Applied Calculus for Business Students

NKS-applied calculus 1

Applied Calculusfor business students

Srinivasan Nenmeli -K


What are Functions?

❖ 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.

y= function of x


Simple functions

1 y= constant = c example: y= 3.0

2 y= mx where m is a constantHere 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 unit3 Combine both functions 1 and 2 given above: y=mx + c This is also a straight line ; when x=0, y=c -----> the y 'intercept'=c


Linear Equations

1 Straight line equations : y = mx+c2 Here m is the slope, c is the y-intercept3 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 $1.5What 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 is m here: m= c= .


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 - b p where p is the price

3. John employs four persons with total wages $500 per day.They make n cakes for $2 each per day.Total cost = 500 + 2 n per day C = 500 + 2n


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 + 2This is not a straight line equation! Watch out!!5 As Q increases, avcost decreases.!


Average Cost

Av-cost = 500/Q +2Find 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?Describe the shape of the curve!


Combined Expression

A function with rational term and linear term is: y= f(x) = a/x + b x + cExample: 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 = 5000xFixed cost=insurance and so on =2000$The total cost = TC = 40000/x +5000x +2000Plot this curveThis equation has a minimum at x= 2.8 years.Would you replace the truck by a new one at 2.9 years?


Quadratic Equation

Add an x.x {x squared term} to a linear equation.y= axx + b x + cThey are shaped as parabolas.Parabolas have maximum or minimum points.!Example 1: Revenue R = price x quantity sold = pQUsing Demand equation: p = c - d Q R = p Q = (c-dQ)Q= cQ - dQ.QThis Revenue Equation is quadratic.Example 2 : Cost to produce Q units of any item decreases with increasing Q.Larger the production scale,lower the cost per unit. Cost Equation: C = aQQ +bQ + cThis curve has a minimum point.


Use of differential calculusA common use is to find the max or min points.revenue equation: R = cQ - dQQFind when this is maximum. dR/dQ = c - 2dQ [see the differentiation rules -slide no:30]Let dR/dQ =0 -----> Q* = c/2dMax 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 Max revenue is reached at 200 cakes.max revenue=4x200 - 0.01x200x200=400$ max revenue = 4x200 - 0.01(40000)=$400


The Profit Equation

What is profit? Profit = Revenue - Cost For John making and selling cakes:revenue equation is: R= 4 Q -0.01QQHis 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$Is it okay to conclude that greater the production,you get more profit ? Think for a moment.


Break-even point

profit = revenue - costprofit =0 ------>when revenue = costFind the quantity sold for zero profit---> break-even pointExample: Linda's 'Home Bakers' bakery sells each cake for $3 to the local store.If Q cakes are sold so far, revenue R = 3QThe cost of making cakes is given by the cost equation: cost = fixed cost + variable cost per cake x QFixed cost per month =500$ [rent, power,equipment cost etc]variable cost = $1.25 [fluor,sugar,mixer power,oven power,labor] Cost C = 500 + 1.25 QWhat is the break even point Q*? Let R = C and solve for Q 3Q* = 500 + 1.25Q* 3Q* - 1.25Q* = 500 Q*=500/(3-1.25)=500/1.75 = 286 cakes.Linda gets profit only if the sells more than 286 cakes per month.Or else, she suffers loss.


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.QSuppose 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 + 0.02QFor 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: Av-cost= 300/200 -0.5 +0.01(200) = 1.5 -0.5 +2 = $3Therefore marginal cost is not the same as average cost.!Think about this. Linda should consider marginal cost for deciding to take extra orders!


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 millionWhat 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) Here r= 0.347 [exp] stands for a strange number:e=2.72Use a calculator and press the key e^xCheck: t=3 s=0.2 x exp(0.347x3)=0.2exp(1.041)=0.566 million


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) = 10000To find k, y(3) = 14000=10000exp(k.3) exp(3k) = 14/10=1.4Take 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 number of newly built houses in 2006? Using the doubling time: 2000+6.19 years=2006.19Approximately 20000new houses.Keep this in mind: doubling time= ln2/k=0.693/k or 0.7/k {approx}


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 life'-the time for the original value to drop by 50% Half-life = ln2/k = 0.693/k


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 dropped by almost 50%.What is the half life for sales? Half-life = 0.693/1.5=0.46=.5 nearly


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)=0y 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%.So he hopes to sell 28000 units there.If the growth rate k=0.12,find the equation for growth and find the time when he would have sold 14000 units.[See the next slide]Plot the graph of this function.


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)=0.5Taking log both sides: -0.12t = ln(0.5)=-0.693 t = 5.8 years


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) =2.0x=10 p(x) = 2.0 +1.5. ln10 = 2+ 3.45 = 5.45x=20 p(x) = 2.0 + 1.5. ln20 = 2+ 4.5 = 6.5Note that if we double the training period, the productivity increases only by 20%.


Other Simple Functions

y(t) = a + b sqroot(t)y(t) = a/t + bty(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 +200xDifferentiate: d(TC)/dx= -8000/x.x +200Set to zero: -8000 + 200xx =0 x.x=40 x=6.3 0r just 6 containersTherefore each order may be 6 containers for this item.


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 minutesFind the rate of change at t= 10 minutes.Using the formula: dy/dt = 5 (2t) = 10 tInstantanous change at 10 minutes : dy/dt at t=10 mins =10x 10 =100 bacterias per minute.That is the growth rate at 10 minutes.[see slide 30 for differentiation rules.]


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 -0.02QSet 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 higher and lower than Q* and find R:R(150)= 4(150)-0.01(150)(150)= $375R(250) = 4(250) - 0.01(250)(250)= $375 So R* =400 is the maximum


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$] Example: C= 4 -0.01Q + 0.0005Q.Qfor making cakes in large quantities.Find the minimum cost with Q* the optimal quantity. Differentiate: dC/dQ = -0.01 +0.0005(2Q) = -0.01 + 0.001Q Q = 10 or 10000 cakes..The min cost : C* = 4 - 0.01x10 + 0.0005(100)= 4 - 0.10+.05 C*= 4.04 x 1000$


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 month--a constant. The sales rate goes to zero.The growth rate is maximium when sales is at [a/2] level--that is half of the maximum sales.Many businesses follow this growth curve.


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 tFind the total profit over the period. Profit = P(t) = R(t) - C(t) =15 -6 + (.23 - 0.05)t P(t) = 9 + 0.225 t Integrate over t :we get: 9t +0.225 t.t/2=9t + 0.1125t^2with limits 0 to 5: Total profit = 45+ 2.75=47.76 million$


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^2Substituting t=15,we get: p =50x15 - 2 (15)(15) p = 750 -450 = 300 [x1000] barrels


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.tFind 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 + t^3Integ{6} = 24 +7.2 +216=247.2Integ{3} = 12+10.8+27=49.8Integ{ [3,6]} = 247.2-49.8 = 197.4Average value in this period= 197.4/3=65.8 or 66 [approx]


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/dtExample: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 after14 years.Find the total profit from the mine over 14 years.Integ{[R'-C']dt=integ{20exp(-0.1t)-4}=-200exp(-.1t)-4tEvaluate at t=14 and at t=0 and take the difference:-200exp(-1.4)-4x14+200=-48.8 - 56 +200= 95.2 million $


A short table of differentiation rules

Function y = f(x) y= constant

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

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

y= exp(x) ----------> y= ln (x)-------------->

y=1/x --------------->

y=sqroot(x) --------->

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

y'(x) =1

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

y' = exp(x)y' = 1/x

y' = - 1/x.x

y' = (1/2)1/sqroot(x)


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).hV 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]If x= 1200 units and y=2000 units, the production= z= 20[ (1200)^0.4][(2000)^0.6]=20 (17)(95)=1615units


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: z= 20(x^0.4)(y^0.6), find the partial derivatives: z'(x) = 20 (.4)x^(-0.6)(y^0.6)=8 (x^(-0.6)y^0.6 z'(y) = 20 (x^0.4) [0.6y^(-0.4)]=12(x^0.4)(y^-0.4)Find the partial derivatives for x= 1200, y=2000:Partial derivatives give the marginal productivity wrt capital and labor .Which one is greater?


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Srinivasan Nenmeli-K Ph D (Columbia) Palo Alto CA 94303

email: nksrinivasan at hotmail dot comMail your feedback and requests for any specific tutorial presentations/publishing interests

Applied Calculus for Business Students

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