Bahan Mata Kuliah Matematika Ekonomi Gratis Terbaru - Kosngosan Situs Anak Kost Mahasiswa Pelajar Course 5
EQUATION IN ECONOMICS
(Course 5)
OLEH
SYAIFUL HADI
DJAIMI BAKCE
JURUSAN AGRIBISNIS
FAKULTAS PERTANIAN
UNIVERSITAS RIAU
QUADRATIC EQUATION
In economic modelling we work with the simples function that can
adequately represent a relationship, but sometimes a curve
rather than a straight line is required.
We used quadratic functions to represent total revenue and
average variable cost
General form : y = ax2 + bx + c
where a, b, and c are constants. When you sketch a quadratic
function you find it has either a hill or U shape and that generally
two values of x give the same value of y.
A quadratic function has a term in x2 but no higher powers of x.
A quadratic equation you can solve it graphically or sometimes
by factorizing it or by using the formula.
We first write it in the form ax2 + bx +c = 0. The value(s) of
x for which this equation is true can be found graphically by
plotting
y= ax2 + bx + c and looking y = 0.
Algebraic methods are more accurate than graphical ones,
but the
squared term means we need a special technique. Sometimes
factori- zing the expression helps. Consider for example:
5x2- 20x = 0
Since each term is divisible by 5x we can factorize the left hand
side and write: 5x(x-4) = 0
There are now two term multiplied together, 5x and (x-4). For
their product to be 0, one of term must be zero. This mean
either 5x = 0 or (x-4) =0.
If 5x = 0, dividing by 5 we find that x = 0 and this one possible
solution to the equation.
If x – 4 = 0, adding 4 to both sides we find that x = 4, which is
the other possible solution.
The graph of the function is plotted It also shows that x = 0
and x = 4 are two solutions to the equation, since they are the
values of x at which at which the curve cuts the x axis
Y = 5x2 – 20x
A more general method for solving the quadratic equation ax2+bx
+c =0 uses a formula:
-b b2 – 4ac
x = ------------------2a
Where a is the coefficient of x2, b is the coefficient of x and c is the
constant term.
Example: 8x2 – 20x + 3 = 0
We identify a= 8, b= -20 and c= 3. Notice that the sign of the
coefficient must be included. Begin the calculating the expression
the square root sign. This gives:
b2 – 4ac = (-20)2 – (4 . 8 . 3) = 400 – 96 = 304
We take the square root of this value and obtain 17.436. substituting
this result , -b and a into the formula gives:
20 17.436
x = -------------16
x = 37.436/16= 2.34 or x = 2.564/16 = 0.16
The solution of the equation is x = 2.34 or 0.16 (cari turunan
pertama ato kedua yang bernilai negatif)
INTERSECTION OF MC WITH MR OR
AVC
Quadratic equation arise in economics when we want to
discover where a quadratic function, say marginal cost,
cuts another quadratic function, say average variable
cost, or cuts a linier function, say marginal revenue.
We equate the two functional expressions, then subtract
the right hand side from both sides so that the value on
the right becomes zero.
After collecting term we solve the quadratic equation
using one of the methods explained above.
A firm has the marginal cost function MC = 3Q2 – 32Q +
96 and marginal revenue function MR = 236 – 16Q. Find
the firm’s profit maximizing output.
To maximize profits the firm chooses to produce where marginal cost
equals marginal revenue. Equating the MC and MR functions we have that:
3Q2 – 32Q + 96 = 236 – 16Q
Subtracting the right hand side from both sides gives:
3Q2 – 32Q + 96 – (236 – 16Q) = 0
Removing the bracket gives
3Q2 – 32Q + 96 – 236 + 16Q =0
And by collecting term we obtain:
3Q2 – 16Q – 140 = 0
-b b2 – 4ac
We now use the formula for solving a quadratic equation: x = ----------------2a
where a= 3, b= -16, and c = -140. Calculating the expression within the
square root sign gives: b2 – 4ac = (-16)2 – (4 . 3. -140) = 256 + 1680 =
1936.
1936 = 44. We have them:
x = (1644) / (2 . 3) = 60/6 or -28/6
So x = 10 or x = -4.67. Only the positive value is economically meaningful,
so profit maximization occurs when x = 10.
SIMULTANEOUS
EQUATIONS
When economists model how market operate, they often
use different equations to represent different aspects of the
market.
For market equilibrium, they values of the variables are
such that are true simultaneously.
Quantity demanded and quantity supplied are function pf
price (P). In equilibrium these quantities are equal.
To solve for equilibrium values, we equate the two
expression in P, thus eliminating Q. We obtain an equation
which we can solve for P.
Another method of eliminating a variable is to subtract (or
add) the left hand sides and the right hand sides of a pair of
equations.
Solve the simultaneous equation 2x + 4y = 20 and 3x + 5y = 28
For ease of reference we number the equation
2x + 4y = 20 ….. (1)
3x + 5y = 28 ……(2)
we choose the variable to be eliminated, say x. We need to get x with the
same coefficient in both equation. Using x’s coefficient in the other
equation, we multiply through equation (1) by 3 and equation (2) by 2. This
give:
6x + 12y = 60 …… (3)
6x + 10y = 56 ….. (4)
Now that x has a coefficient of 6 in both equations we subtract the
corresponding sides of equations (3) and (4). We obtain:
0 + 2y = 4
Since 2y = 4 y = 2 is solution for y. Now substitute it in to either equation,
say (1). We get:
2x + 4(2) = 20 2x + 8 = 20
Subtracting 9 from both sides gives: 2x = 12 x = 6
As a check, substitute x = 6, y = 2 in equation (2). The left hand side is
3(6) + 5(2) = 18 + 10
This equals the right hand side of 28, so the solution x = 6, y = 2 is correct !
SIMULTANEOUS EQUILIBRIUM IN
RELATED MARKETS
Demand and supply in two related markets forms
an example of an economics model using
simultaneous equations.
Demand each market depend both on the price of
the good it self and on the price of the related
good.
To solve the model we use the equilibrium
condition for each market and equate the quantity
supplied to the quantity demanded in that market.
This give two equations in two unknows which we
then solve.
The market for activity holidays is represented by the functions
Demand : Qa = 600 – (Pa/3) + (Pb/4)
Supply
: Qa = -100 + Pa
and the market for beach holidays is represented by the functions
Demand : Qb = 1800 – 3Pb + (Pa/3)
Supply : Qb = -400 + 3Pb
Where Qa and Qb are quantity of activity and beach holidays respec
tively and Pa and Pb are the prices of each type of holiday. Find the
equilibrium prices and quantities of each type of holiday.
Equating the quantity supplied and demanded in the activity holiday
market and substituting, we get:
-100 + Pa = 600 – (Pa/3) + (Pb/4)
or (4Pa/3) – (Pb/4) = 700 …… (1)
And equating the quantity supplied and demanded in the beach holiday market
give:
-400 + 3Pb = 1800 – 3Pb + (Pa/3) or (-Pa/3) + 6Pb = 2200
…… (2)
We now have two simultaneous equation equations to solve for Pa and Pb.
Multiply equation (2) by 4, which gives:
(-4Pa/3) + 24Pb = 8800
…… (3)
Adding equation (1) and (3) we find:
0Pa + 23.75Pb = 9500
so, 23.75Pb = 9500
Pb = 400
We can now find Pa from equation (2)
Pa/3 + 6(400) = 2200
Subtracting 2400 from each sides we have: (-Pa/3) = -200
Multiplying by -3 gives :
Pa = 600
We can find the quantities of holidays most easily from the
supply equations. For activity holidays: Qa = -100 + Pa
Which gives: Qa = -100 + 600 = 500
Using the beach holidays supply equation: Qb = -400 + 3Pb
Which gives: Qb = -400 + 3(400) = 800
The solution is:
Pa = 600, Pb = 400, Qa = 500, Qb = 800
Check by substituting in the demand equations
Activity holiday: Qa = 600 – (Pa/3) + (Pb/4)
The right hand side gives:
600 –(600/3)+(400/4) = 500 = Qa
Beach Holiday: Qb = 1800 – 3Pb + (Pa/3)
The right hand sides gives:
1800 – 3(400) + (600/3) = 800 = Qb
Therefore, the solution is correct
(Course 5)
OLEH
SYAIFUL HADI
DJAIMI BAKCE
JURUSAN AGRIBISNIS
FAKULTAS PERTANIAN
UNIVERSITAS RIAU
QUADRATIC EQUATION
In economic modelling we work with the simples function that can
adequately represent a relationship, but sometimes a curve
rather than a straight line is required.
We used quadratic functions to represent total revenue and
average variable cost
General form : y = ax2 + bx + c
where a, b, and c are constants. When you sketch a quadratic
function you find it has either a hill or U shape and that generally
two values of x give the same value of y.
A quadratic function has a term in x2 but no higher powers of x.
A quadratic equation you can solve it graphically or sometimes
by factorizing it or by using the formula.
We first write it in the form ax2 + bx +c = 0. The value(s) of
x for which this equation is true can be found graphically by
plotting
y= ax2 + bx + c and looking y = 0.
Algebraic methods are more accurate than graphical ones,
but the
squared term means we need a special technique. Sometimes
factori- zing the expression helps. Consider for example:
5x2- 20x = 0
Since each term is divisible by 5x we can factorize the left hand
side and write: 5x(x-4) = 0
There are now two term multiplied together, 5x and (x-4). For
their product to be 0, one of term must be zero. This mean
either 5x = 0 or (x-4) =0.
If 5x = 0, dividing by 5 we find that x = 0 and this one possible
solution to the equation.
If x – 4 = 0, adding 4 to both sides we find that x = 4, which is
the other possible solution.
The graph of the function is plotted It also shows that x = 0
and x = 4 are two solutions to the equation, since they are the
values of x at which at which the curve cuts the x axis
Y = 5x2 – 20x
A more general method for solving the quadratic equation ax2+bx
+c =0 uses a formula:
-b b2 – 4ac
x = ------------------2a
Where a is the coefficient of x2, b is the coefficient of x and c is the
constant term.
Example: 8x2 – 20x + 3 = 0
We identify a= 8, b= -20 and c= 3. Notice that the sign of the
coefficient must be included. Begin the calculating the expression
the square root sign. This gives:
b2 – 4ac = (-20)2 – (4 . 8 . 3) = 400 – 96 = 304
We take the square root of this value and obtain 17.436. substituting
this result , -b and a into the formula gives:
20 17.436
x = -------------16
x = 37.436/16= 2.34 or x = 2.564/16 = 0.16
The solution of the equation is x = 2.34 or 0.16 (cari turunan
pertama ato kedua yang bernilai negatif)
INTERSECTION OF MC WITH MR OR
AVC
Quadratic equation arise in economics when we want to
discover where a quadratic function, say marginal cost,
cuts another quadratic function, say average variable
cost, or cuts a linier function, say marginal revenue.
We equate the two functional expressions, then subtract
the right hand side from both sides so that the value on
the right becomes zero.
After collecting term we solve the quadratic equation
using one of the methods explained above.
A firm has the marginal cost function MC = 3Q2 – 32Q +
96 and marginal revenue function MR = 236 – 16Q. Find
the firm’s profit maximizing output.
To maximize profits the firm chooses to produce where marginal cost
equals marginal revenue. Equating the MC and MR functions we have that:
3Q2 – 32Q + 96 = 236 – 16Q
Subtracting the right hand side from both sides gives:
3Q2 – 32Q + 96 – (236 – 16Q) = 0
Removing the bracket gives
3Q2 – 32Q + 96 – 236 + 16Q =0
And by collecting term we obtain:
3Q2 – 16Q – 140 = 0
-b b2 – 4ac
We now use the formula for solving a quadratic equation: x = ----------------2a
where a= 3, b= -16, and c = -140. Calculating the expression within the
square root sign gives: b2 – 4ac = (-16)2 – (4 . 3. -140) = 256 + 1680 =
1936.
1936 = 44. We have them:
x = (1644) / (2 . 3) = 60/6 or -28/6
So x = 10 or x = -4.67. Only the positive value is economically meaningful,
so profit maximization occurs when x = 10.
SIMULTANEOUS
EQUATIONS
When economists model how market operate, they often
use different equations to represent different aspects of the
market.
For market equilibrium, they values of the variables are
such that are true simultaneously.
Quantity demanded and quantity supplied are function pf
price (P). In equilibrium these quantities are equal.
To solve for equilibrium values, we equate the two
expression in P, thus eliminating Q. We obtain an equation
which we can solve for P.
Another method of eliminating a variable is to subtract (or
add) the left hand sides and the right hand sides of a pair of
equations.
Solve the simultaneous equation 2x + 4y = 20 and 3x + 5y = 28
For ease of reference we number the equation
2x + 4y = 20 ….. (1)
3x + 5y = 28 ……(2)
we choose the variable to be eliminated, say x. We need to get x with the
same coefficient in both equation. Using x’s coefficient in the other
equation, we multiply through equation (1) by 3 and equation (2) by 2. This
give:
6x + 12y = 60 …… (3)
6x + 10y = 56 ….. (4)
Now that x has a coefficient of 6 in both equations we subtract the
corresponding sides of equations (3) and (4). We obtain:
0 + 2y = 4
Since 2y = 4 y = 2 is solution for y. Now substitute it in to either equation,
say (1). We get:
2x + 4(2) = 20 2x + 8 = 20
Subtracting 9 from both sides gives: 2x = 12 x = 6
As a check, substitute x = 6, y = 2 in equation (2). The left hand side is
3(6) + 5(2) = 18 + 10
This equals the right hand side of 28, so the solution x = 6, y = 2 is correct !
SIMULTANEOUS EQUILIBRIUM IN
RELATED MARKETS
Demand and supply in two related markets forms
an example of an economics model using
simultaneous equations.
Demand each market depend both on the price of
the good it self and on the price of the related
good.
To solve the model we use the equilibrium
condition for each market and equate the quantity
supplied to the quantity demanded in that market.
This give two equations in two unknows which we
then solve.
The market for activity holidays is represented by the functions
Demand : Qa = 600 – (Pa/3) + (Pb/4)
Supply
: Qa = -100 + Pa
and the market for beach holidays is represented by the functions
Demand : Qb = 1800 – 3Pb + (Pa/3)
Supply : Qb = -400 + 3Pb
Where Qa and Qb are quantity of activity and beach holidays respec
tively and Pa and Pb are the prices of each type of holiday. Find the
equilibrium prices and quantities of each type of holiday.
Equating the quantity supplied and demanded in the activity holiday
market and substituting, we get:
-100 + Pa = 600 – (Pa/3) + (Pb/4)
or (4Pa/3) – (Pb/4) = 700 …… (1)
And equating the quantity supplied and demanded in the beach holiday market
give:
-400 + 3Pb = 1800 – 3Pb + (Pa/3) or (-Pa/3) + 6Pb = 2200
…… (2)
We now have two simultaneous equation equations to solve for Pa and Pb.
Multiply equation (2) by 4, which gives:
(-4Pa/3) + 24Pb = 8800
…… (3)
Adding equation (1) and (3) we find:
0Pa + 23.75Pb = 9500
so, 23.75Pb = 9500
Pb = 400
We can now find Pa from equation (2)
Pa/3 + 6(400) = 2200
Subtracting 2400 from each sides we have: (-Pa/3) = -200
Multiplying by -3 gives :
Pa = 600
We can find the quantities of holidays most easily from the
supply equations. For activity holidays: Qa = -100 + Pa
Which gives: Qa = -100 + 600 = 500
Using the beach holidays supply equation: Qb = -400 + 3Pb
Which gives: Qb = -400 + 3(400) = 800
The solution is:
Pa = 600, Pb = 400, Qa = 500, Qb = 800
Check by substituting in the demand equations
Activity holiday: Qa = 600 – (Pa/3) + (Pb/4)
The right hand side gives:
600 –(600/3)+(400/4) = 500 = Qa
Beach Holiday: Qb = 1800 – 3Pb + (Pa/3)
The right hand sides gives:
1800 – 3(400) + (600/3) = 800 = Qb
Therefore, the solution is correct