This document discusses marginal analysis in business and economics. It provides an example problem involving the demand and cost functions of producing computer sets. It asks to: (1) Express price as a function of demand, (2) Find marginal cost, (3) Find revenue function and its domain, (4) Find marginal revenue, (5) Interpret marginal revenue at two production levels, (6) Graph cost and revenue and find break-even points, (7) Find profit function, (8) Find marginal profit, and (9) Interpret marginal profit at two production levels. The solutions provide mathematical derivations and economic interpretations.

1. MARGINAL ANALYSIS
IN BUSINESS AND ECONOMICS

2. MARGINAL ANALYSIS
■ An analysis of problems that involves the instantaneous rate
of change of one variable with respect to another.
■ The instantaneous rate of change or the first derivative of
an economic marginal function such as total cost, total
revenue or total profit represents the marginal function,
that is, marginal cost, marginal revenue or marginal profit.
■ When Revenue exceeds Cost, greater sales yields greater
profits.
■ When Revenue falls below Cost, greater sales will lessen
profits.

3. Example
The price-demand equation and the cost
function for the production of computer sets are
given, respectively by
x=9000-30p and C(x)=150000+30x
Where x is the number of sets that can be sold at a
price of $p per set and C(x) is the total cost (in
dollars) of producing x sets.

4. Find the following:
A. Express the price p as a function of the demand x and find the
domain of this function.
B. Find the marginal cost.
C. Find the Revenue function and state its domain.
D. Find the marginal revenue.
E. Find R’(3000) and R’(6000) and interpret the results.
F. Graph the cost function and the revenue function on the same
coordinate system for 0≤x ≤9000. Find the break-even points and
indicate regions of loss and profit.
G. Find the profit function in terms of x.
H. Find the marginal profit.
I. Find P’(1500) and P’(4500) and interpret the results.

5. A. Express the price p as a function of
the demand x and find the domain of
this function.
SOLUTION:
x=9000-30p
30p=9000-x
p=
9000−𝑥
30
p=300 --
𝟏
𝟑𝟎
x
Domain: p≥0
300 --
1
30
x ≥0
--
1
30
x ≥-300
-- x ≥
−300
−
1
30
x≤9000 units
Therefore, the domain is:
0 ≤x ≤9000 or
[0, 9000]or
{x│ 0 ≤x ≤9000}

6. B. Find the marginal cost.
Solution:
C(x)=150000+30x
C’(x)=30

7. C. Find the Revenue function and
state its domain.
Solution:
R(x)=xp
R(x)=x(300 --
1
30
x)
R(x)=300x --
𝟏
𝟑𝟎
𝒙𝟐
Domain: R(x)≥0
x(300 --
1
30
x) ≥0
● x≥0
300 --
1
30
x ≥0
--
1
30
x ≥-300
-- x ≥
−300
−
1
30
● x≤9000
Therefore, the domain is:
0 ≤x ≤9000 or
[0, 9000]or
{x│ 0 ≤x ≤9000}

8. D. Find the marginal revenue.
Solution:
R(x)=300x --
1
30
𝑥2
R’(x)=300--
𝟏
𝟏𝟓
𝒙

9. E. Find R’(3000) and R’(6000) and
interpret the results.
Solution:
* R’(3000)
R’(3000)=300--
𝟏
𝟏𝟓
𝟑𝟎𝟎𝟎
R’(3000)=300-- 𝟐𝟎𝟎
R’(3000)=100
Interpretation:
R(x) is increasing when production level is at 3000 units.

10. E. Find R’(3000) and R’(6000) and
interpret the results.
Solution:
* R’(6000)
R’(6000)=300--
𝟏
𝟏𝟓
𝟔𝟎𝟎𝟎
R’(6000)=300-- 4𝟎𝟎
R’(6000)=-100
Interpretation:
R(x) is decreasing when production level is at 6000 units.

11. F. Graph the cost function and the
revenue function on the same coordinate
system for 0≤x ≤9000. Find the break-
even points and indicate regions of loss
and profit.
Solution:
• BREAK-EVEN POINTS: C(x)=R(x)
C(x)=150000+ 30x
R(x)=300x --
1
30
𝑥2
So, C(x)=R(x)
150000+ 30x=300x --
1
30
𝑥2
--
1
30
𝑥2+270x – 150000=0
Using your SciCal:
𝑥1 = 600
𝑥2 = 7500
Therefore, BEPs are:
BEP1 (600, $168000)
BEP2 (7500, $375000)

12. Graph
x 0 600 3000 6000 7500 9000
R(x) 0 168000 600000 600000 375000 0
C(x) 150000 168000 240000 330000 375000 420000

13. G. Find the profit function in terms
of x.
Solution:
P(x)=R(x)-C(x)
P(x)= (300x --
𝟏
𝟑𝟎
𝒙𝟐) – (150000+30x)
P(x)= 300x --
𝟏
𝟑𝟎
𝒙𝟐
-- 150000 –30x
P(x)=
−𝟏
𝟑𝟎
𝒙𝟐
+270x-150000

14. H. Find the marginal profit.
Solution:
P(x)=
−𝟏
𝟑𝟎
𝒙𝟐+270x-150000
P’(x)=
−𝟏
𝟏𝟓
𝒙+270

15. I. Find P’(1500) and P’(4500) and
interpret the results.
Solution:
P’(1500)
P’(1500)=
−𝟏
𝟏𝟓
(𝟏𝟓𝟎𝟎)+270
P’(1500)=170
Interpretation:
P(x) increases at x=1500 units.

16. I. Find P’(1500) and P’(4500) and
interpret the results.
Solution:
P’(4500)
P’(4500)=
−𝟏
𝟏𝟓
(𝟒𝟓𝟎𝟎)+270
P’(4500)=-30
Interpretation:
P(x) decreases at x=4500 units.