Differentiation

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DIFFERENTIATION Dr. T.K. Jain.

AFTERSCHO☺OL Centre for social entrepreneurship Bikaner M: 9414430763 [email protected] www.afterschool.tk, www.afterschoool.tk www.afterschoool.tk 11/28/09 www.afterschoool.tk

AFTERSCHO☺OL's AFTERSCHO☺OL's MATERIAL MATERIAL FOR PGPSE PARTICIPANTS

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DERIVATIVE OF X^N • N * X ^ (N-1)

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DERIVATIVE OF X^10 • = 10 X ^ 9

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DERIVATIVE OF 5X^3 – 2X^2 + 10 X -3 • 5*3 X^(3-1) -4X + 10 • =15X^2 – 4X +10

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DERIVATIVE OF (3X^2 -1)/(X+1) • • • •

[(X+1) * (6X ) - (3X^2 -1) (1) ] / (X+1)^2 (6x^2 + 6x - 3x^2 + 1) / ( x^2 + 2x + 1) =(3x^2 + 6x +1) / (X^2 + 2x +1) Answer.

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Differentiate e^x / x^2 • • • • •

= [(e^x * 2X) - (X^2 * e^x) ]/ (X^4) = (2x e^x - X^2e^x ) / X ^4 X (2e^x - Xe^x) / X^4 = (2e^x - Xe^x) / X^3 =e^x (2 - X) / X ^3 answer.

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Differentiate 2 ^(X^3 +X^2 +1) = 2^ (X^3 + X^2 + 1) * log2 * differentiation of (X^3 + X^2 + 1) = 2^ ( X ^3 + X ^2 + 1) * log 2 * ( 3x^2 + 2x) Answer.

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Differentiate (2x^3 + x^2 + 1) ^(3/2) • = 3/2 * ( 2x^3 + X^2 +1)^(1/2) * differentiation of (2x^3 + X^2 +1) • =3/2 * ( 2x^3 + X^2 +1)^(1/2) * (6x^2 + 2x) • Answer.

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Find differentiation of log (X^2 + 1) • = 1/ (x^2 + 1) * differentiation of (X^2 +1) • = 1/ (x^2 + 1) * 2X • Answer.

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Differentiate (log x)^2 • = 2 * log X * diffferentiation of log x • = 2 * log x * 1/x answer

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Find the total cost when output is 15000, C = (X^2 / 10000) + 500 • • • •

X = 15000 Thus C =[ (15000)^2 / 10000] + 500 C = 22500 + 500 Total cost = 23000 answer.

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Find the marginal cost when output is 15000, C = (X^2 / 10000) + 500 • Differentiate the function : • = 2X / 10000 X = 15000 = 30000/ 10000 = Rs. 3 answer.

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Find the average cost when output is 15000, C = (X^2 / 10000) + 500 • Average is obtained by dividing the total cost by the number of units. • The number of units is denoted by X. • Thus divide total cost by X • = [(X^2 / 10000) + 500 ] / x • =(x/10000) + (500/x) • = 1.5 + 500/15000 • = 1.53 answer. 11/28/09 www.afterschoool.tk

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During a period, a retail shop can sell X units at price P. What is the total revenue, if P = 20 - .03X and X = 50 ? • Total revenue is total units multiplied by price. • = (20 -.03X) X • =(20 – 1.5) * 50 • = 925 answer. 11/28/09 www.afterschoool.tk

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During a period, a retail shop can sell X units at price P. What is the marginal revenue, if P = 20 - .03X and X = 100 ? • • • • •

Differentiate total revenue Total revenue = P * X= (price * number of units) =(20 - .03x) X Differentiated = 20 - .06X (put x = 100) = 14 Answer

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During a period, a retail shop can sell X units at price P. What is the average revenue, if P = 20 - .03X and X = 100 ? • • • • •

Average revenue = total revenue / no. of units =(20X - .03X^2) / X = 20 - .03x = 20 – 3 = 17 answer.

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Find the maximum value of the function : f(x) = 2x^3 + 3x^2 -12x +1 • Maximum value is when first order differentiation is zero & second order differentiation is less than zero. • First differentiation : • 6X^2 + 6x -12 = 0 • X^2 +x – 2 = 0 • X^2 +2x –x -2 =0 • X (X +2) -1(X +2) •11/28/09 Thus X = 1 or -2. AFTERSCHO☺OL's MATERIAL 18 www.afterschoool.tk

FOR PGPSE PARTICIPANTS

Solution…. • Second differentiation: • 12X + 6 • Here if we put the value of X = -2, we get value in negative. • f(x) = 2x^3 + 3x^2 -12x +1 • = 2 * ( -8) + 3 * 4 +24 +1 • =-16 +12 +24 +1 • =21 answer. 11/28/09 www.afterschoool.tk

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Find the minimum value of the function : f(x) = 2x^3 + 3x^2 -12x +1 • Minimum value is when first order differentiation is zero & second order differentiation is less than zero. • First differentiation : • 6X^2 + 6x -12 = 0 • X^2 +x – 2 = 0 • X^2 +2x –x -2 =0 • X (X +2) -1(X +2) • Thus X = 1 or -2. 11/28/09 www.afterschoool.tk

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Solution… • Second differentiation: • 12X + 6 • Here if we put the value of X = 1, we get value in positive, so X = 1 • f(x) = 2x^3 + 3x^2 -12x +1 • =-6 answer. ( minimum value).

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Differentials • Definition – (dy/dx = f (x)) – dy = f (x) Δx

• Use: – Δy ≈ dy

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Indefinite Integral • Anti-derivative of f(x) is F(x) such that – F'(x) = f(x) – dF(x) = f(x) dx

• Indefinite integral ∫f(x) dx = F(x) + C Integral sign

x is the variable on integration

Integrand

Constant of integration

Technique 1: Guess the anti-derivative 11/28/09 www.afterschoool.tk

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Indefinite Integral • Basic integration formulas – Page 625 of text

• Examples from notes

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Integrations with initial conditions • y = ∫f(x) dx = F(x) + C • If I know y(x0) = y0, I can find C – C = y0 – F(x0) – Example from notes

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EVPI • Expected Value of Perfect Information • the maximum amount which the decision maker can spend for obtaining the perfect information as to which event would occur is called EVPI. • EVPI = expected profit with perfect information - expected monetary value of optimal act. 11/28/09 www.afterschoool.tk

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EXAMPLE… • A physician purchases a particular medicine on Monday of each week. The medicine must be used within the week following, otherwise it becomes worthless. The medicine costs Rs. 2 per dose and Physician charges Rs. 4 per dose. The past 50 weeks record of uses are as foliows • Dose per week 20 25 40 60 • No. of weeks 5 15 25 5 11/28/09 www.afterschoool.tk

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Solution – payoff matrix Demand Probabilit 20 y

25

40

60

20 25

5/50 = .1 40 15/50=.3 40

30 50

0 20

-40 -20

40 60

25/50=.5 40 5/50=.1 40

50 50

80 80

40 120

40

48

54

22

EMV Optimal action

Perfect information

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SOLUTION…. • • • •

EMPI = with perfect information: 40*.1 + 50*.3 + 80*.5 + 120*.1 = 71 EMV = 54 EVPI = 71 – 54 = 17 answer.

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MAXIMIN CRITERION • The decision maker tries to maximise the minimum gains or he tries to pick the best of the worst.

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MINIMAX REGRET CRETERION • The decision maker selects that act, which will give the minimum of the maximum opportunity losses. Hence prepare opportunity loss table for each option (act).

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Example = solve it using Maximin and Minimax criterion Action

A

B

C

D

S1

8

0

10

6

S2

-4

12

18

-2

S3

14

6

9

8

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MAXIMIN Action

A

B

C

D

MIN.

S1

8

0

10

6

0

S2

-4

12

18

-2

-4

S3

14

6

9

8

6

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SOLUTION • Maximin is 6, because it is maximum out of the minimums.

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MINIMAX OPPORTUNITY LOSS TABLE Action

A

B

C

D

max regret

S1

6

12

8

2

12

S2

18

0

0

10

18

S3

0

6

9

0

9

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Solution … • We will first identify the regrets for each value, then select maximum regret for each row and then pick up the minimum regret out of all the regrets (which are maximum from their rows). For calculating regrets, subtract each value from the maximum value from its column.

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Decision tree… • An organisation has two packaging machines old and new. the new machine is more efficient if the materials are of good quality; on the other hand the old machine performs better if the materials are of poor quality. The following information is available • (i) In the previous batches 80% of the materials have been of good quality and 20% of poor quality • (ii) The profit position is as under • (a) Using old machine • If the materials are good Profit of Rs. 2,000 If the materials are poor Profit of Rs. 1,600 • (b) Using new Machine • If the materials are good Profit of Rs. 2,400 • If the materials are poor Profit of Rs. 800 You have to decide which machine should be used under the condition that quality of the materials is not known at this stage.

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Decision tree…. 2000 P = .8

Good old

+ 320 = 1920

poor

1600 p =.2

Good

NEW

2400 p =.8 1920+160

Poor

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1600

AFTERSCHO☺OL's MATERIAL FOR PGPSE PARTICIPANTS

=2080 800 P = .2 38

Solution …. • When we prepare a decision tree, we are preparing a list of all the options with their respective payoffs along with their probabilities. The option, which has the highest payoff possibilities is selected based on the computations.

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• A client asks an estate agent to sell three properties A, B and C:. The agent has to sell A first within 60 days. If he is successful he can either backout or try to sell B or C. After selling second again either he can backout or try to sell third. He receives 5% commission on each sale. The detail is given in next slide: 11/28/09 www.afterschoool.tk

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Continued.. Property

Cost

Probability

Selling price

A

400

.7

12000

B

225

.6

25000

C

450

.5

50000

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Decision tree .. 800

765 Sold b 884

Sell C . 5

Sell B

200 Sell A .7

.6 Fail -225.4

1063 Sell C 1288

Final expected value = 764 11/28/09 www.afterschoool.tk

backout Try B

.5 .3 Fail -400

400

Sold C

so ld c

Sold b

525 Backo ut

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fail

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Tabular presentation …. Sell Suc Fail Sell Fail • We will try for each eed C option. We will look at B all the possibilities and multiply those with respective probabilities to find their expected payoffs. First option : sell B : 11/28/09 www.afterschoool.tk

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Total cost of a firm is as under: c = 25 +2x +.01x^2 the firm can sell as much as it wishes at 30 per unit. How much should it sell to maximise profits? • • • •

Profit = revenue – cost; revenue = 30x =30x – (25 +2x +.01x^2) =-.01x^2 +28X -25 To maximise, first differentiation must be 0 and second differentiation must be negative.

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Solution… • • • •

=-.02x+28 = 0 X = 1400 Second differentiation : -.02 (which is negative), thus profit is maximum at x = 1400 answer. • Profit: • =-.01*1400*1400 +28*1400 -25 • =19575 answer. 11/28/09 www.afterschoool.tk

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A manufacturer can sell x items per week at price (one item) p=20-.001x and cost is c = 5x +2000 to produce x items. How much should it produce to maximise profits? • • • •

Profit = (20-.001x)x - (5x+2000)) =20x - .001x^2 – 5x – 2000 = -.001x^2 +15x -2000 Differentiate it :

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Solution… • • • • •

-.001x^2 +15x -2000 Differentiate it : -.002x+15 = 0 X = 7500 Second differentiation is -.002 (which is negative). • The firm should make 7500 items. 11/28/09 www.afterschoool.tk

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A company charges Rs. 200 for each box of tools on orders of 150 or fewer boxes. The cost to the buyer on every box is reduced by Re. 1 for each order in excess of 150. For what size order is the revenue maximum ? • Price per box = (suppose order size is x) • =200 – 1(x-150) OR –x + 350 • Total revenue = (-x+ 350)x 11/28/09 www.afterschoool.tk

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Solution… • • • • • • •

Total revenue = (-x+ 350)x Differentiate it : -2x +350 = 0 X = 175 Second differentiation: -2 (which is negative). Thus order size of 175 will maximize the revenue.

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Demand function of a firm is p=700-.3x and cost function (total) is c=15x+20000, where p is price per unit and x is number of units produced. Find the production for maximum profit? • • • •

Profit = revenue – cost Revenue = (700 -.3x) X = 700x -.3x^2 Cost= c=15x+20000 Profit = 700x -.3x^2 – (15x+20000)

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Solution • • • • • • •

Profit = 700x -.3x^2 – (15x+20000) =685x - .3x^2 – 20000 Differentiate it : 685 - .6x =0 X = 1141.67 Second differentiation : = -.6 (this is negative). Thus the production should be 1142 answer.

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A manufacturing firm requires 45000 units of a particular item as raw material. The item is packed in boxes of 10 items each. The ordering cost is rupees 45 per order. If the carrying charges are Rs. 2 per box, how many boxes should be ordered for minimum cost ? • Number of orders = 4500/x • X = number of boxes per order • Cost = carrying cost + ordering cost.

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Solution … • Cost = carrying cost + ordering cost. • Carrying cost = 2 * average inventory • Average inventory = x/2, thus carrying cost=2x/2 • Ordering cost = 45 * ( 4500/x) • Total cost = 202500/x + 2x/2

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Solution… • First Differentation • Total cost = 202500/x + 2x/2 • • • • • •

( x * differentiation of 202500)- (202500*1)/ x^2 + 2/2 =-202500/x^2 +1 = 0 OR -202500/x^2 = -1 X^2 = 202500 or X = 450 Second differentiation : (x^2*differentiation of -202500)-(-202500*2x)/x^4 =405000x / x^4 OR 405000/X^3 (this is positive, therefore the number of boxes should be 450 per order). Answer.

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Remember … • For maximisation, first differentiation should be zero and second differentiation must be negative. • For minimisation, first differentiation should be zero and second differentiation must be positive.

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About AFTERSCHO☺OL • PGPSE - World’s most comprehensive programme on social entrepreneurship – after class 12th • Flexible – fast changing to meet the requirements • Admission open throughout the year • Complete support from beginning to the end – from idea generation to making the project viable. 11/28/09 www.afterschoool.tk

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Branches of AFTERSCHO☺OL • PGPSE programme is open all over the world as free online programme. • Those who complete PSPSE have the freedom to start branches of AFTERSCHO☺OL • A few branches have already started one such branch is at KOTA (Rajasthan).

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Workshop on social entrepreneurship • We conduct workshop on social entrepreneurship – all over India and out of India also - in school, college, club, association or any such place - just send us a call and we will come to conduct the workshop on social entrepreeurship. • These workshops are great moments of learning, sharing, and commitments. 11/28/09 www.afterschoool.tk

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FREE ONLINE PROGRAMME • AFTERSCHO☺OL is absolutely free programme available online – any person can join it. The programme has four components : • 1. case studies – writing and analysing – using latest tools of management • 2. articles / reports writing & presentation of them in conferences / seminars • 3. Study material / books / ebooks / audio / audio visual material to support the study • 4. business plan preparation and presentations of those plans in conferences / seminars 11/28/09 www.afterschoool.tk

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100% placement / entrepreneurship • AFTERSCHO☺OL has the record of 100% placement / entrepreneurship till date • Be assured of a bright career – if you join AFTERSCHO☺OL

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Pursue professional courses along with PGPSE • AFTERSCHO☺OL permits you to pursue distance education based professional / vocational courses and gives you support for that also. Many students are doing CA / CS/ ICWA / CMA / FRM / CFP / CFA and other courses along with PGPSE. • Come and join AFTERSCHO☺OL

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