Calculus (Part-II) for Undergraduates

Calculus (Part-II) for Undergraduates By Dr. Anju Gupta Director, NCWEB, University of Delhi Ms. Surbhi Jain Assistant Professor, NCWEB, University of Delhi Elasticity of Demand and Supply In economics, the term elasticity refers to the responsiveness of one economic variable to change sin another economic variable. Theelasticityismeasuredintermsofpercentage changes instead of absolute changes. This means we measure the change in a variable as a percentage of the original amount of the variable. For instance, the percent change in a variable X is defined as Percentage change in x = Change in the variable X Original value of X This definition can be symbolized in a compact form by symbolizing the change as X. Percentage change in x = X/X Suppose that the value of X changes from 20to 30.Thepercentagechangeis Percentage change in X = 30 20 = 0.5 20 Written as a percentage, this is a percentage change in X of 50%. To find the elasticity of Y with respect to X, we need to calculate percentage change in Y due to percentage change in X. The elasticity of Y with respect to X is Elasticity of Y with respect to X = Percentage change in Y Percentage change in X 1

Now, we will use above stated concept to derive price elasticity of demand.price elasticity of demand is an important economic concept. It is based on law of demand. Law of demand states the quantitative relationship between price and quantity demanded i.e. if price is high, quantity demanded is low and vice-versa. The same quantitative relationships are quantified by price elasticity of demand. It is always negative because of negative relationships between price and quantity stated by law of demand. Say, thepriceof coffeeincreased from Rs1.65to Rs1.70.Asexpected,thedemandfor coffee dropped from440units per day to436unitsperday.thepercentchangeinpricep is P/P= 1.70-1.65/1.65 = 0.0303or about 3%.Thepercentchangeinthedemand Qis Q/Q= 436-440/440 = -0.009 or about - 0.9%. An 3% increaseinpricecorrespondsto a0.9% decreaseinthequantityofcoffee sold. Theelasticityofdemandwithrespectto priceorpriceelasticityofdemandcan be derived as E= Percentage change in Q = -.0009 Percentage change in P.03 Iftherelationshipbetweendemandandpriceisgivenby function,wecanutilizethederivativeofthedemandfunctiontocalculatethepriceelasticityofdeman d. Change in price being P to P + Pand corresponding change in demand as Q to Q + Q the corresponding percent changes in price and quantity is P/P and Q/Q. It is defined as ratio of percentage change in the demand of a commodity to a given percentage change in price. Thedefinitionofelasticity of demand leadsto a QQ QQ E = - PP PP = - QQ QQ PP PP = PP QQ QQ PP ŋ= PP QQ = PP QQ PP QQ E= PercentagechangeinQ Percentagechangein P Elasticity is denoted by ŋ also. (Note that since demand is a decreasing function of p, the derivative is negative. If E <1, we say demand is inelastic. In this case, rising prices increase revenue. If E >1, we say demand is elastic. In this case, rising prices decreases revenue. If E=1, we say demand is unitary. In this case increase or decrease in price result into no change in revenue. 2

Example 1:- A smalltechnologycompanyinnortherncaliforniaisdeveloping atablet PC to competewithapple sipad.basedonmarketsurveys,thecompanybelievesthequantityq(inthousandsof units)thatwillbedemandedbyconsumersisrelatedtotheprice P(inthousandsofrupees)bytherelationship Q= 4000 250P 2 FindthepriceelasticityofdemandwhenthepriceofthetabletPCis Rs.3000. Sol:- Tocomputethepriceelasticityofdemand,weneedtofindthederivative ofthedemandfunctionq= 4000 250P 2 = 4000 250PP2 = 2(250P) = 500P Substitutetheexpression for the derivative and the Q=4000 ddpp 250P2 into the elasticity formula yields E = PP QQ = PP ( 500PP) = 3 4000 250PP 2 4000 250(3) 2 500(3) = 2.57 Since the elasticity E=-2.57 is more than -1, the demand is elastic. Example 2 :- Using data from market years 2000 through 2010, the relationship between the price per bushel of oats and the number of bushels of oats sold is given byq= 152.07P 0.543 Find the price elasticity of demand when the price per bushel is Rs.1.75. where P is price and Q is quantity demanded. Sol:- Since the relationship between the quantity and price is given as a function of price, we may compute the price elasticity of demand from E = PP QQ QQ PP = PP QQ ddpp The derivate dq/dp is found from the relationship between quantity and price. Using the Product Rule and Power Rule we get, = 152.07( 0.543PP 0.543 ) = 82.574PP 0.543 Substitute the expression for the derivative and the Q=152.07P 0.543 into the elasticity formula yields E = PP QQ QQ PP = 1.75 152.07PP 0.543 82.574PP 0.543 = 0.543 Price elasticity of Supply:- It is ratio of percentage change in the supply of a commodity to a given percentage change in price. ŋ(elasticity of supply)= PP QQ QQ PP = PP QQ Since both price and quantity supplied go up and down together, it has a positive sign. 3

Example 3 :- The supply function for a commodity is = 2pp 2 + 5. Find the elasticity of supply when p=3. Sol:- E = PP = pp (4pp) = 2pp 2 +5 4pp2 / 2pp 2 + 5 E= 4(3) 2 /2(3) 2 + 5 =36/23 Measurement of Price elasticity of demand and supply at equilibrium price In case demand and supply functions for a commodity are given, it is possible to measure both price elasticity of demand and supply at equilibrium price using simple differentiation. The first step is the determination of equilibrium price and quantity using demand and supply function. Equilibrium price is the price at which both quantity demanded and supplied are equal and corresponding quantity is known as equilibrium quantity, so X = f(p) (demand function) and X = g(p) (Supply function) Then put f(p) = g(p), this gives equilibrium price, putting price in either demand function or supply function and solving for xgives equilibrium quantity. At this equilibrium price, price elasticity of demand is obtained by using demand function and price elasticity of supply is obtained by using supply function. The only difference in this application and the previous two is that in this elasticities are calculated at equilibrium price only whereas in earlier two at any arbitrary price (given). Example 4:- Find the elasticities of demand and supply at equilibrium point for the demand function pp = 100 2 and the supply function x=2p-10, where p is price and x is quantity. Sol. We are given: pp = 100 2 x=2p-10 or p= (x+10)/2 at equilibrium demand=supply : demand function : supply function 100 2 = ( + 10)/2 oooo 4(100 2 ) = ( + 10) 2 2 + 4 60 = 0 ii. ee. ( 6)( + 10) = 0 = 6 oooo = 10 we reject the since = 10 since x represents quantity. Substituting x=6 into the demand function or the supply function,we obtain p=8.thus equilibrium point is(6,8) Computation of ŋ(elasticity of demand)= pp = 100 2 = = 100 2 = = 100 2 ŋ = PP = 100 2 x 100 2 QQ when x=6, ŋ=100-36/36 = 64/36 =16/9 Computation of ŋ(elasticity of supply) = (100 2 )/ 2 4

= 2pp 10 ŋ = PP QQ = pp 2pp 10 when p=8,ŋ=8/3 = = 2 x 2= pp pp 5 Exercise Q1.The demand function for a commodity is p=10+5x-2 22.Find the elasticity of demand when x=2. Sol. 2 Q2.The The demand function for a commodity is x= 500-40p+pp 2.Find the elasticity of demand when p=15. Sol 1.2 Q3. The supply curve for a commodity is p= 9 +.Find the elasticity of supply when p=4. Sol. 32/7 Q4. :- Find the elasticities of demand and supply at equilibrium point for the demand function pp = 16 2 and the supply function pp = 2 2 + 4, where p is price and x is quantity. Sol. elasticity of demand=3/2 and elasticity of supply=3/4 5

As we have already studied earlier that if y = f(x), we can calculate and it is called first order derivative. In the same way, we can calculate second order derivates i.e. dd2 yy dd 2. For example. Lety=15x 4 3x 2 Now = 603 6 Second order derivative = dd = dd2 yy = dd dd 2 (603 6) = 180 2 6 MaximaandMinima Lety=f(x),afunctionofx,becontinuousneary=xsovalueoff(x)iscalledtheminimumormaximumvalu eoff(x)accordinglyasvalueoff(x+h)islessthanorgreaterthanf(x).valueofh (increaseordecrease)istakentobesmallinmagnitudebutitisneverequaltozero. Steps to find out whether a function is maxima or minima:- 1.Find i.e. f (x) for the function f(x). 2. Put f (x) = 0 and solve this equation to obtain various values of x ( say a1, a2, a3 ) these are the only points at which f (x) will have minimum or maximum value. 3. Find dd 2 yy 2 i.e. f (x) and find its value by substituting the values of a1, a2, a3 in f (x). 4. If dd 2 yy is negative then f(x) is maximum at x = a1. If dd 2 yy 2 2 is positive then f(x) is minimum at x = a1. Similarly, we can check for other values a2, a3 Example 5. Find the local maximum and minimum for the following functions using the first derivative test only. ff() = 3 6 2 + 9 + 15 Sol. f(x) = 3 6 2 + 9 + 15 f (x) = 3 2 12 + 9 = 3( 2 4 + 3) = 3( 1)( 3) f (x)=0 3( 1)( 3) = 0 = 3,1 Now, dd 2 yy 2 = 3(2 4) Put value of x in above At x=1, dd 2 yy 2 = 3(2(1) 4) = 6 AAAA = 3, dd2 yy = 3(2(3) 4) = 6 2 Hence the above function is maxima at x=1 and minima at x=3 6

Some important functions:- Cost Function The total cost C of producing and marketing x units of a product depends upon the number of units (x). So the function relating C and x is called Cost-function and is written as C = C (x). The total cost of producing x units of the product consists of two parts (i) (ii) Fixed Cost = F Variable Cost =V(x) Therefore, C (x) = F + V (x) Fixed Cost : The fixed cost consists of all types of costs which do not change with the level ofproduction. For example, the rent of the premises, the insurance, taxes, etc. Variable Cost : The variable cost is the sum of all costs that are dependent on the level ofproduction. For example, the cost of material, labour cost, cost of packaging, etc. Demand Function An equation that relates price per unit and quantity demanded at that price is called a demand function. If 'p' is the price per unit of a certain product and x is the number of units demanded, then we can write the demand function as x=f(p) or p = g (x) i.e., price (p) expressed as a function of x. Revenue function If x is the number of units of certain product sold at a rate of Rs. 'p' per unit, then the amount derived from the sale of x units of a product is the total revenue. Thus, if R represents the total revenue from x units of the product at the rate of Rs. 'p' per unit then R(x)= (p)(x) is the total revenue function Profit Function The profit is calculated by subtracting the total cost from the total revenue obtained by selling x units of a product. Thus, if P (x) is the profit function, then P(x) = R(x) C(x) Average and Marginal Functions If two quantities x and y are related as y =f(x), then the average function may be defined as f(x)/x and the marginal function is the instantaneous rate of change of y with respect to x. i.e. 7

Given Total Cost as C(x). hence Marginal Cost = dd CC() cc() and Average Cost = Given Total Revenue = as R(x). hence Marginal Revenue = dd RR() RR() and Average Revenue = Applications of Maxima and Minima (a) Maximisation of total revenue function : If a firm's demand function is given, it is possible to determine the corresponding level of output whose total revenue is maximum using maxima and minima. But it will be possible only if firm is operating under imperfect competition (where price decreases as firm sells more units). In case of perfect competition, price is constant for a firm resulting in total revenue as an increasing function. The total revenue is maximum when marginal revenue is zero. R=p.x is the revenue function, where R is total revenue and p is per unit and x is quantity sold. (b) Minimisation of Average cost function Maxima and minima can also be used for determining the level of output where per unit cost is minimum corresponding to given total or average cost function. It is often used in economic analysis. It gives the optimum level of output. For a given cost function C(x) its Average Cost = CC() It is very important to find the level of output for which the average cost is minimum. Using calculus this can be calculated as dd dd2 (AAAA) = 0 and to get the value of x for which (AAAA) > 0 dd 2 Similarly to find the level of output for which the total revenue is maximum. Using calculus this can be calculated as dd dd2 (RR()) = 0 and to get the value of x for which (RR()) < 0 dd 2 Similarly to find the level of output for which the profit is maximum. Using calculus this can be calculated as dd dd2 (PP()) = 0 and to get the value of x for which (PP()) < 0 dd 2 Example 6:- The manufacturing cost of an item consists of Rs.6000 as over heads, material cost Rs. 5 per unit and labour cost Rs 2 for x units produced. Find how many units must be 60 produced so that the average cost is minimum. Sol. Total cost C(x)= 6000 + 5 + 2 60 8

Average cost = 6000 Now dd + 5 + 60 AAAA = 6000 2 + 1 60 dd 6000 AAAA = 0 = 2 + 1 60 = 0 2 = 3,60,000 x= 600 dd 2 12000 dd 2 AAAA = > 0 at x=600. 3 Hence AC is minimum when x=600. Example 7. Total cost function is given as C(x) where x is the number of units produced. Determine the number of units that must be produced to minimize the total cost. Sol: CC() = 3 615 2 /2+15750x+180000 = dd CC() = 32 615+15750 Put dd CC() = 0 = 3 2 615+15750=0 = ( 175)( 30) x=175, 30 dd 2 = 6 615 dd2 At x=175, dd2 dd2 = 435 and at x= 30 = 435 dd 2 dd 2 Hence cost is minimum at when 175units are produced. Example 8:- The cost function for x units of a product produced and sold by the company is CC() = 250 + 0.005 2 and the total revenue is given as R=4x. Find how many items should be produced to maximize the profit. What is the total profit? Sol:- CC() = 250 + 0.005 2 and R(x) =4x P(x) = R(x) C(x) and P(x) = 4 250 + 0.005 2 = dd PP() = 4 0.001 PPPPPP dd PP() = 0 4 0.001 = 0 x=400 dd 2 PP dd 2 = 0.001 < 0 Hence maximum profit is when x=400 Maximum Profit = P(x) = 4 250 + 0.005 2 P(x) = 4(400) 250 + 0.005(400) 2 =550 9

Exercise Q1. A company s total revenue function is RR() = 40,00,000 ( 2000) 2,where R is the total revenue and x is the number of units sold. (i) Find what number of units sold will maximizes the revenue. (ii) What is the amount of maximum revenue? (iii) What would be the total revenue if 2500 units are sold. Sol (i) 2000 (ii) 40,00,000 (iii) 37,50,000 Q2.A company charges Rs15000 for a television set on orders of 60 or less sets. the charges are reduced on every set by Rs100 per set ordered in excess of 60 sets. Find out the largest order size to maximize revenue. Sol 80 sets Q3. A demand equation for a manufacturer product is p=(80-x)/4, where x is the number of units and p is price per unit. At what value of x will there be maximum revenue? what is the maximum revenue? Sol. 40, 400 Q4. A manufacturer s total cost function is given as 2 + 3 + 400, where x is the number of 4 units produced. At what level of output will average cost be minimum. What is this minimum Sol x=40, AC=23 Q5. let the cost function be C=300x-10 2 +1/3 3, where c is total cost and x is output. Calculate (i)the output for which the marginal cost is minimum. (ii)the output for which the average cost is minimum. Sol 1 and 3/2 10

Unit 2- Topic 2.4 Integration Now we will consider the inverse process of differentiation. In differentiation, we find the differential co-efficient of a given function while in integration if we are given the differential co-efficient of a function, we have to find the function. That is why integration is called antiderivative i.e. in differentiation if y = f(x) we find dx dy. In integration, we are given dx and we have to find y. This integration is also called indefinite integral. dy Given a function f,if Fis a function such that FF () = ff() then F is called a antiderative of f. Thus the antiderative of f is simply a function whose derivate is f. Defination: if the function F(x)the antiderative of f(x),then the simple integral of f(x) is defined as F(x)+c, where c is the arbitrary constant. It is denoted as f(x)dx. Example 1.d(x 2 )=2x dx 2xdx=x 2 Example 2. Find 5x 6 dx 5x 6.dx = 5 x 6 dx= 5 x 7 +C =5 x 7 +C 7 7 Standard Formulas 1. x n dx= x n+1 + C n+1 2. 1 dx= log x + C X 3. e x dx = e x + C 4. e ax+b dx = e ax+b + C a 5. (ax+ b) n dx = (ax+ b) n + C a(n+1) 11

Example 3. Find the integrals of the following (i) 3 dx (ii) 1 dx (iii) (5 3)3 (iv) ee 5 dx Solution (i) 3 dx= 3+1 / 3+1 +c = 4 /4 +c (ii) 1 1 dx = 2 = 1 2 + CC = 2 + cc 1 +1 2 (iii) (5 3) 3 = (5 3)3+1. 1 = 1 (5 3+1 5 20 3)4 + cc (iv) ee 5 =)ee 5 / 5 + cc Methods of Integration 1. Integration by substitution Example 4.Evaluate 2 + 1 2 Solution. Let uu = 2 + 1 ; = 2 Using the rule of substitution, we get Replacing uu bbbb 2 + 1, we get 2 + 1 2 = uu 1 2 = 2 3 uu3 2 + cc 2 + 1 2 = 2 3 (2 + 1) 3 2 + cc 2. Integration by parts uuuu = uu vv dduu. vv Example 5.Evaluate the following integral (i) ee Integrating by parts taking uu =, we get 3. Integration by Partial Fractions = ee ee = ee ee = ee ( 1) + cc NN() RR() = QQ() + DD() DD() 12

Example 6.Find (+1)( 2) ( + 1)( 2) = AA 1 + BB 2 AA( 2) + BB( + 1) = ( + 1)( 2) ( + 1)( 2) = AA( 2) + BB( + 1) = (AA + BB) 2AA + BB We can write the above equation in the following manner; + 0 = (AA + BB) 2AA + BB We get two necessary conditions to solve the above equation & By solving (i) and (ii) we get = (AA + BB) (i) AA + BB = 1 2AA + BB = 0 (ii) 2AA = BB AA = 1 3 & BB = 2 3 1/3 = ( + 1)( 2) + 1 + 2/3 2 = 1 3 log( + 1) + 2 log( 2) + cc 3 = 1 3 log( + 1) + 1 3 log( 2)2 + cc = 1 3 [log( + 1) + log( 2)2 ] + cc 13

Applications of Integral in Business and Economics (a)determination of cost function from marginal cost function If C denotes the total cost and MC= is the marginal cost, then we can write C = C ( x )= ( MC ) dx + k, where k is the constant of integration, k, being the constant, isthe fixed cost. Example 3 :- The marginal cost function of manufacturing x units of a product is 5 + 16x 3x 2. The total cost of producing 5 items is Rs. 500. Find the total cost function. Solution :- MC = 5+ 16x -3x 2 C(x)= (5 + 16x -3x2 ) dx CC() = 5 + 162 2 33 3 + kk C (x)= 5x +8x 2 - x 3 +k When x = 5, C(x) = C(5) = 500 500=25+200-125+k k=400 Hence, CC() = 5 + 8 2 3 = 400 (b) Determination of Total Revenue Function from marginal revenue function If R(x) denotes the total revenue function and MR is the marginal revenue function, then MR= dd (RR()) RR() = (MMMM) + kk,where k is the constant of integration. Also, when R (x) is known, the demand function can be found aspp = RR() Example 4:- The marginal revenue function of a commodity is given asmr = 12 32 + 4. Find the total revenue and the corresponding demand function. Solution RR() = (MMMM) + kk RR() = (12 3 2 + 4) + kk 14

RR() = 12 3 + 2 2 PP = RR() = 12 + 2 2 (constant of integration is zero here) (c) Derivation of demand function : As we knowŋ= PP where P is price per unit and Q is QQ quantity demanded. Using techniques of integration, we can find out the demand function using elasticity of demand. Example 5:- Elasticity of demand for a commodity is p/ 3.Find the demand function if the demand is 3 when the price is 2. Sol:ŋ= PP = p = PP = p QQ 3 3 = 2 + = 0 Integrating we get 3 + pp = kk,where k is the arbitrary constant. However it is given that x=3 and 3 p=2. (3) 3 + 2 = kk = kk = 11 3 hence 3 3 + pp = 11 aaaaaa pp = 11, which is the required demand function. 3 3 (d) Measurement of consumer's surplus: In economics, consumer's surplus is defined as the difference between the price consumer is willing to pay and the price which he actually pays. If the price consumer is willing to pay, is more than the price he actually pays, the difference is consumer's surplus. Otherwise there will be no consumer's surplus. The price which the consumer is willing to pay is given by demand function and the price which he pays is given by market demand and supply. Diagrammatically it is shown in figure below. Let P = f(q) is a price function stating the price consumer is willing to pay for different values of Q and P* is the price fixed by market demand and supply at which consumer is purchasing Q* units. Then total price consumer is willing to pay, is Pwhereas price actually paid is P*. So consumer is getting same quantity at a lower price than what he is willing to pay and hence the total of consumer benefits, i.e. shaded area is the consumer's surplus. QQ CS= DD(QQ) QQ PP, where Q * is the equilibrium quantity and PP is the equilibrium price 0 (price where market demand equals market supply) 15

An alternative method is to measure consumer's surplus with the help of integrating demand function(as function of p) with respect to p: pp CS= DD(ffffffffffffffff oooo pp) pp the equilibrium price, where p is the price when quantity demanded i.e. Q=0 and PP is Example:- The demand law for a commodity is pp = 20 2 2. Find the consumer surplus when the market demand is 3. Sol: Market demand is equals to equilibrium demand i.e. x=3.substituting this value in demand function pp = 20 2(3) (3) 2 = 5 CS= QQ 0 DD(QQ) QQ PP 3 = (20 2 2 ) 3 5 0 = 20 2 3 /3 3 0 15 = 20(3) (3) 2 (3)3 3 20(0) (0) 2 (0)3 15 3 = 27 Hence the consumer surplus is 27. (e) Measurement of Producer's Surplus: As in the case of consumer's surplus, producer's surplus is the difference between the price received by the producer and the price at which producer is willing to supply. The price producer is willing to accept is given by his supply curve and the price he actually receives is determined by market demand and supply. Diagrammatically it is shown in figure below. Let P = f(q) is a supply function stating the price producer is willing to sell for different values of Q and P* is the price fixed by market demand and supply at which producer is selling Q* units. Then total price producer is willing to receive, is 0whereas price actually received is P*.So producer more price for the same quantity than what he is willing to receive and hence the total of producer benefits, i.e. shaded area is the producer s surplus. 16

Therefore Producer's Surplus QQ PS=QQ PP SS(QQ), where Q * is the equilibrium quantity and PP is the equilibrium price 0 (price where market demand equals market supply) An alternative method is to measure producer's surplus with the help of integrating supply function(as function of p) with respect to p: PS= pp pp SS( aaaa aa ffffffffffffffff oooo pp) where pp is the price when quantity supplied i.e. Q=0 and PP is the equilibrium price, Example:-The supply function for a commodity is 100pp = ( + 20) 2. Find the producer surplus when the market price is 25. Sol: : Market price is equals to equilibrium price i.e. p*=25. Substituting this value in suppl function 100pp = ( + 20) 2 = pp = (+20)2 100 we obtain x*=30. PS =QQ PP QQ 0 SS(QQ) 30 0 = 25 30 (+20)2 100 = 750 1/300( + 20) 3 30 0 750 1/300{(50) 3 (20) 3 } = 360 Hence the producer surplus is 360. Exercise Q1The marginal cost function of manufacturing x units of a product is MMMM = 4 2 + 2. The fixed cost is Rs100. Find the total cost function. (Sol: CC = 4 2 + 3 /3+100) Q2 The marginal cost of production is found to be MMMM = 1000 20 + 2, where x is the number of units produced. The fixed cost of production is Rs9000. Find the total cost function. (Sol: CC = 1000 10 2 + 3 /3+9000) 17

Q3The marginal revenue function of a commodity is given asmr = 2 6. Find Total revenue function.(sol: pp = 2 3 2 ) Q4The marginal revenue function of a commodity is given asmr = 9 42. Find demand function.(sol: pp = 9 4 3 2 ) Q5 The price elasticity of demand for a commodity is pp/.find the demand function if the demand is 3 when the price is 1. (Sol: 4 ) Q6 The price elasticity of demand for a commodity is pp/(pp 1)(pp 2) where p is the price.find the demand function if the quantity demand is 5 when the price is 3. [Sol: x= 5(pp+3) ] 6(pp 2) Q7The demand law for a commodity is pp = 35 2 2. Find the consumer surplus when the equilibrium price is p = 20. (Sol: CS=27) Q8The demand law for a commodity is pp = 4 2. Find the consumer surplus when the market demand is 1. (Sol: CS=2/3) Q9 The supply function for a commodity is pp = 4 +. Find the producer surplus if 12 units of goods are sold. (Sol: PS=72) Q10 The supply function for a commodity is pp = 1 2 ( + 3) Find the producer surplus when the market price is 2. (Sol: PS=1/4) 18