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Transcript
```MATH103
Economics - I
Chapter 1 Applications of
Equation and inequality
Procedure for Solving
Word Problems
1.
2.
3.
4.
5.
Read the problem carefully and introduce a variable to
represent an unknown quantity in the problem.
Identify other quantities in the problem (known or
unknown) and express unknown quantities in terms of the
variable you introduced in the first step.
Write a verbal statement using the conditions stated in the
problem and then write an equivalent mathematical
statement (equation or inequality.)
Solve the equation or inequality and answer the questions
posed in the problem.
Check that the solution solves the original problem.
Section
EXAMPLE 1

PROFIT
Anderson Company produces a product for which the variable
cost per unit is \$6 and fixed cost is \$80,000. Each unit has a
selling price of \$10. How many units must be sold for the
company to earn a profit of \$60,000?
EXAMPLE 2
Analyzing Investments
EXAMPLE 3
Analyzing Investments
Tyrick invests \$15,000, some in stocks and the rest in
bonds. If he invests twice as much in stocks as he does in
bonds, how much does he invest in each?
Solution
Step 2
Let x = the amount invested in stocks. The rest
of the \$15,000 investment (\$15,000 – x) is
invested in bonds. We have one more
important piece of information to use:
Twice
the
amount
Amount invested
=
invested in bonds,
in stocks, x
15,000 – x
EXAMPLE 3
Step 3
Analyzing Investments
x  2 15, 000  x 
Replace the verbal
description with algebraic
expressions.
Step 4
x  30, 000  2x
3x  30, 000
x  10, 000
Distributive property
Divide both sides by 3.
Step 5
Tyrick invests \$10,000 in stocks and \$15,000
– \$10,000 = \$5000 in bonds.
Step 6
Tyrick’s total investment is \$10,000 + \$5000 =
\$15,000, and \$10,000 (stocks) is twice \$5,000
(bonds).
EXAMPLE 4
Solving Problems Involving Simple Interest
Ms. Sharpy invests a total of \$10,000 in blue-chip and
technology stocks. At the end of a year, the blue-chips
returned 12% and the technology stocks returned 8% on
the original investments. How much was invested in each
type of stock if the total interest earned was \$1060?
Solution
Step 1
Step 2
We are asked to find two amounts: that
invested in blue-chip stocks and that invested
in technology stocks.
If we know how much was invested in blue-chip
stocks, then we know that the rest of the
\$10,000 was invested in technology stocks.
Let x = amount invested in blue-chip stocks. Then
10,000 – x = amount invested in technology
stocks.
Invest
P
r
t
I = Prt
Blue
x
0.12
1
0.12x
Tech
10,000 – x
0.08
1
0.08(10,000 – x)
Interest from
Blue-chip
Step 3
Step 4
Interest from
+ technology
=
Total
Interest
0.12x  0.08 10, 000  x   1060
12x  8 10, 000  x   106, 000
12x  80, 000  8x  106, 000
4x  26, 000
\$ in blue-chip stocks x  6500
10, 000  x  10, 000  6500
\$ in technology stocks
 3500
Step 5
Ms. Sharpy invests \$3500 in technology stocks
and \$6500 in blue- chip stocks.
Step 6
\$3500  \$6500  \$10,000
12% of \$6500  \$780
8% of \$3500  \$280
Total interest earned = \$1060.
EXAMPLES] A fence is to enclose a rectangular area of 800 square feet. The
length of the plot is twice the width. How much fencing must be used?
Solution: Let w = width of the plot then 2w = length of the plot
So the width of the plot is 20 ft and the length of the plot is 40ft.
So 20+20+40+40 = 120 ft of fencing must be used.
EXAMPLES] A builder makes concrete by mixing 1 part clay, 3 parts sand and
5 parts crushed stone. If 765 cubic feet of concrete are needed, how much of
each ingredient is needed?
Solution: Let x = amount of clay, 3x = amount of sand and 5x = amount of crushed
stone.
So 85 cubic feet of clay, 255 cubic feet of sand and 425 cubic feet of stone.
EXAMPLES] A rectangular plot 4 meters by 8 meters is to be used for a
garden. The owner wants a border so that 12 square meters of the plot is left
for flowers. How wide should the border be?
Solution: Let X = the width of the border then 4-2x = the width of the plot and
8-2x = the length of the plot
So the border should be 1m wide
EXAMPLES] A company wants to know the total sales units required to earn
a profit of \$100,000. The unit selling price is \$20, the variable cost per unit is
\$15, and the total fixed costs are \$600,000. Find the sales units required.
Solution:Let q = the number of units then the cost = 15q+600000 and
revenue = 20q
So 140,000 units would need to be sold.
EXAMPLES] Some workers went on strike for 46 days. Before this they
earned \$7.50 per hour and worked 260 eight-hour days a year. What percent
increase is needed in yearly income to make up for the lost time in 1 year?
Solution: Let x=the percent increase then x/100 is the decimal form of the
percentage
So about 17.7% increase will be needed.
EXAMPLES]\$20000 is invested in two enterprises so that the total income
per year will be \$1440.One enterprise pays 6% annually the other 7%. How
much was invested into each enterprise?
Solution: Let x = amount invested in 6% enterprise and 20000-x = amount in 7%
enterprise
So \$4000 was invested at 6% and \$16000 was invested at 7%
EXAMPLES]The price of a product is p dollars each. Suppose a manufacturer
will supply 3p2-4p units of product to the market and consumers will demand
24-p2 units. Find the value of p for which the supply will equal the demand.
Solution:
So the price is \$3 per product unit.
EXAMPLES] A circular ventilation duct with diameter 140 mm is attached to a
square duct system. To ensure smooth air flow the areas of the circle and
square sections must be equal. Find the length of the square side of the
section to the nearest millimeter.
Solution: Let x = the length of the side of the square then As= x2 and As=Ac
So the length of the side of the square section is 124mm.
Section 1.2
An inequality is a statement that one algebraic expression is less
than, or is less than or equal to, another algebraic expression.
If the equality symbol = in a linear equation is replaced by an
inequality symbol (<, >, ≤, or ≥), the resulting expression is called a
first-degree, or linear, inequality. For example
x
5  1  3x  2 
2
is a linear inequality.
A linear inequality in one variable is an inequality that is
equivalent to one of the forms
ax + b < 0 or ax + b ≤ 0,
where a and b represent real numbers and a ≠ 0.
Linear Inequalities in One Variable
Interval Notation is used to write solution sets of inequalities.
Note: A parenthesis is used to indicate an endpoint in not
included. A square bracket indicates the endpoint is included.
Interval Notation
Type of Interval
Set
Graph
Interval Notation
{x| a < x }
(a, )
(
a
{x | a < x < b}
(a, b)
(
a
{x | x < b}
(-, b)
{x | x is a real number}
(-, )
Open Interval
)
b
)
b
Interval Notation
Type of Interval
Set
Interval Notation
Graph
{x| a  x}
[a, )
[
a
{x | a < x  b}
(a, b]
(
a
]
b
{x | a  x  b}
[a, b)
[
a
)
b
{x | x  b}
(-, b]
Half-open
Interval
]
b
Interval Notation
Type of Interval
Closed Interval
Set
{x| a  x  b}
Interval Notation
[a, b]
Graph
[
a
]
b
• Example 1: Solve k – 5 > 1
k – 5 +5 > 1 + 5
k>6
Solution set: (6,)
• Example 2: Solve 5x + 3  4x – 1 and graph the solution set.
5x – 4x  -1 –3
x  -4
[
Solution set: [-4,)
-4
• Example 3: Solve -2x < 10
x > -5
Solution set: (-5,)
• Example 4: Solve 2x < -10
x < -5
Solution set: (-,-5)
• Example 5: Solve –9m < -81 and graph the solution set
m>9
(
Solution set: (9, )
9
Solving and Graphing Linear Inequalities
EXAMPLE 6
Solve the inequality
3(x-1) < 5(x + 2) - 5
Solution:
3(x-1) < 5(x + 2) - 5
3x - 3 < 5x + 10 - 5
Distribute the 3 and the 5
3x - 3 < 5x + 5
Combine like terms.
-2x < 8
Subtract 5x from both sides,
and add 3 to both sides
x > -4
Notice that the sense of the inequality
reverses when we divide both sides by -2.
x > -4 is equivalent to (-4, ∞)
(
-4
x
EXAMPLE 7
Solving and Graphing Linear Inequalities
Solve the inequality
2(x-3) < 4
Line Notation
Solution Set = { x: x < 5 }
Set Notation
(-∞, 5)
Interval Notation
EXAMPLE 7
Solving and Graphing Linear Inequalities
Solve the inequality
3-2x  6
Set Notation
Solution Set = { x: x >= 5 }
Line Notation
EXAMPLE 8
Solving and Graphing Linear Inequalities
Solve the inequality
2(x-4)-3 > 2x-1
Example 9: Solve 6(x-1) + 3x  -x –3(x + 2) and graph the solution set
Step 1: 6x-6 + 3x  -x –3x – 6
9x - 6  –4x – 6
Step 2:
13x  0
[
Step 3:
x  0 Solution set: [0, )
0
Example 10: Solve
1
3
(m  3)  2  ( m  8)
4
4
1
3
3
m 2 m6
4
4
4
1
11 3
24
m  m
4
4 4
4
1
3
24 11
m m

4
4
4
4
2
13
 m
4
4
13
m
2
Solution set: [  13 , )
2
[
-13/2
and graph the solution set
EXAMPLE 8
Solving and Graphing Linear Inequalities
Solve the inequality
Line Notation
3
 s  2   1  2  s  4 
2
Section 1.3 Applications of Inequalities
EXAMPLES] A company manufactures a product that has a unit selling price
of \$20 and a unit cost of \$15. If fixed costs are \$600,000, determine the least
number of units that must be sold for the company to have a profit.
Solution: Let the x be the number of units then 20x = total revenue and
15x+600000 = total cost
So we need at least 120001 units to make a profit
EXAMPLES] A company invests \$30,000 of surplus funds at two annual rates:
5% and 6%. The company wishes to have an annual yield of no less than
6%.What is the least amount of money the company can invest at 6%?
Solution: Let x = amount at 6% then 30000-x= amount at 5%
So need to invest at least \$25714.29
EXAMPLES]The cost of publication of each copy of a magazine is \$0.65. It is
sold to dealers for \$0.60 each and the amount received for advertising is 10%
of the amount received for all magazines issued beyond 10000. What is the
least number of magazines that can be published without loss?
Solution:Let x = the number of units then 0.65x = total cost and
0.6x+0.1(0.6)(x-10000) = Total revenue
So at least 60,000 magazines must
be published.
EXAMPLES]At present a manufacturer has 2500 units of a product in stock.
This month, the product sells for \$4 a unit. Next month the unit price will
increase by \$0.50. The manufacturer wants the total revenue received from
the sale of 2500 units to be no less than \$10,750. What is the maximum
number of units that can be sold this month?
Solution:Let x = the number of units then the total revenue is 4x+4.5(2500-x)
So need to sell a max of 1000 units this month.
Given
Find the roots by equating to zero,
=0
Solve by using either Factoring or Quadratic formula
Suppose
and
Suppose
=
are the roots
are the roots
Ex: Solve
Solution :
Solution set
2
x
Example : Consider the inequality  x  6  0
Solution: We can find the values where the quadratic equals zero
by solving the equation,
x2  x  6  0
x  3x  2  0
x  3  0 or x  2  0
x  3 or x  2
2
x
For the quadratic inequality,  x  6  0
we found zeros 3 and –2 by solving the equation
x 2  x  6  0 . Put these values on a number line and we can
see three intervals that we will test in the inequality. We will
test one value from each interval.
-2
3
Interval
 ,2
 2, 3
 3, 
Test Point
x  3
Evaluate in the inequality
True/False
x2  x  6  0
 3   3  6  9  3  6  6  0 True
2
x2  x  6  0
x0
02  0  6  0  0  6  6 < 0
False
x2  x  6  0
x4
42  4  6  16  4  6  6  0
True
Thus the intervals  ,2 or 3,  make up the solution set for
the quadratic inequality, x 2  x  6  0 .
In summary, one way to solve quadratic inequalities is to find the
zeros and test a value from each of the intervals surrounding the
zeros to determine which intervals make the inequality true.
Example:Solve 2 x 2  3x  1  0
Solution: First find the zeros by solving the equation,
2 x 2  3x  1  0
2 x 2  3x  1  0
2x 1x 1  0
2 x  1  0 or x  1  0
1
x  or x  1
2
Now consider the intervals around the zeros and test a value
from each interval in the inequality.
The intervals can be seen by putting the zeros on a number
line.
1/2
1
Interval
Test Point
Evaluate in Inequality
True/False
2 x 2  3x  1 < 0
1

  , 
2

x0
20  30  1  0  0  1  1  0
2
False
2 x 2  3x  1 < 0
1 
 ,1
2 
9 9
1
3 3
2   3   1    1  < 0
8 4
8
4 4
2
x
3
4
True
2 x 2  3x  1 < 0
1, 
x2
Thus the interval
the inequality
22  32  1  8  6  1  3  0
2
1 
 ,1
2 
False
makes up the solution set for
2 x 2  3x  1  0 .
Ex: Solve
Solution :
There are no real roots, thus
Solution set is empty set, {}
Ex: Solve
Solution :
No real roots
Solution set =
Practice Problems
x 2  5 x  24  0
12  x  x 2  0
3x 2  5 x  2 < 0
5 x 2  13x  6 < 0
9  x2  0
2 x 2  5x  1 < 0
16 x 2  1  0
x 2  5 x < 4
3x 2  2 x  1  0
x2  2x  4
```