MATH Review Examples PDF

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EXAMPLE 1 Demand function: Pd = 100 – 0.5Qd Supply function: Ps = 10 + 0.5Qs (a) Calculate the equilibrium price and quantity. Use a diagram to show the equilibrium point (b) Assume that the government imposes a fixed tax of £6 per unit sold. (i) Write down the equation of the supply function, adjusted for tax. (ii) Find the new equilibrium price and quantity. (iii) Outline the distribution of the tax, (calculate the tax paid by the consumer and the producer.) Solution (a) The equilibrium quantity and price are 90 units and £55 respectively. (b) (i) The tax of £6 per unit sold means that the effective price received by the producer is (Ps — 6). The equation of the supply function adjusted for tax is Ps – 6 = 10 + 0.5Q Ps = 16 + 0.5Q (ii) The new equilibrium price and quantity are calculated by equating the original demand function, equation, and the supply function adjusted for tax, equation: Pd = Ps 100 – 0.5Q = 16 + 0.5Q Q = 84 Substitute the new equilibrium quantity, Q = 84 into either equation and solve for the new equilibrium price: P = 100 – 0 5(84) substituting Q = 84 we get P = 58 (iii) The consumer always pays the equilibrium price, therefore the consumer pays £58, an increase of £3 on the original equilibrium price with no tax, which was £55. The producer receives the new equilibrium price, minus the tax, so the producer receives £58 — £6 = £52, a reduction of £3 on the original equilibrium price of £55.

1

EXAMPLE 2 The demand and supply functions for a good (£P per ton of potatoes) are given as Demand function: Pd = 450 – 2Qd Supply function: Ps = 100 + 5Qs (a) Calculate the equilibrium price and quantity. (b) The government provides a subsidy of £70 per unit (ton) sold: (i) Write down the equation of the supply function, adjusted for the subsidy, (ii) Find the new equilibrium price and quantity, Solution (a) The equilibrium quantity and price are 50 units and £350 respectively. (b) (i) With a subsidy of £70 per unit sold, the producer receives (Ps + 70). The equation of the supply function adjusted for subsidy is Ps + 70 = 100 + 5Q Ps = 30 + 5Q (ii) The new equilibrium price and quantity are calculated by equating the original demand function and the supply function adjusted for the subsidy, equation: Pd = (Ps + subsidy) 450 – 2Q = 30 + 5Q Q = 60 The new equilibrium price: 2

P = 450 - 2Q P = 450 - 2(60) P = 330

EXAMPLE 3 Demand function: Pd = 200 – 3Qd Supply function: Ps = 50 + 2Qs (a) Calculate the equilibrium price and quantity. (b) Assume that the government imposes a fixed tax of £10 per unit sold. (i) Write down the equation of the supply function, adjusted for tax. (ii) Find the new equilibrium price and quantity. Solution (a) The equilibrium quantity and price are 30 units and £110 respectively. (b) (i) The tax of £10 per unit sold means that the effective price received by the producer is (Ps – 10). The equation of the supply function adjusted for tax is Ps – 10 = 50 + 2Q Ps = 60 + 2Q (ii) The new equilibrium price and quantity are calculated by equating the original demand function, equation, and the supply function adjusted for tax, equation: Pd = Ps 3

200 – 3Qd = 60 + 2Qs Q = 28 Substitute the new equilibrium quantity, Q = 28 into either equation and solve for the new equilibrium price: P = 200 – 3(28) substituting Q = 28 we get P = 116 EXAMPLE 4 (a) Find the inverse of the matrix A=

[1610 128 ]

by Gauss-Jordan elimination. (b) Solve the system of equations through the inverse matrix method. 3Y = 7 – 4X 4Y = 4 – 5X (c) Solve the system by using Cramer’s rule P = 75 – 20Q P = 18 – 12Q Solution (a) Through the use of Gauss-Jordan elimination we get

[

| ]

|

[

A 16 12 1 0 A /16=C 1 0,75 0,0625 0 = B 10 8 0 1 B 10 8 0 1

|

[

]

¿

C 1 0,75 0,0625 0 B−10 C=D 0 0,5 −0,625 1

¿

C 1 0,75 0,0625 0 1 −1,25 2 D/0,5=E 0

¿

C−0,75 E 1 0 1 −1,5 0 1 −1,25 2 E

[

[ |

|

]

] ]

(b) 3Y = 7 – 4X 4Y = 4 – 5X The system can also be written as: 4

16X + 12Y = 28 10X + 8Y = 8 According to matrices’ properties AX = B => A-1AX = A-1B => IX = A-1B => X = A-1B Therefore 16 12 X 28 = 10 8 Y 8

[

][ ] [ ]

1 [YX ]= [−1,25

][ ] [ ]

16 −1,5 28 = 8 −19 2

(c) P = 75 – 20Q P = 18 – 12Q P + 20Q = 75 P + 12Q = 18 1 75 | 1 18| −57 = Q= 1 20 −8 |1 12| 75 20 | 18 12| 540 = P= 1 20 |1 12| −8

EXAMPLE 5 (a) Find the inverse of the matrix A=

[57 46 ]

by Gauss-Jordan elimination. (b) Solve the system of equations through the inverse matrix method. 4Y = 12 – 5X 6Y = 20 – 7X 5

(c) Solve the system by using Cramer’s rule P = 80 + 2Q P = 60 – 7Q Solution (a) Through the use of Gauss-Jordan elimination we get

[ | ] [

[

|

A 5 4 1 0 A /5=C 1 0,8 0,2 0 = B 7 60 1 B 7 6 0 1 C 1 0,8 0,2 0 ¿ B−7 C=D 0 0,4 −1,4 1

[

|

]

|

¿

C 1 0,8 0,2 0 D/0,4=E 0 1 −3,5 2,5

¿

C−0,8 E 1 0 3 −2 0 1 −3,5 2,5 E

[ |

]

]

]

(b) 12Y = 36 – 15X 24Y = 80 – 28X The system can also be written as: 5X + 4Y = 12 7X + 6Y = 20 According to matrices’ properties AX = B => A-1AX = A-1B => IX = A-1B => X = A-1B Therefore 5 4 X 12 = 7 6 Y 20

[ ][ ] [ ] 3 −2 12 −4 = [YX ]= [−3,5 2,5 ][ 20 ] [ 8 ]

(c) P = 80 + 2Q P = 60 – 7Q

6

P – 2Q = 80 P + 7Q = 60 1 80 | 1 60| −20 Q= = 1 −2 |1 7 | 9 80 −2 | 60 7 | 680 = P= 1 −2 |1 7 | 9 EXAMPLE 6 (a) Find the inverse of the matrix A=

[2015 1510 ]

by Gauss-Jordan elimination. (b) Solve the system of equations through the inverse matrix method. 7,5Y = 30 – 10X 20Y = 40 – 30X (c) Solve the system by using Cramer’s rule P = 45 – 6Q P = 22 – 3Q Solution (a) Through the use of Gauss-Jordan elimination we get

[

| ]

|

[

A 20 15 1 0 = A /20=C 1 0,75 0,05 0 B 15 10 0 1 B 15 10 0 1

|

[

]

¿

C 1 0,75 0,05 0 B−15 C=D 0 −1,25 −0,75 1

¿

C 1 0,75 0,05 0 1 0,6 − 0,8 D/−1,25=E 0

[

|

]

] 7

¿

[ |

C−0,75 E 1 0 −0,4 0,6 0 1 0,6 −0,8 E

]

(b) Solve the system of equations through the inverse matrix method. 7,5Y = 30 – 10X 20Y = 40 – 30X The system can also be written as: 20X + 15Y = 60 15X + 10Y = 20 According to matrices’ properties AX = B => A-1AX = A-1B => IX = A-1B => X = A-1B Therefore 20 15 X 60 = 15 10 Y 20

[

][ ] [ ]

[YX ]= [−0,4 0,6

][ ] [ ]

0,6 60 −12 = 20 −0,8 20

(c) P = 45 – 6Q P = 22 – 3Q P + 6Q = 45 P + 3Q = 22 1 45 | 1 22| −23 = Q= 1 6 |1 3| −3 45 6 | 22 3| 3 = P= 1 6 −3 |1 3|

EXAMPLE 7 8

(a) You have placed your money at the bank with an annual interest of 3% which is compounded monthly. How much must you deposit in order to have 50,000€ at the end of the year? Solution We use the formula

( )

Pt =P 0 1+

r m

m×t

Where Pt = 50,000 and r = 0,03, m = 12, t = 1 P0 = 48,524.093 EXAMPLE 8 You decide to invest 45,000€ for 4 years at 10% per annum. Calculate the total value of the investment if it is compounded (a) semi-annually (b) monthly (c) daily Solution (a) We use the formula

( )

Pt =P 0 1+

r m

m×t

Where P0 = 45,000 and r = 0,10, m = 2, t = 4 ( 2 ) × ( 4) 0,10 Pt =45,000 1+ 2

(

)

Pt = 66,485.50

(b) We use the formula m×t

( )

Pt =P 0 1+

r m

Where P0 = 45,000 and r = 0,10, m = 12, t = 4

9

(

Pt =45,000 1+

0,10 12

( 12) ×(4)

)

Pt = 67,020.93 (c) We use the formula

( )

Pt =P 0 1+

r m

m×t

Where P0 = 45,000 and r = 0,10, m = 365, t = 4

(

Pt =45,000 1+

0,10 365

)

(365)× (4 )

Pt = 67,128.43

EXAMPLE 9 You want to invest the above mentioned sum (45,000€). Will you choose to invest the money in an account with an annual rate of 5% that is compounded semi-annually or will you choose to invest the money in another project which will result in you earning 58,000€ after 5 years? Solution (a) We use the formula

( )

Pt =P 0 1+

r m

m×t

Where P0 = 45,000 and r = 0,05, m = 2, t = 5

(

Pt =45,000 1+

0,05 2

)

( 2 ) × ( 5)

Pt = 57,603.8 You will choose to invest the money in the project EXAMPLE 10 You have decided to deposit 350€ each month into a fund for your son. Your son is 1 year old and you want the money to be available to him when he turns 21. The annual interest rate is 6% and it is compounded monthly. How much money will your son 10

have available at the age of 21 if he decides to withdraw all the money from the trust fund. Solution

( ( ) ) ( ( 1− 1+

P0 = A 0

−t ×m

i m

1− 1+

=350

i m

0,06 12

0,06 12

) ) =48,853.27 −12 × 20

The Future Value will be:

(

FV =48,853,27 × 1+

)

0,06 12 × 20 =161,714.31 12

EXAMPLE 11 You are given the demand function P = 120 – 5Q for a monopolist: (a) Give the expression for Marginal Revenues (b) What is the level of production that maximizes revenues? (c) Which are the prices that correspond to the inelastic part of the demand curve. (d) Compute the elasticity of demand when P = 20 (e) Present graphically that when we are in the inelastic part of the demand curve a price decrease results in a decrease in revenues whereas if we are in the elastic part of the demand curve a price increase results in an increase in revenues Solution (a) P = 120 – 5Q => TR = P*Q = 120Q - 5Q2

MR=

∂TR =120−10 Q ∂Q

(b) MR = 0 => MR = 120 – 10Q = 0 => Q = 12

(c) P = 120 – 5*12 => P = 60 these are the prices below 60

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(d) We construct a table with the respective changes. For Q = 12, P = 60. For P = 20, Q = 20 % Change ∈Q ΔQ P ∂Q P 60 e= =0,2 × =1 = = Q 12 % Change∈P ΔP Q ∂ P (e) Diagram

EXAMPLE 12 You are given the demand function P = 70 – 3,5Q for a monopolist: (a) Give the expression for Marginal Revenues (b) What is the level of production that maximizes revenues? (c) Which are the prices that correspond to the inelastic part of the demand curve. (d) Compute the elasticity of demand when P = 70 (e) Present graphically that when we are in the inelastic part of the demand curve a price decrease results in a decrease in revenues whereas if we are in the elastic part of the demand curve a price increase results in an increase in revenues Solution 12

(a) P = 70 – 3,5Q => TR = P*Q = 70Q - 3,5Q2

MR=

∂TR =70−7 Q ∂Q

(b) MR = 0 => MR = 70 – 7Q = 0 => Q = 10

(c) P = 70 – 3,5*10 => P = 35 these are the prices below 35

(d) We construct a table with the respective changes. For Q = 10, P = 35. For P = 70, Q = 20 % Change ∈Q ΔQ P ∂Q P 35 = = e= =0,6 × =2,3 10 % Change∈P ΔP Q ∂ P Q

(e) Diagram

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