MANAGERIAL ECONIMIC IN GLOBAL ECONOMY EDITION 15 SALVATORE

Q no.1

WHAT FACTORS SHOULD BE CONSIDERED IN DETERMINING WHETHER PROFIT LEVEL IS EXCESSIVE IN A PARTICULAR INDUSTRY?

ANS: Following are the factors which should be considered in determining whether profit level is excessive in a particular industry.
HIGH PRODUCTION:
High production also lead to excessive profits when the production is at peak then this means that the demand of the product is high so high demand leads to increase in sales and sales have an effect on profits.... See More
INCREASE IN SALE:
Profit of a firm or industry depends upon high sales. If sales are high profit also goes up so increase in sales is another factor which tells about the excessive profits.
LESS COMPITITION:
Businesses survive by eliminating their competition. By having less competition more consumers have to buy their goods and services, thusly increasing profit margins, and allowing for expansion. Industry can earn excessive profits if they have very less competitors.
Excessive profits coming out of monopoly:
If a industry or a firm has a monopoly then they have a chance to earn excessive profits. Because they have no competitors in the market so they fix the prices of there own choices so another factor of excessive profits is monopoly.
SOME OTHER FACTORS
1.) Cost of production

2.)Profit margin that the entrepreneur wants to keep

3.) The marketing strategy

4.) Competitor’s price

2.) GIVE TWO REASONS WHY IN GENERAL MAXIMIZING PROFITS BENEFITS SOCIETY?
In general maximizing profits benefits society in a lot of ways because in business every thing depends upon profit.
1.) In general profit maximizing profit benefits society because maximizing profits leads to the reduction in poverty as the profit of a firm increases automatically the wages of employee will rise which will increase the purchasing power of individuals, which is beneficial for society.
2.) The price charged to individuals equals the marginal cost of production to each firm. In other words, one can say sellers charge buyers a reasonable or fair price.
3.) In general, output produced under a perfectly competitive market structure is larger than other market organizations due to maximizing profits. Thus, this becomes desirable also for the amount of the product supplied to consumers as a whole.
4.) Profits are earned by allocating resources in a best way which gives the best output at very less cost due to which the consumer gets the goods at fair price.

3.) WHY ECONOMIC PROFIT IS ZERO IN LONG RUN?
Economic profit can be seen as excess profit that an owner earns from an activity that is higher than one can expect to earn in doing a similar type of activity with similar characteristics and risks. Example. An economy has three types of industries: making clocks, making pans, and making toasters. Let's say these are very similar in terms of risk and in terms of the kinds of resources you need to operate in these industries. But at the moment, the industries have different returns of investment:

clocks return 10%
pans return 10%
toasters return 13%

Since these industries are competitive markets ,so in the long run there is NO barrier to entry. Some of the clock makers and pan makers may start making toasters instead, so they shift their resources to making this. So eventually, with the entry of new competition, you may have returns that normalize across the industries where all industries now have returns of 11%

When economic profit is zero, this is JUST enough to keep a factor of production in its current use. In competitive markets, because there are no barriers to entry, new firms will keep entering the market while there are still economic profits to be made. Once economic profit is zero, there is no incentive for more new firms to enter the market and there is no reason for the firms in the market to leave it because zero economic profit is just enough to keep them producing what they are producing rather than making something else.

Question: 07.

Find the derivative of the following:

(a) Y = f (X) = a
TC = f (Q) = 182
(b) Y = 2X2
Y = -1X3
Y = ½(X-2)



TR = f (Q) = 10Q



TR = f (Q) = -Q2

Solution:

(1) Y = f (X) = a DIFFERENTIATING W.R.T "X"



dY/dX =f’ (X) = 0

(2) TC = f (Q) = 182 DIFFERENTIATING W.R.T "Q"
d (TC)/dX = f’ (q) = 0

(3) Y = 2X2 DIFFERENTIATING W.R.T "X"



dY/dX = 4X

(4) Y = -1X3 DIFFERENTIATING W.R.T "X"
dY/dX = -1(3) X3-1
dY/dX = -3X2

(5) Y = ½ (X-2) DIFFERENTIATING W.R.T "X"
dY/dX = ½ (-2) (X-2-1)
dY/dX = -X-3

(6) TR = f (Q) = 10Q DIFFERENTIATING W.R.T "Q"
dTR/dX = f’ (Q) = 10

(7) TR = f (Q) = -Q2 DIFFERENTIATING W.R.T "Q"
dTR/dX = f’ (Q) = -2Q

Question: 8.

Find the derivative of the following functions:

(a) Y = 45X – 0.05X2
Y = X3 – 8X2 + 57X + 2



TR = 100Q – 10Q2 (Equation 2-1)



TC = 182 + 56Q (Equation 2-2)
(b) Y = X3 - 2X2
Y = 8X4 – 20X3
Y = 4X3 (2X - 5)

Solution:

(1) Y = 45X – 0.05X2 DIFFERENTIATING W.R.T "X"
dY/dX = 45 – 0.1X

(2) Y = X3 – 8X2 + 57X + 2 DIFFERENTIATING W.R.T "X"
dY/dX = 3X2 – 16X + 57

(3) = TR – TC
= 100Q – 10Q2 – (182 + 56Q)
= 100Q - 10Q2 – 182 – 56Q
= 44Q – 10Q2 – 182 DIFFERENTIATING W.R.T ""
d/dQ = 44 – 20Q

(4) Y = X3 – 2X2 DIFFERENTIATING W.R.T "X"
dY/dX = 3X3-2 – 2(2) X2-1
dY/dX = 3X2 – 4X

(5) Y = 8X4 – 20X3 DIFFERENTIATING W.R.T "X"



dY/dX = 8(4) X4-1 – 20(3) X3-1
dY/dX = 32X3 – 60X2

(6) Y = 4X3 (2X-5) DIFFERENTIATING W.R.T "X"
USING PRODUCT RULE
dY/dX = 4(3) X3-1(2X-5) + 4X3 (2)
dY/dX = 12X2 (2X-5) + 8X3
dY/dX = 24X3 – 60X2 + 8X3
dY/dX = 32X3 – 60X2

Question: 09.

Find the derivative of the following functions:

(a) Y = 3X3
X2
Y = 5X3
4X+3
(b) Y = U5 and U = 2X3 + 3
Y = U3 + 3U and U = -X2 + 10X
Y = (2X3 + 5)2

Solution:
(1) Y = 3X3 DIFFERENTIATING W.R.T "X"
X2
Using Quotient Rule
dY/dX = (X2) (9X2) – (2X) (3X3)
(X2)2
dY/dX = 9X4 – 6X4
X4
dY/dX = X4. (9-6)
X4
dY/dX = 3
(2) Y = 5X3 DIFFERENTIATING W.R.T "X"
(4X+3)
Using Quotient Rule
dY/dX = (4X+3) (15X2) – (5X3) (4)
(4X + 3)2
dY/dX = 60X3 + 45X2 – 20X3
(4X + 3)2
dY/dX = 40X3 + 45X2
(4X+3)2

(3) Y = U5 and U = 2X3 + 3
DIFFERENTIATING W.R.T "U" DIFFERENTIATING W.R.T "X"




dY/dU = 5U4 dU/dX = 6X2

dY = dY . dU
dX dU dX




= (5U4)*(6X2)



= {5(2X3 + 3)4}*(6X2)



dY/dX = 30X2(2X3 + 3)4

(4) Y = U3 + 3U and U = 10X – X2
DIFFERENTIATING W.R.T "U" DIFFERENTIATING W.R.T "X"
dY/dX = 3U2 + 3 dU/dX = 10 – 2X
dY = dY . dU
dX dU dX



= (3U2 + 3)*(10 – 2X)
= {3(10X – X2)2 + 3}*(10 – 2X)
= {3 (100X2 + X4 – 20X3) + 3}*(10 – 2X)
= {300X2 + 3X4 – 60X3 +3}* (10 – 2X)
= 3000X2 + 30X4 – 600X3 + 30 – 600X3 – 6X5 + 120X4 – 6X



dY/dX = 30 – 6X + 3000X2 – 1200X3 + 150X4 – 6X5

(5) Y = (2X3 + 5)2 DIFFERENTIATING W.R.T "X"

dY/dX = 2(2X3 + 5)2-1 *(6X2)
= 12X2 *(2X3 + 5)
dY/dX = 24X5 + 60X2

QUESTION: 10

. Given the following total-revenue and total cost function of a firm:
TR = 22Q – 0.5Q2
TC = 1/3Q3 – 8.5Q2 + 50Q + 90
... See More
Determine:
(a) The level of output at which the firm maximize its total profit.
(b) The maximum profit that the firm could earn.
Solution:
TR – TC



= 22Q – 0.5Q2 – (1/3Q3 – 8.5Q2 + 50Q + 90)



= 22Q – 0.5Q2 – 1/3Q3 + 8.5Q2 – 50Q – 90



= -1/3Q3 + 8Q2 – 28Q – 90 -----------------------(I)
DIFFERENTIATING W.R.T "Q"
ddQ = -Q2 + 16Q – 28 -----------------------(II)
To Maximize We Take ddQ = 0
So, -Q2 + 16Q – 28 = 0
-Q2 + 14Q + 2Q – 28 = 0
Q (-Q + 14) –2 (14 – Q) = 0
(Q – 2)*(14 – Q) = 0
Q – 2 = 0 and 14 – Q = 0
Q = 2 and Q = 14
AGAIN DIFFERENTIATING W.R.T "Q" Eq (II)
d2dQ2 = 16 – 2Q
NOW
AT Q = 2,
d2dQ2 = 16 – 2(2)
= 12 > 0 Is Minimum.
AT Q = 14,
d2dQ2 = 16 – 2(14)
= - 12 Is Maximum.
FIRM CAN EARN MAXIMUM PROFIT AT Q = 14
PUTTING Q = 14 In Eq (I)
= -1/3Q3 + 8Q2 – 28Q – 90



= -1/3(14)3 + 8(14)2 – 28(14) – 90



= -914.67 + 1568 – 392 – 90
= 171.33

QUESTION: 11.

A firm’s total-revenue and total cost functions are
TR = 4Q
TC = 0.04Q3 – 0.9Q2 + 10Q + 5
Determine:
(c) Determine the best level of output.
(d) Determine the total profit of the firm at its best level of output.
Solution:
TR – TC



= 4Q – (0.04Q3 – 0.9Q2 + 10Q + 5)



= 4Q – 0.04Q3 + 0.9Q2 – 10Q – 5



= -0.04Q3 + 0.9Q2 – 6Q – 5 ----------------------- (I)
DIFFERENTIATING W.R.T "Q"
ddQ = -0.12Q2 + 1.8Q – 6 ----------------------- (II)
To Maximize We Take ddQ = 0
So, -0.12Q2 + 1.8Q – 6 = 0 dividing by 0.12
-Q2 + 15Q – 50 = 0
-Q2 + 10Q + 5Q – 50 =0
Q (-Q + 10) –5 (10 – Q) = 0
(Q – 5)*(10 – Q) = 0
Q – 5 = 0 and 10 – Q = 0
Q = 5 and Q = 10
AGAIN DIFFERENTIATING W.R.T "Q" Eq (II)
d2dQ2 = 1.8 – 0.24Q
NOW
AT Q = 5,
d2dQ2 = 1.8 – 0.24(5)
= 0.6 > 0 Is Minimum.
AT Q = 10,
d2dQ2 = 1.8 – 0.24(10)
= - 0.6 Is Maximum.
FIRM CAN EARN MAXIMUM PROFIT AT Q = 10
PUTTING Q = 10 In Eq (I)
= -0.04Q3 + 0.9Q2 – 6Q – 5



= -0.04(10)3 + 0.9(10)2 – 6(10) – 5



= -40 + 90 – 60 – 5
= -15