Applied Calculus For Business Students

Applied Calculus for business students Srinivasan Nenmeli -K

NKS-applied calculus

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What are Functions? y= function of x ❖ Functions are the raw materials for Calculus ❖ A function is an "input-output' relation ❖ Write y = function of x or y= f(x) ❖ x the input,y the output ❖ For every input,there is only one output in most cases ❖ Function is mapping x space to y space. NKS-applied calculus

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Simple functions

1 y= constant = c example: y= 3.0 2 y= mx where m is a constant Here this is a straight line passing through the origin. When x= 0 , y=0. Example: Revenue R = 4 (number sold)= 4x $; x is the quantity sold at $4 each unit 3 Combine both functions 1 and 2 given above: NKS-applied calculus

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Linear Equations 1 Straight line equations : y = mx+c 2 Here m is the slope, c is the y-intercept 3 This is the most useful equation of all. Example: John runs a bakery in his garage.The initial cost [ovens,mixers,trays] is $300.The cost for making each cake is $1.5 What is the Cost Equation? 4 Total cost TC=fixed cost + variable cost TC = 300 + 1.5 Q where Q is the quantity of cakes made. 5 Is this a straight line equation? y= mx+c ? what NKS-applied calculus

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Examples of linear equations 1. Revenue

R = price x quantity ;If price is constant, R (Q)= p Q 2. The demand for an item Q,in general,decreases with increasing price.The "demand equation": Q=a-bp where p is the price 3. John employs four persons with total wages $500 per day.They make n cakes NKS-applied calculus

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Rational expressions

1 We get x values in the denominator too.--called 'Rational expressions' 2 The total cost for John: TC = 500 + 2 Q per day to produce Q number of cakes. $500 is the labor cost. 3 What is the cost for making one cake at this production level? 4 Average cost = total cost/Q Avcost = 500/Q + 2 This is not a straight line equation! Watch NKS-applied calculus

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Average Cost

Av-cost = 500/Q +2 Find the average cost for 500 cakes per day and 1000 cakes per day. Q = 500 -----> av-cost = 500/500 +2 =3$ Q= 1000 -----> av-cost = 500/1000 +2 = 2$ When Q increases, av-cost decreases. Plot avcost versus Q. What do you find? NKS-applied calculus

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Combined Expression A function with rational term and linear term is: y= f(x) = a/x + b x + c Example: Joseph finds that his truck is getting old.The capital cost of the truck decreases each year: CC=40000/x where x is the year of usage. The operating and maintenace cost keeps increasing: O&M cost = 5000x Fixed cost=insurance and so on =2000$ The total cost = TC = 40000/x +5000x +2000 Plot this curve This equation has a minimum at x= 2.8 years. Would you replace the truck by a new one at 2.9 NKS-applied calculus 8 years?

Quadratic Equation Add an x.x {x squared term} to a linear equation. y= axx + b x + c They are shaped as parabolas. Parabolas have maximum or minimum points.! Example 1: Revenue R = price x quantity sold = pQ Using Demand equation: p = c - d Q R = p Q = (c-dQ)Q= cQ - dQ.Q This Revenue Equation is quadratic. Example 2 : Cost to produce Q units of any item decreases with increasing Q.Larger the NKS-applied calculus

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Use of differential calculus A common use is to find the max or min points. revenue equation: R = cQ - dQQ Find when this is maximum. dR/dQ = c - 2dQ [see the differentiation rules -slide no:30] Let dR/dQ =0 -----> Q* = c/2d Max revenue R* = cQ* - dQ*Q* Example: The revenue equation for cakes is: [price p = 4 - 0.01Q ] Revenue= pQ= 4Q -0.01QQ Here c= 4, d=0.01 Q*=c/2d=4/0.02=200 NKS-applied calculus

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The Profit Equation What is profit? Profit = Revenue - Cost For John making and selling cakes: revenue equation is: R= 4 Q -0.01QQ His cost equation: C = 100 + 0.2 Q Profit equation becomes: Profit = Pr = R - C= -100+(4-0.2)Q-0.01QQ This is also a quadratic equation.The max profit is found as before: d (Pr)/dQ = 3.8 - 0.02Q =0 Q*= 3.8/0.02=190 cakes. Max profit = -100 + 3.8x190 -0.01(190)(190) =-100 + 722-361= 261$ NKS-applied calculus Is it okay to conclude that greater the production,you

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Break-even point profit = revenue - cost profit =0 ------>when revenue = cost Find the quantity sold for zero profit---> break-even point Example: Linda's 'Home Bakers' bakery sells each cake for $3 to the local store.If Q cakes are sold so far, revenue R = 3Q The cost of making cakes is given by the cost equation: cost = fixed cost + variable cost per cake x Q Fixed cost per month =500$ [rent, power,equipment cost etc] variable cost = $1.25 [fluor,sugar,mixer power,oven power,labor] Cost C = 500 + 1.25 Q What is the break even point Q*? Let R = C and solve for Q 3Q* = 500 + 1.25Q* 3Q* - 1.25Q*NKS-applied = 500calculus 12 Q*=500/(3-1.25)=500/1.75 = 286 cakes.

Marginal Costing Linda is producing cakes for local store at the rate of 200 cakes per day.Her cost equation is: C = 300 - 0.5 Q + 0.01Q.Q Suppose her demand increases to 250 cakes per day.What would be the additional cost for the 50 cakes at her current level of production. The marginal cost is the change in cost for unit increase or one cake at that level. Differentiate first: dC/dQ= -0.5 + 0.02Q For one cake , dC/dQ = -0.5+0.02 (200) = -0.5 +4 = 3.5$ For making 50 more cakes, she has to spend : 3.5x 50 = 165 $ If Linda sells the cakes at $5, she gets a revenue of 5x50 =250 $ ;she gets a profit of 250-165= 85 $ Note that average cost of making donuts at the level of 200 per day: NKS-applied calculus 13 Av-cost= 300/200 -0.5 +0.01(200) = 1.5 -0.5 +2 = $3

Exponential Function This function is an extension of the compound interest formula.---useful for rapid growth or rapid decay of anything. The growth in sales of mobile phones in the past few years is found to be: t =0 [say 2002] sales s=0.2 million t=2 2004 s= 0.4 million t =4 2006 s= 0.8 million t =6 2008 s = 1.6 million What pattern you find in this data? Every two years, the sales in doubling! Doubling time=2 years. This is a case of exponential growth: S= 0.2exp(rt) NKS-applied calculus 14 Here r= 0.347 [exp] stands for a strange

Exponential Function The number of houses built in Sonoma county was 10000 in theyear 2000. Take 2000 as t=0.The number of houses built in 2003 was 14000.If the growth is exponential,what is the exponential function for the housing growth? y(t) = y(0) exp (kt) y(0) = 10000 To find k, y(3) = 14000=10000exp(k.3) exp(3k) = 14/10=1.4 Take logarithms both sides: 3k = ln (1.4)=0.336 k =0.112 y(t) = 10000 exp(0.112t) What is the doubling time,t* : t* = ln2/k =0.693/0.112=6.19 years. What would be the numberNKS-applied of newly built houses in 2006? calculus 15 Using the doubling time: 2000+6.19 years=2006.19

Exponential Drop

Many variables decrease rapidly.--use exponential drop equations: y(t) = y(0)exp(-kt) Note the (-kt) in the exponential function. Example: A text book has market value of $100 in the year of publication.Let that year be t=0. Amanda, a bookstore manager learns that the value y(t) decreases rapidly and follows the equation: y(t) = y(0)exp(-t) where t is the year. Find its value after 1 year, after 2years, after 3 years. y(1) =100 exp(-1) = 100/e = 100/2.72=36.8$ y(2) = 100exp(-2) = 100/(2.72^2)=13.53$ y(3) = 100exp(-3) = 100/ (2.72)^3 =4.97$ Note the rapid drop in market value of the book in 3 years.! Like doubling time,for exponential drop or decay, we use 'Half NKS-applied calculus life'-the time for the original value to drop by 50%

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Exponential Drop Linda, the advertising executive for Skonda motor bikes find that the present sales figure of 50 million$ will drop if advertising is totally stopped for next three months according to this model: Sales S(t) = S(0)exp(-1.5t) where t is time in years. Find the predicted sales after two quarters: t=0.5 years S(t) = 50 (exp ( -1.5x0.5)= 50 exp( -.75)= Using the calculator: exp(0.75)=2.12 exp(-.75) = 1/2.12=0.472 S(1/2) = 50 x 0.472 =23.62 million. The sales has droppedNKS-applied by almost 50%. calculus What is the half life for sales? Half-life =

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Limited Growth Many growth processes do not grow indefinitely.Soon the growth rate falls and the growth comes to a stop--mainly due to limited resources. 1Population growth may get slowed due to lack of food or space or epidemics or wars.[as Malthus discussed.] 2Markets reach a saturation value because potential buyers have already bought the item. Here we use the exponential function in a modified way: y(t) = a [1- exp(-kt)] At t=0, exp(-kt) =exp(0)=1 y(t)=0 y increases gradually; when t is large, exp(-kt) goes to 0; y(t) approaches 'a', the saturation value. Example: From experience, a sales manager estimates that there are 40000 customers in Santa Clara county for large fridges.The market share of his company nationwide is 70%.So18 NKS-applied calculus he hopes to sell 28000 units there.If the growth rate k=0.12,find

Limited Growth -2

y(t) = 28000 [ 1- exp( -0.12 t)] To find the time when y= 14000, 14 = 28[ 1-exp(-.12t)] 1-exp(-.12t)= 0.5 exp(-.12t)=0.5 Taking log both sides: -0.12t = ln(0.5)=-0.693 t = 5.8 years

NKS-applied calculus

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Logarithmic Function

This function is the inverse function of exponential function: If y=exp(x) then ln y = x. While exponential is useful for rapid growth, log function is useful for slow growth. Linda is a training manager at Toy-train works.Inc and finds that the productivity of new employees increases with training period for x hours following this equation: prod p(x) = a + b ln(x)=2.0 + 1.5 ln (x) Find the productivity figures for: no training,for 10 hours and 20 hours of training: x=0 p(x) =2.0 NKS-applied calculus 20

Other Simple Functions y(t) = a + b sqroot(t) y(t) = a/t + bt y(t)= at/(t+b) Example: The cost of transportation of corn flakes is given by :c1=8000/x where x is the number of containers used;cost of storing goes linearly with number of containers: c2=200x Find the number of containers for minimum total cost: TC= 8000/x +200x Differentiate: d(TC)/dx= -8000/x.x +200 Set to zero: -8000 + 200xx =0 x.x=40 NKS-applied calculus 21 x=6.3 0r just 6 containers

What is differentiation?

Differentiation is a process of finding the instantaneous rate of change of y for a small change of x at given value of x. The number of bacterias in a broth is given by : y= 5 t.t where t is in minutes Find the rate of change at t= 10 minutes. Using the formula: dy/dt = 5 (2t) = 10 t Instantanous change at 10 minutes : dy/dt at t=10 mins =10x 10 =100 bacterias per minute.That is the growth rate at 10 NKS-applied calculus

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Finding the maximum revenue The revenue function for selling cakes is given by: R = price x quantity = p Q The price varies with quantity sold: p = 4 - 0.01Q for the quantity between min order of 100 cakes and max order of 250 cakes. For 100 cakes, price = 4 - 0.01 x 100= 4-1 =3$ For 200 cakes ,price = 4-0.01x 200= 2$ and so on. Revenue R = pQ = (4-0.01Q)Q= 4Q-0.01Q.Q To find max revenue: differentiate R with respect to Q,the quantity sold. dR/dQ =4 -0.02Q Set dR/dQ=0 to get the quantity for max revenue. 4 - 0.02Q*=0 ------> Q*=4/0.02 = 200 cakes. Max revenue: R* = 4 x200 - 0.01 (200)(200)= 400$ To check whether this is maximum: take two values of Q higher23 NKS-applied calculus and lower than Q* and find R:

Finding the minimum cost

The cost function in simplest form: Total cost = fixed cost + variable cost = a + b Q where q is the quantity produced. To this we add one more term cQQ which includes a slight increase in cost at higher quantities.[Q is in thousands of cakes,, C is also in 1000$] Example: C= 4 -0.01Q + 0.0005Q.Q for making cakes in large quantities. Find the minimum cost with Q* the optimal quantity. Differentiate: dC/dQ = -0.01 +0.0005(2Q) = -0.01 + 0.001Q Q = 10 or 10000 cakes.. The min cost : C* = 4 - 0.01x10 + 0.0005(100)= 4 - 0.10+.05 C*= 4.04 x 1000$ NKS-applied calculus

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Logistic Growth Curve New activities and businesses grow slowly at first ,then become rapid,later grow slowly and come to zero growth.Suppose you analyse the sale of mobile phones in a region : S(t) = a/ ( 1+ b exp(-ct)) where t is in months. We have used the exponential function exp(-ct) which is placed in the denominator. This function is called Logistic curve or sigmoidal curve. After a long time ,exp(-ct) goes to zero, S(t) reaches 'a' ,the saturation value. The market is then saturated.Still sales continue at 'a' mobile phones per 25 NKS-applied calculus month--a constant. The sales rate goes to zero.

Integrating Profit If you plot profit versus time,the curve will be rarely smooth.The curve will be zig-zag.We can use functions to smoothen it and then integrate over some time-period. Mr Chou Chan found that his chip manufacture showed a steady increase in revenue from 2000 to 2005,with the equation: R(t) = 15 + 0.23 t where t is the time.[take t=0 for 2000 and t=5 for 2005] During the same period ,the cost of making chips also went up: C(t) = 6 + 0.05 t Find the total profit over the period. Profit = P(t) =NKS-applied R(t) calculus - C(t) =15 -6 + (.23 26 0.05)t

Using the integrals Example: An oil well produces 50 x1000 barrels of oil in the beginning [t=0] and then the pumping rate decreases as follows: pumping rate=p'=dp/dt = 50 -4t where t is in years. Using integration find the total production over 15 years. Integ {p'dt} from 1 to 15. p = 50 t - 4t^2/2 = 50t -2t^2 Substituting t=15,we get: p =50x15 - 2 (15)(15) p = 750 -450 = 300 [x1000] barrels NKS-applied calculus

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Integration-Finding the Average

Suppose you have a zig-zag function.How do you find the average value of the function? Find the area under the curve between limits x=a and x=b and then divide by the interval d=b-a. To find the area,integrate the function from a [the lower limit] to b [the upper limit.] Example: The inventory of motor bikes of a particular model has been varying over 12 month period according to the function: y(t) = 4+2.4 t +3t.t Find the average inventory between months t=3 to t=6. Integ{y(t)} = 4t +2.4t^2/2+3t^3/3 = 4t + 1.2t.t + NKS-applied t^3 calculus 28 Integ{6} = 24 +7.2 +216=247.2

Useful Life The useful life or a product line or a mine is reached when the rate of revenue =rate of cost as a function of time: R' = dr/dt = C'=dc/dt Example:A copper mine was operated for a few years. R'=20exp(-0.1t) C'= 4 [value in millions of $] Equating the two, we get: exp(-.1t)=4/20= 0.25 taking logarithms: -0.1t = ln(.25)=-1.39 t= 13.9 or nearly 14 years. Note: A wise manager will shut down the mine after 14 years. Find the total profit from the mine over 14 years. NKS-applied calculus 29 Integ{[R'-C']dt=integ{20exp(-0.1t)-4}=-200exp(-.1t)-4t

A short table of differentiation rules Function y = f(x) y= constant

dy/dx =y'(x) y' (x) =0

y =x -------------->

y'(x) =1

y=x^2------------> y=x ^3-----------> y=x^n ----------->

y'(x) = 2x y'(x) =3x^2 y'(x) = nx^(n-1)

y= exp(x) ---------- y' = exp(x) > y' = 1/x y= ln y' = - 1/x.x (x)--------------> y=1/x --------------->

y' = (1/2)1/sqroot(x) NKS-applied calculus

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Functions of two variables Consider a function z which is a function of two variables: x and y. z=f(x,y) Example: Volume of a right circular cylinder V = pi(r^2).h V (r,h) = (pi r^2).h V is a function of two variables ,radius r and height,h Example : The production function of a petroleum plant is a function of utilization of capital (x) and utilization of labor (y). A well-known model,called "Cobb-Douglas model" applied here: production = z=f(x,y) = 20 [x^0.4][y^0.6]31 NKS-applied calculus If x= 1200 units and y=2000 units, the production=

Partial Derivatives you are given the function : z = f(x,y) Differentiate z with respect to x, keeping y constant: f'(x) Differentiate z with respect to y,keeping x constant: f'(y) These are called partial derivatives. f'(x) gives the rate of change of z for small changes in x, at some level of x and y. Take again: V(r,h) = pi (r.r)h v'(r) = pi (2r)h = 2 pi rh v'(h) = pi (r.r)1=pi(r.r) Example: For Cobb-Douglas model for production function: NKS-applied calculus 32 z= 20(x^0.4)(y^0.6), find the partial derivatives:

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Srinivasan Nenmeli-K Ph D (Columbia) Palo Alto CA 94303 email: nksrinivasan at hotmail dot com Mail your feedback and requests for any specific tutorial presentations/publishing interests NKS-applied calculus

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