Mangerial Economics(solution of Assignment 2)

Managerial Economics

Assignment 2

Q1. The demand equation faced by DuMont Electronics for its personal computers is given by

P = 10,000 – 4Q

SOLUTION:

a)                Write the marginal revenue equation.

We know that

TR                   =          P  x  Q

           

TR                   =          (10,000-4Q) x Q

           

TR                   =          10,000Q-4Q²

Now taking derivative of TR

dTR/dQ           =         10,000Q-4Q²  

MR                  =          10,000-8Q

b)    At what price and quantity will marginal revenue be zero?

MR                  =          0

10,000 – 8Q    =          0

10,000             =          8Q

Q                     =          1000/8

Q                     =          1,250

Putting  the value in Price equation,

P          =          10,000 – 4(1250)

P          =          5,000

c)                If the price is increased from 6,000 to 7,000, what will be the effect on total revenue? What does this imply about price elasticity?   

When p = 6,000, then Q will be

6,000   =          10,000 – 4Q

4Q       =          10,000 – 6,000

Q         =          1,000

TR will be

TR       =          10,000(1,000) – 4(1,000)²

TR       =          10000000 -4000000

TR       =          6,000,000

When p = 7,000, then Q will be

7,000   =          10,000 – 4Q

4Q       =          10,000 – 7,000

Q         =          750

TR will be

TR       =          10,000(750) – 4(750)²

TR       =          7500000-2220000

TR       =          5,250,000

It tells that there is  elastic price elasticity of demand

Q2. The Inquiry club at Jefferson University has compiled a book that exposes the private lives of many of the professors on campus. Economics majors in the club estimate that total revenue from sales of the book is given by the equation

            TR = 120Q – 0.1Q³

.

SOLUTION:

a)    Over what output range is demand elastic?

By taking derivative of TR

dTR/dQ=120Q-0.1Q3

0          =          120 – 0.3Q²

120      =          0.3Q²

Q²        =          120/0.3          

Q²        =          400

Q         =          20

We know that

TR       =          PxQ

P          =          TR/Q

P          =          120 – 0.1Q²

P          =          120 – 0.1(20)²

P          =          120 – 0.1(400)

P          =          120 - 40

P          =          80

Q = 20 demand is elast

b)   Initially, the price is set at $71.6. To maximize total revenue should the price be increased or decreased? Explain

If price is 80, then TR

TR       =          120(20) – 0.1(20)²

TR       =          2,400 – 800

TR       =          1,600

If price is set to 71.6, then TR becomes

TR       =          20 x 71.6

TR       =          143

2This clearly indicate that price should be increased from 71.6 to 80 in order to get maximum Total revenue

Q3. The coefficient of income in a regression of quantity demanded of commodity on price, income and other variables is 10.

SOLUTION:

a)             Calculate the income elasticity of demand for this commodity at income of 10,000 and sales of 80, 000 units. Interpret your answer.

The income elasticity of demand is

E I          =dQ/dI X I/Q

So we have

dQ/dI=           10

So by putting its value in E I, we get

E I          =          10   X 10,000/80,000

E I          =          1.25

Interpretation:

1% increase the Quantity will lead to 1.25% increase . Keeping other thing constant

b)     What would be the income elasticity of demand if sales increased from 80,000 to 90,000 units and income rose from 10, 000 to 11,000? What type of good is this commodity?

E I          =Q_2-Q₁/I₂-I₁ X I₂+I₁/Q₂+Q₁

E I          =90,000-80,000/11,000  X  11,000+10,000/90,000+80,000

E I          =          1.24

Value is positive and greater than 1, so goods are luxurious.

]

Q5. The coefficient of the price of gasoline in the regression of the quantity demanded of automobiles (in millions of units) on the price of gasoline (in dollars) and other variables is -14.

SOLUTION:

a)    Calculate the cross price elasticity of demand between automobiles and gasoline at the gasoline price of $1per gallon and sales of automobiles of 8 (million units). Interpret your answer

The cross elasticity of demand is

Exy         =        dQ_x/dP_y X P_y/Q_x

Wher

dQ_x/dP_y=         -14

Then

Exy       =         -14(  1/8)

Exy       =         -1.75

 1 % increase in price of automobiles would result the 1.75% decrease in gasoline

b)   What would be the cross price elasticity of demand between automobiles and gasoline if sales of automobiles decline from 8 to 6 with an increase in gasoline price from $1 to $1.20 per gallon? Interpret you answer.,

Exy       =            Q_x1–Q_x1 /P_y2-P_y1  X  P_y2+P_y1 /Q_x2+Q_x1

Exy        =           6-8/1.2-1 X 1.2 +1/6+8

Exy         =         -1.57

So Exy is –ve and the goods are complimentary

Q6. The Profit function of a firm selling two products is

                        Π = 50 Q₁ – Q₁² + 100Q₂ – 4Q₂²

where Q1 and Q2 represent the output rates of product 1 and product 2, respectively. Determine the profit maximizing output rates for the two products and the profit associated with the two output rates.

SOLUTION:

a)     

First taking partial derivative w.r.t Q₁, and equate the equation to zero

∂∏/∂Q₁=         50Q₁-Q₁²+100Q₂-4Q₂²          

0          =          50 – 2Q₁

Q₁       =          25

Now taking derivative w.r.t Q₂ and equate the equation to zero

∂∏/∂Q₂            =50Q₁-Q₁²+100Q₂-4Q₂²       

0          =          100 – 8Q₂

Q₂       =          12.5

Profit will maximum when it sell 25 units of product 1 and 12.5 units of product 2.

b)    

Putting the value of Q₁ and Q₂ in profit equation,

                       

Π = 50 (25) – (25⁾² + 100(12.5) – 4(12.5)₂

Π = 1,250

So firm will earn 1250 by producing 25 units of product 1 and 12.5 units of product 2

Q7. Given the Total Cost function

                        TC = 100Q – Q² +0.3Q³

SOLUTION:

a)    The marginal and average cost functions

MC Function

MC                  =         

dTC/dQ        =          100Q – Q² +0.3Q³

MC                  =          100 – 2Q + 0.9Q²

AC                  =          TC/Q

AC                  =          100Q – Q² +0.3Q³

           

AC                  =          100 – Q + 0.3Q²

b)   The rate of output that results in minimum average cost.

AC                  =          100 – Q + 0.3Q²

Taking derivative

dAC/dQ        =          100 – Q + 0.3Q²

 0         =          -1 +0.6Q

1          =          0.6Q

Q         =          1.667

Average cost will be min at the rate of output of 1.667 units