~ 0 ~

INTRODUCTION

TO

DERIVATIVES

CHAPTER

1

This chapter aims at

making you JUGGLE

with derivatives!

Understand

derivatives as a

concept which grew

from a small idea to a

standalone branch.

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DERIVATIVES – THE CONCEPT Now that we know what a function is, let us try to measure the way a function changes with respect to the change in inputs. If the function can be represented as a straight line, it is obviously easy to calculate the rate of change of the function, which otherwise is the slope of the straight line! But, what if the function represents a curve, instead of a straight line? Then, how do we determine the rate of change? For a curve, as we all know that the slope changes over the entire stretch and if the slope varies and is not constant then, it becomes difficult to determine the change in function at any given point using a general form of equation. This limitation is overcome by using derivatives. While computing derivative, we utilize the differentiability and dependability of every point in a function on the function itself. In other words, first looking at the property of the entire function, we attempt to find the property of a single point in the function. A similar activity is sometimes followed in our day-to-day lives to. Suppose, you are a part of an MBA class which is known to be loud, noisy, and talkative then this image of the entire class would be seen even on you by your faculties. The idea behind finding derivative of a function is very similar. As we try to determine the changes in a function with respect to the changing independent variable, we start with the slope of a larger portion on the function (here, a curve) and reduce the distance between the two points to the lowest possible value. Look at the curve given here. The line drawn tangentially the point A is shown. Will this tangential line remain same over the entire stretch of this curve? Obviously not! Thus, finding the rate of change of the function here is much more complicated than just determining the slope of a straight line. Now, look at the figure given here. Our aim is to differentiate the curve AB which here is represented by the function y = f(x) So, as per the earlier argument, I will start with a larger portion of the curve (Here shown by the arc CD) and will reduce the distance between C and D to such an extent that it finally becomes almost one and the same point, thus giving us the slope at only that point. We join the points C and D with a straight line. The point C has coordinates (x, f(x)) and the point D has coordinates (x+h, f(x+h)). The slope of this line will be found in terms of unit change in Y-coordinate for this line for every unit change in X-coordinates from the points D to C. Thus, Slope = y1 – y2 x1 – x2 If point C is (x2,y2) = (x, f(x)) and D is (x1,y1) = (x+h, f(x+h)); the slope can be written as: Slope = f(x+h) – f(x) (x+h) – (x)

A

B

C

D

A

B

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We bring the points C and D close so that the arc CD and the secant CD are one and the same, at this stage the slope of secant CD will obviously equal the slope of arc CD! To minimize the distance between C and D on the x axis (in terms of ‘h’), we will use the concept of limits which will be discussed in detail further. Slope, as such, physically represents rate of change. Thus if the function represents a distance between two points, then slope or the rate of change of speed will be the speed! Similarly, if the function stands for the speed, then obviously its slope given as rate of change of speed would be the acceleration. It is imperative to understand here that the rate of change is measured for speed and acceleration with respect to time (which is perhaps the most definitely cited independent variable!).

DERIVATIVE OF A FUNCTION In addition to determining the continuity of a function, limits are also required to estimate the derivative of a function. Let us first enumerate the properties that we debated in the beginning of this chapter which lead to the concept of derivative. Firstly, the derivative is found for a curve because the slope changes at every point on the curve. Secondly, the slope represents rate of change which essentially is not constant over the entire curve (unless it is a straight line). In order to approximate the rate of change at every point on the curve (which is represented by a function) with respect to the input to the function, we start with the slope of the line joining points on a larger stretch of the curve as shown in this figure and then reducing the distance between these two points to the minimum (by applying the limits) so that the derivative is found. I hope the figure shown here to the right seems familiar to you. We had used this to determine the slope between the points (x, f(x)) and (x+h, f(x+h)). We have already established that, given the two points on a curve, the slope would be

Slope = f(x+h) – f(x) (x+h) – (x) Or, Slope = f(x+h) – f(x) h where ‘h’, as shown in figure, represents the difference between the inputs to the function at the two points taken. If h ‘tends to’ or ‘moves close to’ zero, we know that the two points will be almost one and the same, and thus the supposed slope between those two practically same points would give us the rate of change for the locus of the function! In other words, the rate of change here would be lim f(x+h) – f(x) ! h→0 h .

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Definition of Derivatives: This brings us to a formal definition of derivatives. If a function ‘f’ is defined by y = f(x) at a point (x, f (x)) then, lim f(x+h) – f(x) or lim f(x) – f(a) when it exists, is called the h→0 h x→a x - a derivative of ‘f’ at x & this derivative is symbolically denoted by f ‘(x) , y’ ,

, or Df.

When f‘(x) exists, we say that ‘f’ is derivable (differentiable) at x. This definition is also called as differentiation using the First Principle.

Interpretation Of Derivative: 1. The derivative of a function at a point is the slope of the tangent line at that point. It may be thought of the limiting position of the secant line to a curve y = f (x) at a point is the tangent at the point. Thus the limit of the slope of the secant line through a fixed point on the curve and any other points on the curve that gets closer and closer to the fixed point when this limit exists. It is the slope of the tangent line at that fixed point (x, f (x)) on the curve y = f (x). 2. The derivative is the instantaneous velocity of a function representing the position of a particle along a line at time ‘t’ where y = f (x). The derivative may be thought of as the limit of the average velocities between a fixed time and other times that get near and nearer to this fixed time. When this limit exists, the instantaneous velocity at time t for y = f (t). 3. The derivative is the instantaneous rate of change of a function at a point. It may be thought of as the limit of the average rates of change between a fixed point and other point on the curve that get closer and closer to that fixed point. When this limits exists, it is known as the instantaneous rate of change of the function y = f (x) at the fixed point (x, f (x)).

Example~ Using the first principle, differentiate the following

1. y = f(x) = 2x+4

2. y = f(x) = 3x2

+ 4x

3. y = f(x) =

Solution 1~

Step 1: Write down f(x) and f(x+h)

f(x) = 2x+4

f(x+h) = 2(x+h) + 4

Step 2: Simplify the form f(x+h) – f(x)

h

f(x+h) – f(x) = 2(x+h) + 4 – [2x+4] = 2h

h h h

Step 3: Apply the limit -

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= 2

Solution 2~

Step 1: Write down f(x) and f(x+h)

f(x) = 3x2 + 4x

f(x+h) = 3(x+h)2 + 4(x+h)

Step 2: Simplify the form f(x+h) – f(x)

h

f(x+h) – f(x) = 3(x+h)2 + 4(x+h) – [3x

2 + 4x]

h h

= 3x2+3h

2+6xh + 4x+4h – [3x

2 + 4x]

h

= 3h2+6xh+4h = 3h + 6x + 4

h

Step 3: Apply the limit -

Solution 3~

Step 1: Write down f(x) and f(x+h)

f(x) =

f(x+h) =

Step 2: Simplify the form f(x+h) – f(x)

h

=

= 8+x – [8+x+h]

h(8+x)(8+x+h)

= =

Step 3: Apply the limit -

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LIFELINES FOR DIFFERENTIATION: Now that we are familiar with differentiation (the act of finding the derivative!), let us explore some short-cut methods and rules that help us in differentiating.

1. ZEROED CONSTANTS RULE: The derivative of a constant is zero.

(c) = 0 (when c is a constant)

e.g. if y = f(x) = 10

then, (f(x)) = 0

2. I’VE-GOT-THE-POWER RULE: Given a function y = f(x) = xn, where n is any real number, then

y’= (f(x)) = nxn-1

e.g.. a) if y = x5

then, y’= = 5x5-1 = 5x4

b) if y = x3

then, y’= = 3x3-1 = 3x2

3. CONSTANT-MULTIPLE RULE: A function multiplied by a constant can be differentiated separately and then multiplied by the constant. Given y = f(x) = c f(x) , then

y’= (c f(x)) = c f(x)

4. SOME-SUM RULE: The derivative of the sum (or difference) of two functions is equal to the sum (or difference) of their individual derivatives Given two functions u = f(x) and v = g(x), and given that y = u + v = f(x) + g(x), then

e.g.. a) if y = x2+ x, then

y’= = (5x2+ 3x) = (5x2) + x = 5 (x2) + 3 (x)

= 5.2x2-1 + 3.1x1-1 = 10x + 3

b) if y = x4 – , then

y’= = (x4 – ) = (x4) + x-1)

= 4x4-1 + (-1)x-1-1 = 4x3 – 2

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PRACTICE CORNER: 1. Differentiate the following to find the “rate of change”

a. y = 1.5x2 + 18x1/3

b. y = 10 +

2. Given that the distance covered by two trucks A and B are A(t) = -t3 + 96t2 +195t + 5 B(t) = -2t4 + 16t3 + 5t2 + 8t

Find the speed at any time ‘t’. (Hint: speed = rate of change of distance, so just differentiate with respect to ‘t’!)

3. If f(x) = 4x5/4 + 3x4/3 + 2x3/2; Find f’(0), f’(7) and f’(10) 4. As per recent research, typical car’s fuel economy depends on the speed its driven

and is approximated by the function F(x) = 0.000003103x4 – 0.000455174x3 + 0.00287869x2 + 1.25986x Where x is measured in mph and f(x) is measured in miles per gallon.

a) Find the rate of change of ‘f’ b) Find f’(x) when x=20 and x=50

(Source: US Department of Energy and Shell Development Company) 5. The estimated number of children newly infected with HIV through mother-to-child

contact worldwide is given by f(t) = -0.2083t3 +3.0357t2 +44.067t +200.2857

where f(t) is measured in thousands and t is measured in years with t=0 corresponding to start of 1990. How fast was estimated number of children infected through this mean increasing at the beginning of year 2000? (Hint: t = 10)

(Source: United Nations)

THE PRODUCT RULE: The derivative of product of two functions is the first function times derivative of second function plus the second function times derivative of first function. Given two functions u = f(x) and v = g(x) then,

Or

Example~ Differentiate:

1. (x2+1)(3x4-3)

2. (x2+5x+2)(x+

Solution 1~ y = (x2+1)(3x4-3) (A product of two functions u = x2+1 and v = 3x4-3 )

Therefore,

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And thus,

(x2 +1) (3x4-3) + (3x4-3) (x2 +1)

(x2 +1) (3.4x4-1-0) + (3x4-3) (2x2-1 +0) = 12x3(x2 +1) + 2x(3x4-3) Solution 2~

y = (x2+5x+2)(x+ (A product of two functions u = x2+5x+2 and v = x+ = x +

2x-1) Therefore,

And thus,

(x2+5x+2) (x+ ) + (x+ ) (x2+5x+2)

(x2+5x+2) (1x1-1+ 2(-1)x-1-1) + (x+ ) (2x2-1 +5x1-1 +0)

= (x2+5x+2) (1 - 2) + (2x+5)(x+ )

QUOTIENT RULE: The derivative of division of two functions u=f(x) and v=g(x) is given by the following rule:

Or, simply,

Example~ Differentiate:

1.

2.

Solution 1~

(A division of two functions u = and v = x)

Therefore,

And thus,

Solution 2~

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(A division of two functions u = and v = )

Therefore,

And thus,

PRACTICE CORNER: i) Find derivative of following functions

1. f(t) = (2t2+1)(t+5)

2. f(x) =

3.

4.

5.

ii) On usage of a certain performance-enhancing drug by athletes, its concentration remnant inside the body follows the function

1. Find the rate at which the concentration changes inside the body 2. How fast is the concentration changing 1/2hr, 1hr and 2 hrs after taking the drug?

iii) A study conducted by 20th century fox concluded that the total box office collections of the super hero movies depend on the period for which the movie runs in theatres worldwide. If this function was approximated as

Where C(x) is measured in millions of dollars and x is the number of years since movies

release. How fast are the collections coming in after 1 year, 3 years and 6 years of movies release?

SOME MORE COMMON DERIVATIVES: We already saw some rules and lifelines that make the whole task of calculating the derivatives simpler for us. In addition to these rules there are a few more derivatives that we need to know based on some common functions since we may come across such functions even in business problems:

Derivative of a square root: If a function is given as y =

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Then

Derivative of an exponential function: if a function is given as y =

Then itself!

Derivative of a logarithmic function: if a function is given as y = logx

Then

CHAIN RULE: Sometimes the function you come across might not be as simple as a sum, product or division of two functions. It might be a composite function or a function of a function! In such cases, we use the CHAIN RULE. E.g. of such functions are where the power

(4) function is a function of another function ( or where the square root (√) function is a function of ( . To generalize, if a function of a function is given as h(x) = f{g(x)}, then

As we see here g(x) is an inner function and f(x) is outer function. The rule is to differentiate the outer function first by taking inner function as a constant and multiplying it then with the derivative of the inner function. Might sound a little confusing, but if you go through the following examples and practice problems then you will get a better idea. Example~ Differentiate:

1.

2.

3.

Solution~

1. (Inner function g = and outer f = power 24) Therefore,

And thus,

2. (Inner function g = and outer f = square root)

Therefore,

And thus,

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3. (Inner function g = and outer f = power 5)

Therefore,

And thus,

Use the Quotient Rule to differentiate the inner function.

PRACTICE CORNER: i) Find derivative of the following:

a.

b. c. d.

e.

ii) ‘KYA AAP PANCHVI PASS SE TEZ HAIN’ has seen the number of viewers following a function given by

With V(x) given in terms of millions of viewers in the xth week. How many viewers were there in week 3 and week 4? At what rate was the weekly audience increasing at the end of week 2 and week 8?

iii) Luxury tax

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APPLICATION

OF

DERIVATIVES - I

CHAPTER

2

This chapter aims at

making you JUGGLE

with derivatives!

Understand impact of

derivatives on

economics .

M ZAKRIYA

Page | 12

APPLICATIONS OF DERIVATIVES – I It is clear to us now that the essence of derivative lies in the fact that it measures the rate of change of functions. Hence, in order to measure the rate of change of a dependent variable with respect to any changes in an independent variable, we require derivatives. Numerous functions are modeled in Businesses, some of which we have already seen, for e.g. cost functions, revenue functions, demand functions, price functions etc. It evidently makes sense for us to find the rate of changes of each of these functions because they help understand the relationship between the variables and the extent of impact of changes in independent variable on the dependent variable. (I know that might have sounded a little confusing. Read it once more, and it might make sense) Most of the applications of derivatives that we are going to see here would be related to concepts that you would have already stumbled upon in your economics classes. This results from the fact that microeconomics deals a lot with study of demand curves, cost curves and revenue curves.

ELASTICTY – RESPONSIVENESS OF DEMAND: When we think elastic, it brings to our mind an image of a rubber band. Elasticity represents stretchability. Not all the goods that are available in market have similar demand curves (which are a plot of price v/s quantity demanded). It depends on the essentiality of the goods and their substitutability. For example, an increase in the fuel prices (which has been the talk of the town, and the world for a few years now), has negligible effect on the demand (which is ever increasing despite price rises!). Alternately, if we consider the demand of consumer durables, the demand increases during festive seasons because there are price discounts available on them. These changes between demands of various goods are effectively measured using elasticity. Elasticity is defined as the responsiveness of demand to the visibly changing environment which is represented by factors such as price, income etc. Hence, the names price elasticity and income elasticity. (I remember somebody getting carried away by the nomenclature used here and defining elasticity as “a city which is elastic”. I hope none of you apply such exemplary logics!)

PRICE ELASTICITY: It is defined as the responsiveness of a proportionate change in demand to the proportionate change in price. The word “proportionate” here is extremely important, because we don’t consider the absolute changes in demand to the absolute changes in price ratio. Instead we find the changes for each factor (whether demand or price) with respect to a point of reference and then measure the elasticity. In order to elaborate more on “proportionate change”, consider this situation. Suppose a company has sold 100 units of a product in previous year and was able to sell 140 units this year. How much has been the proportionate change in sales or percentage increase in sales? The calculations can be shown as follows

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Increase in sales

Here, we have found the increase with respect to the previous year (which inherently is our point of reference) Now, coming back to price elasticity; as I mentioned, it is the measure of proportionate change in demand with respect to the proportionate change in price. Thus if we consider P as the reference price and P1 as the present price, the proportionate change in price can be shown as

Similarly, if x is the quantity demanded, the proportionate change in x can be shown as

Utilizing these measures and the definition of elasticity, we can thus derive price elasticity (εp) as

The quantity measures the rate of change of demand with respect to changing price, and

hence can be represented using the derivative . Thus, we can write price elasticity as:

In general the demand curve is downward sloping, and hence price elasticity is usually negative. Price elasticity can lead to categorization of demands as follows: Elastic demand: When a small proportionate change in price leads to a large change proportionate change in demand, it is called elastic demand. The price elasticity follows the condition | εp|> 1 Inelastic demand: When a large proportionate change in price leads to a small change proportionate change in demand, it is called inelastic demand. The price elasticity follows the condition | εp|< 1 Unit elasticity: If the changes in price and demand are linear and vary proportionately, the demand is Unit elastic. The price elasticity follows the condition | εp|= 1 Examples~ Find price elasticity for the following demand/price functions (given in terms of price ‘p’ and quantity demanded ‘x’) and comment on the same for p = 10:

1.

2.

3.

Solution 1~ Given demand is

Step 1: Find rate of change of demand

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Step 2: Substitute the value found in above step into the elasticity formula

Step 3: Substitute x in terms of p ( or p in terms of x and simplify

Now, when p = 10;

or,

Since, here | εp|> 1, this product has an elastic demand. Solution 2~

Given demand is

Step 1: Find rate of change of demand

Step 2: Substitute the value found in above step into the elasticity formula

Step 3: Substitute x in terms of p ( or p in terms of x and simplify

Now, when p = 10;

or,

Since, here | εp|< 1, this product has an inelastic demand. Solution 3~

Given price is

Step 1: Find rate of change of demand , here it will be found by first finding ,

Now,

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Step 2: Substitute the value found in above step into the elasticity formula

Step 3: Substitute p in terms of x ) and simplify

Now, when p = 10;

or

Which means

Squaring both sides, we get We substitute x = -4

and,

Since, here | εp|< 1, this product has an inelastic demand.

CROSS ELASTICITY: “Cross” can be confusing as well as controversial sometimes (I am not hinting at cross-dressers! Brazilian Football star Ronaldo would know better about this ). Sometimes the demand for one product might be sensitive towards the price of some other related or unrelated product. This responsiveness in proportionate change of demand of one product with respect to change in another product is called as cross elasticity. If x1 denotes the quantity demanded for one product and p2 denotes the price of another product, then cross elasticity represented by c is given by

Cross elasticity helps us determine if any two given products are complements or substitutes. For example, if we study the demand pattern of tea with respect to change in price of coffee (which obviously are substitute products), we can notice that as the price of coffee increases, the demand for tea increases since, tea becomes more affordable to consumers with increasing price of coffee. We see that there is a direct relationship here as increase in price of coffee results in increase in demand for tea, or in other terms, the cross elasticity is positive. Now let us look at the demand for car stereos vis-à-vis the price of cars. Obviously cars and car stereos would be complimentary products. Hence, if the price of cars increases, the

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demand for cars falls, consequently causing the fall in demand for car stereos. So we see that there is an inverse relationship here as increase in price of one product causes decrease in demand for another product, thus, the cross elasticity is negative. To summarize, if cross elasticity is c; For substitutes, c > 0 (Positive) For compliments, c < 0 (Negative) Examples~ Find Cross elasticity for the following demand function (given in terms of price ‘p2’ of soft drinks and quantity demanded ‘x1’ of fresh fruit juices) and comment on the same.

23

Solution~ Given demand is 2

3 Step 1:

23 = 6 2

2

Step 2:

6 2

2 .

= 6 23

Step 3: Substitute x1 in terms of p2 ( 23

or,

Since, here εc> 0, these products are substitutes.

PRACTICE CORNER: 1. Consider the demand equation

P(x) = -0.02x + 400

Which describes the relationship between the unit price and the quantity demanded x of

Ultrasonic Loudspeaker systems.

a. Find elasticity of demand EP.

b. Compute EP for x=100 and x=300.

2. The demand equation for the Balds-R-Beautiful hair dryer is given by:

x = (225-p2)(1/5)

Where p denotes the unit price in dollars and x denotes the quantity demanded per week.

a. Is the demand elastic or inelastic when p=8 and when p=10.

b. At what price is the demand function unit elastic?

3. The quantity demanded per week x (in hundreds) of Hide-Me-Shoot-„Em Miniature camera

is related to the unit price p as follows:

X2 = 400 – 5p

Where x represents the number of units produced.

a. Is the demand elastic or inelastic when p=40? and when p=60?

b. At what price is the demand function unit elastic?

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4. Research findings by MacCrack has shown relationship between demand for the Hair

conditioner products and unit price of shampoo products of UnoLiver Limited.

Xhair conditioner = 90p-1shampoo

Based on cross-elasticity, check if shampoo and conditioners are complimentary products.

MARGINALISM - IMPACT OF ONE EXTRA UNIT A wise man once said, “Ask the importance of one year, to a student who failed and wasted one year; the importance of a month, to a mother who just had a premature caesarian delivery; the importance of a day, to a person who was born on 29th of February and has to wait for 4 years to celebrate his birthday; and the importance of one hour, to those who adapt daylight saving time!” Marginalism studies the effect of one extra unit on an economic quantity or in other words, it measures the rate of change. When the economist, studies parameters like National Income or Per Capita Income; he not only studies them at a given time period but, also determines he rate at which it is increasing on declining. Similarly, a manufacturer will not limit him to finding the Total revenue or Total cost he expects or incurs in a production, it is equally important for him to determine the rate of change of Total revenue and Total cost with respect to number of units sold or produced.

COST FUNCTIONS: We have already seen that the total cost (TC) is given by TC = FC +VC (Fixed Cost plus Variable Cost) Average Cost (AC), would be cost incurred per unit produced, which is found by dividing Total Cost by number of units produced or demanded ‘x’.

Marginal Cost (MC), as we have already seen would be measured by the rate of change of Total Cost with respect to the number of units produced or demanded ‘x’.

REVENUE FUNCTIONS: We have already seen that the total revenue (TR) is given by TR = P * x (Price times quantity demanded) Average Revenue (AR), would be Revenue generated per unit sold, which is found by dividing Total Revenue by number of units demanded ‘x’.

Marginal Revenue (MR) calculations are similar to how we found MC, it is measured by the rate of change of Total Revenue with respect to the number of units demanded ‘x’.

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PROFIT FUNCTIONS: Profit depends on the Revenue and the Cost. Given Total Cost (TC) and Total Revenue (TR), Total profit (TP) is TP = TR - TC (Difference of Revenue and Cost) Average Profit (AP), would be Profits made per unit produced and sold, which is found by dividing Total Profit by number of units produced and demanded ‘x’.

Marginal Profit (MP) calculations are similar to how we found MC and MR

Example~ The Samdung refrigerators were found to incur the weekly costs given by C(x) and generated the weekly revenue R(x) for ‘x’ number of refrigerators manufactured and sold.

a. What is the actual cost incurred for manufacturing 25 refrigerators. b. What is the marginal cost? c. What is the average cost and average revenue? d. Find profit function and determine profits at x = 200. e. What is marginal profit?

Solution~ Cost incurred in for 25 refrigerators;

Marginal cost will be;

Average cost and average revenue are found as follows;

Profit Function will be shown as;

Marginal Profit will be;

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RELATIONSHIP BETWEEN AVERAGE AND MARGINAL COSTS: We have learnt what the average cost and the marginal costs signify. It is important for organizations to minimize the per unit cost or the average cost. The property of Average Cost is such that, at minimum average cost the average cost equals the marginal cost. If we look at the figure shown alongside, the marginal cost curve (MC) intersects the average cost curve (AC) at only one point, which incidentally is also the point which representing the minimum Average cost! Thus, it can be concluded in short that; At minimum AC; AC = MC This property can be used when you face a situation where you are asked to minimize the average cost.

PRACTICE CORNER: 1. The total weekly cost in dollars incurred by the Virgins Record Company in pressing x

audio discs is given by the total cost function:

C(x) = 2500 + 2.2x

a. What is the marginal cost when x=1000? When x=2000?

b. Find the average cost function.

c. Using result in “b” show that when the level of production is very high, the

Average cost approaches $2.20 per disc.

d. Find the level x at which average cost is minimum (Hint: Use AC=MC at Min AC)

2. The marketing department of Telecon corporation has determined that the demand for

their cordless phones obeys the relationship

P = -0.02x +600

Where p denotes the phone‟s unit price (in dollars) and x denotes the quantity demanded.

a. Find revenue function

b. Find marginal revenue function

c. Compute marginal revenue at x=10000 and interpret your result.

3. The weekly demand for the Lectro photocopying machine is given by the demand

equation:

P = 2000 – 0.04x

Where p denotes the wholesale unit price in dollars and x denotes the quantity demanded.

The weekly total cost function of manufacturing these copiers is given by:

C(x) = 0.000002x3 – 0.02x2 +1000x +120,000

a. Find revenue function, profit function P and average cost function.

b. Find marginal revenue function, marginal profit function, marginal cost function

and marginal average cost function.

c. Compute marginal cost at x=3000, marginal revenue at x= 3000, and marginal

profit at x= 3000.

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APPLICATION

OF

DERIVATIVES -

II

CHAPTER

3

This chapter aims at

making you JUGGLE

with derivatives!

Understand impact of

derivatives on

economics .

M ZAKRIYA

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APPLICATION OF DERIVATIVES – II In this chapter, we look at another important application of derivative; to locate the extrema (the extreme points) of a function. These extrema can be either maxima (maximum points) or minima (minimum points). Before we proceed further, let us first understand the concepts of higher order derivatives like 2nd order derivatives.

SECOND ORDER DERIVATIVES: As already stated, the second order derivative represents the rate of rate of change. That is, if rate of change (first derivative) is speed; then the rate of change of speed (second derivative) would represent the acceleration. If a function is given as y = f(x), then the second derivative is found on differentiating the first derivative and can be represented in following ways:

MAXIMA AND MINIMA: The Maxima and Minima are the largest or smallest values that a given function can take at a point within a given proximity or over the entire function. If we consider a local proximity (local maximum or minimum) then these would be the relative extrema (maxima or minima). If we consider the function on its entirety, then it would be absolute extrema (i.e. global maxima or minima). Derivatives help us locate the relative extrema. As the first derivative represents the slope of the tangential line to a curve, if we notice along the curve given alongside, the slope changes throughout. But, at the points of local minimum and local maximum, the tangential lines would be parallel to the X-axis, thus reducing the value of slope to ‘0’. While reaching the maximum point, the slope changes from positive (increasing curve) to zero (highest point) to negative (declining curve). Similarly, while reaching the minimum point, the slope changes from negative (falling curve) to zero (minimum point) to the positive value (increasing curve). Hence, we can locate the extrema, from the fact that the derivative at those points equals zero. But, the problem now would be to differentiate between the two extrema. First derivative, only gives us the pints of extrema but it doesn’t tell us if the point located is a maxima or a minima. This is where the second derivative comes to our rescue. Second derivative will help us determine the direction of the curve once it reaches the extrema. Thus, if we find the second derivative at the extrema, and if it is negative, that means the curve is falling down after reaching this extrema, hence the point is Maximum. Alternately, if we find that

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second derivative at an extrema is positive; then the curve is rising after reaching the extrema, which is logically the Minimum value. To summarize the argument, we can jot down the steps required in locating Maxima and/or Minima for a given function: Step 1: Find first derivative of the given function

Step 2: Equate the first derivative to zero and solve for the unknowns. The solutions would be the extrema for the function.

Step 3: Find second derivative of the function Step 4: Substitute the extrema values found in Step 2 in the second derivative function and check:

If the second derivative of the extrema is +ive, it is MINIMA If the second derivative of the extrema is –ive, it is MAXIMA

Example~ Find maxima and/or minima for the following functions:

1.

2.

Solution 1~

Step 1:

Step 2:

Step 3:

Step 4: at ,

Since (negative), is a MAXIMA

at ,

Since (positive), is a MINIMA

Solution 2~

Step 1:

Step 2:

Step 3:

Step 4: at ,

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Since (negative), is a MAXIMA

at ,

Since (positive), is a MINIMA

PRACTICE CORNER: 1. Honeymoon West, an apartment complex, has 100 two-bedroom units. The monthly profit

realized from renting out x apartments is given by

P(x) = -10x2 + 1760x – 50,000

a. How many units should be rented out in order to maximize the profit?

b. What is the maximum monthly profit realizable?

2. The weekly demand for the DITTO photocopying machine is given by the demand

equation:

P = -0.05X + 600

Where p denotes the wholesale unit price in dollars and x denotes the quantity demanded.

The weekly total cost function of manufacturing these copiers is given by:

C(x) = 0.000002x3 – 0.03x2 +400x + 80,000

a. Compute Profit at x=3000.

b. Find level of production that will yield a maximum profit for manufacturer.

3. Suppose total cost function for Artificial Body manufacturing is

C(x) = 0.2(0.01x2 + 120)

Where x represents the number of units produced.

a. Find the level of production that will minimize the average cost.

b. Find the level of production that will minimize the total cost.

4. The daily average cost function of Ek Akela Electronics Company is given by:

AC(x) = 0.0001x2 -0.08x + 40 + (5000/x)

Where x is number of programmable calculators Ek Akela produces. Show that a

production level of 500 units per day results in minimum average cost for the company.