is a constant’.
• The function f(x) to be integrated is called the integrand and the symbol ∫, the
integral sign.
• Some formulae to find indefinite integrals are listed below.
For any constants k and c,
(a) ∫kdx = kx + c
(b) ∫xn dx = xn+1/(n+1) + c
(c) ∫kf(x)dx = k∫f(x)dx
(d) ∫[f(x) + g(x)]dx = ∫f(x)dx +∫g(x)dx
(e) ∫[f(x) - g(x)]dx = ∫f(x)dx -∫g(x)dx
Example 21
Integrate the following.
(a) 9
(b) x4
(c) 1/x2
(d) (x – 1/x)2
( e) (x3 + x2 + x) /x
(f) 2x1/2 – 3x2
Solution
(a) ∫9 dx = ∫9x0dx = 9x0 + 1 + c = 9x + c
(b) ∫x4dx = x4 + 1 /5 + c
= x5 /5 + c
(c) ∫1/x2 dx = ∫x–2dx
= x–2+1 /(-1) + c
= -x –1
+c
= 1/x + c
(d) ∫(x – 1/x) 2dx = ∫(x2 + x–2 –2)dx
= x3/3– 1/x - 2x + c
(e) ∫[(x3 + x2 + x)/x ]dx = ∫(x2 + x + 1)dx = x3/3 + x2/2 + x + c
(f) ∫(2x1/2 – 3x2) = 4/3(x3/2 ) – x3 + c
Definite integral
• The integral of f(x) from x = a to x = b is denoted
by ∫a b f(x)dx and is defined by
∫a b f(x)dx = F(b) – F(a).
Example 22
Evaluate
(a) ∫12 x3dx
(b) ∫
1
3
(x4 + 1)dx
(c) ∫23(x + 2)2dx
(d) ∫23 1/x2 dx
Solution
(a) ∫12 x3 dx = [x4/4]12 = 24/4 – 14/4 = 3.75
(b) ∫13 (x4 + 1)dx = [x5/5 + x ]13
= [35/5 + 3] – [15/5 + 1]
= 50.4
(c) ∫23 (x + 2)2dx = ∫23 (x2 + 4x + 4) dx
= [x3/3 + 2x2 + 4x]23
= [33/3 + 2(3)2 + 4(3)] – [23/3 + 2(2)2 + 4(2)]
= 20 1/3
(d) ∫23 1/x2 dx = ∫23 x-2 dx
= [-1/x]23
= [-1/3] – [-1/2]
= 1/6
4.8 Applications of calculus in
economics and business
• Calculus is used extensively in business and economics.
• For instance, a company may want to maximize profit,
maximize revenue or minimize cost.
• An enterprise may want to determine the best price for its
products so as to optimize its sales or profits.
• A business entity may want to optimize its advertising
expenses to achieve maximum market exposure.
• Calculus can also be used to determine the optimum
density of cultured shrimps and fish in ponds for an
optimum output.
Total cost, average cost and
marginal cost
•
The total cost of production comprises the fixed costs and the variable costs, that is
total cost = fixed costs + variable costs
•
Variable costs are costs that depend on the level of production. Labour and material costs are examples of variable
costs.
•
Fixed costs are costs incurred before the products are produced. An example of a fixed cost is the cost of
machinery.
•
If C(x) represents the total cost of producing x units of a product, then
average cost = AC(x) = C(x)/x
marginal cost = MC(x) = C' (x)
•
Marginal cost is the rate of change in cost per unit change in production
•
at an output level of x units.
•
Simply stated, it is the cost of producing an additional unit of the product.
•
If C' (x) is the marginal cost of producing x units of a product, then the
total cost of producing x units is given by ∫C' (x)dx.
Example 23
A firm produces x units of a product per month and the
total cost per month in ringgit is given by C(x) = 150 +
0.02x2. Find
(a) the total cost when 20 units are produced,
(b) the average cost function,
(c) the average cost when 20 units are produced,
(d) the marginal cost function,
(e) the marginal cost when 20 units are produced.
Interpret the answer.
Solution
(a) C(x) = 150 + 0.02x2
C(20) = 150 + 0.02(20)2 = RM158
The cost of producing 20 units is RM158.
(b)
AC(x) = (150 + 0.02x)/x
= 150/x + 0.02x
(c)
(d)
AC(20) = 150/20 + 0.02(20) = RM7.90
C' (x) = d/dx(150 + 0.02x2)
C’(x) = 0.04x
(e)
C' (20) = 0.04(20) = RM0.80
It implies that the 21st unit will cost approximately RM0.80 to produce. Note that the marginal
analysis is an approximate. The actual cost of producing the 21st unit is
C(21) – C(20) = 158.82 – 158 = RM0.82
Example 24
The marginal cost, C'(x) for a certain product is given as
C'(x) = 40 – 0.07x where x is the level of output. Find
(a) the total cost function if fixed costs are RM4,000,
(b) the cost to produce 200 units.
Solution
(a) C(x) = ∫C'(x)dx
= ∫(40 – 0.07x)dx
= 40x – 0.035x2 + c,
where c is a constant
To determine c, we use the information that when x = 0, C(0) = RM4,000. Substituting these two
values into the cost expression, we get
RM4,000 = 40(0) – 0.035(0)2 + c
c = 4,000
Hence, the total cost function is C(x) = 40x – 0.035x2 + 4,000
(b) C(200) = 40(200) – 0.035(200)2 + 4,000
= RM10,600
Hence, the total cost incurred when 200 units are produced is
RM10,600.
Example 25
The total cost in producing 50 units of a product is RM3,000. If the marginal cost, C'(x) for the
product is given as C'(x) = RM(20 + 0.4x), where x is the level of output, find the total cost function.
•
Solution
C(x) = ∫C'(x)dx
= ∫(20 + 0.4x)dx
= 20x + 0.2x2 + c
When x = 50, C(50) = 3,000. Substituting these two values into the cost
expression, we get
3,000 = 20(50) + 0.2(50)2 + c
c = 1,500
Hence, the total cost function is C(x) = 20x + 0.2x2 + 1,500.
Total revenue, average revenue
and marginal revenue
• For a given demand function p = f(x) where p is the price per
unit and x is the number of units demanded, the total revenue,
R(x) is the product of unit price and the quantity x; that is R(x)
= px.
• If the total revenue function R(x) is known, then the demand
function can be obtained by
x = R(x) /p
• Marginal revenue is the rate of change in revenue per unit
change in production at an output level of x units of the
product.
• In other words, marginal revenue is the additional revenue
obtained when an additional unit of the product is produced
• Average revenue, AR(x) is the total revenue divided by
the total quantity sold, that is
AR(x) = R(x)/quantity x
• If marginal revenue function R'(x) is known, the total
revenue function R(x) can be obtained by integrating the
marginal revenue function, that is
R(x) = ∫R'(x)dx
The arbitrary constant of the integration (c) is usually
zero since when x = 0, revenue is also nil.
Example 26
If the total revenue per month in ringgit, R(x) is given by
R(x) = 20x –0.03x2
where x is the number of units produced and sold per month, find
(a) the total revenue when 30 units are produced and sold,
(b) the average revenue function,
(c) the average revenue when 30 units are produced and sold,
(d) the marginal revenue function,
(e) the marginal revenue function when 30 units are produced and
sold. Interpret the answer.
Solution
(a)
R(x) = 20x – 0.03x2
R(30) = 20(30) – 0.03(30)2 = RM573
(b) AR(x) = R(x)/x = (20x – 0.03x2)/x
= 20 – 0.03x
(c) AR(30) = 20 – 0.03(30) = RM19.10
(d) R' (x) = d/dx(20x – 0.03x2 ) = 20 – 0.06x
(e) R' (30) = 20 – 0.06(30) = RM18.20
It implies that the revenue from the sale of the 31st unit is approximately RM18.20.
The actual revenue from the sale of the 31st unit is R(31) – R(30) = RM591.17 –
RM573
= RM18.17
Example 27
The marginal revenue, R'(x) of a certain product is given as
R'(x) = 40 – 0.02x where x is the level of output. Find
(a) the total revenue function,
(b) the demand function.
•
Solution
(a) R(x) = ∫R'(x)dx
= ∫(40 – 0.02x)dx
= 40x – 0.01x2 + c
where c is a constant
To determine c, we use the information that revenue is nil when no units are sold, that is, when x = 0, R(0) = 0.
Substituting these two values into the revenue expression, we get
0 = 40(0) – 0.01(0)2 + c
c=0
Hence, the revenue function is R(x) = 40x – 0.01x2.
Solution (continue)
(b) R(x) = px
40x – 0.01x2 = px
(40x – 0.01x2)/x = p
p = 40 – 0.01x
Example 28
The marginal revenue function, R'(x) of a product is given as R'(x)=
60 – 0.5x where x is the level of output. Find the change in total
revenue when production level increases from 60 units to 70 units.
• Solution
Change in total revenue
= ∫6070 (60 – 0.5x)dx
= [60x – 0.25x2]6070
= [60(70) – 0.25(70)2] – [60(60) – 0.25(60)2]
= RM275
Profit, breakeven analysis and
optimisation
• If the total revenue function, R(x) and the total cost function, C(x) are known, then
the total profit function, P(x) is given by
P(x) = R(x) – C(x)
• The breakeven level of production is given by
R(x) = C(x) or P(x) = 0
• Differentiating the total profit function P(x) = R(x) – C(x), we get marginal profit
P’(x)
P' (x) = R' (x) – C' (x)
• For maximum profit, P' (x) = 0. Thus,
R' (x) – C' (x) = 0 or R' (x) = C' (x)
• For a firm to obtain maximum profit, marginal revenue must be equal to marginal
cost.
Example 29
The demand for an item produced by Weelux is given by p + 0.2x
= 100 where p is the price per unit and x is the quantity demanded.
The total cost, C(x) of producing x units of the item is given by
C(x) = 800 + 30x where x is the level of output. Find
(a) the total revenue function,
(b) the total profit function,
(c) the total profit when 100 units are sold,
(d) the marginal profit function,
(e) the marginal profit when (i) 90 units are sold (ii) 300 units are
sold.
Interpret the answers.
Solution
From p + 0.2x = 100, we get
p = 100 – 0.2x
(a) R(x) = px
= (100 – 0.2x)x
= 100x – 0.2x2
(b) P(x) = R(x) – C(x)
= (100x – 0.2x2) – (800 + 30x)
= 70x – 0.2x2 – 800
(c) P(100) = 70(100) – 0.2(100)2 – 800
= RM4,200
(d) P'(x) = d/dx(70x – 0.2x2 )
P’(x) = 70 – 0.4x
Solution (continue)
(e) i. P' (90) = 70 – 0.4(90) = RM34
It implies that when 90 units are sold, the marginal profit
(extra profit per extra item when production is increased
by a small quantity) is RM34 per additional unit.
ii. P' (300) = 70 – 0.4(300) = –RM50
It implies that when 300 units are sold, a small increase in
the production results in a loss (negative profit) of RM50
per extra unit.
Example 30
The revenue and cost functions in ringgit of a product are given as
R(x) = 600x – x2,
C(x) = 200 + 6x + 2x2
where R(x) = total revenue function
C(x) = total cost function
x = output produced and sold
Currently the company is producing 100 units of the product. Should the company increase or
decrease production?
Solution
P(x) = (600x – x2) – (200 + 6x +2x2)
= 594x – 3x2 – 200
P' (x) = 594 – 6x
P' (100) = 594 – 6(100) = –RM6
The company should reduce production as marginal profit is negative.
Example 31
A product is sold at RM5 a unit. The total cost function, C(x) in ringgit for the product when x units are produced
and sold is given as C(x) = 60 + 1.2x + 0.001x2 where x is the level of output. Find the level of output which will
maximise profit. What is this maximum profit?
•
Solution
C(x) = 60 + 1.2x + 0.001x2
R(x) = 5x
P(x) = 5x – (60 + 1.2x + 0.001x2)
P(x)= 3.8x – 0.001x2 – 60
P'(x) = 3.8 – 0.002x and P''(x) = –0.002
Setting P'(x) = 0, we get
3.8 – 0.002x = 0
x = 1,900 units
Since P''(x) is negative, at x = 1,900, there is a local maximum.
P(maximum) = 3.8(1,900) – 0.001(1,900)2 – 60
= RM3,550
Example 32
The following is given for a product of a company.
Marginal cost function: C'(x) = RM(20 + 0.2x) where x is the quantity sold
Fixed cost : RM500
Demand function is linear:
At the price of RM78, 30 units are sold.
At the price of RM58, 50 units are sold.
Find
(a) the demand function,
(b) the total cost function,
(c) the total revenue function,
(d) te level of production in which profit is maximised,
(e) the maximum profit, price per unit, total revenue and total cost
when profit is maximised,
(f) the level of production in which total revenue is maximised,
(g) the profit, price per unit, total revenue and total cost when total
revenue is maximised.
Solution
(a) Let the demand function be p = mx + c. Hence,
78 = 30m + c (1)
58 = 50m + c (2)
Solving for m and c, we get
m = –1 and c = 108
Therefore, the demand function is p = –x + 108 or p = 108 – x.
(b) C(x) = ∫C'(x)dx
= ∫(20 + 0.2x)dx
= 20x + 0.1x2 + c
Substituting x = 0 and C(0) = 500 into the cost expression, we get c = 500
Therefore, the cost function is
C(x) = 20x + 0.1x2 + 500.
(c) R(x) = px
= (108 – x)x
Revenue function is R(x) = 108x – x2
Solution (continue)
(d) There are two methods to use.
Method 1
P(x) = R(x) – C(x)
P(x) = [108x – x2]– [20x + 0.1x2 + 500]
P(x) = 88x – 1.1x2 – 500
Differentiating the total profit function and setting it to zero, we get
P' (x) = 88 – 2.2x
88 – 2.2x = 0
x = 40 units
Note that P''(x) is negative. This indicates that profit is maximised at x = 40.
Solution (continue)
Method 2
R' (x) = 108 – 2x
C' (x) = 20 + 0.2x
Setting R'(x) = C'(x),
108 – 2x = 20 + 0.2x
x = 40 units.
(e) To find maximum profit, substitute x = 40 into the profit function
P(40) = 88(40) – 1.1(40)2 – 500 = RM1,260
To find the price per unit, substitute x = 40 into the demand function
p = 108 – 40 = RM68
To find the total revenue, substitute x = 40 into the total revenue function.
R(40) = 108(40) – (40)2
= RM2,720
To find the total cost, substitute x = 40 into the total cost function.
C(40) = 20(40) + 0.1(40)2 + 500
= RM1,460
Solution (continue)
(f) R(x) = 108x – x2
R'(x) = 108 – 2x
Setting R9(x) = 0, we get
108 – 2x = 0
x = 54
Thus, the revenue is maximum when x = 54.
(g) To find the profit, substitute x = 54 into the profit function
P(54) = 88(54) – 1.1(54)2 – 500 = RM1,044.4
To find the price per unit, substitute x = 54 into the demand function
p = 108 – 54 = RM54
To find the total revenue, substitute x = 54 into the total revenue function
R(54) = 108(54) – (54)2
= RM2,916
To find the total cost, substitute x = 54 into the total cost function
C(54) = 20(54) + 0.1(54)2 + 500
= RM1,871.60
Note that to maximise revenue, the price is set at RM54, which is lower than the price of RM68 set for profit maximisation.