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Topic 6: Optimization I

Maximisation and Minimisation

Jacques (4th Edition):

Chapter 4.6 & 4.7

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For a straight line Y=a+bX Y= f (X) = a + bX

First Derivative dY/dX = f = b

constant slope b

Second Derivative d2Y/dX2 = f = 0

constant rate of change - the change in the slope is zero - (i.e. change in Y due to change in X does not depend on X)

Y

X

a

3

Y

X

Y=X >1

X

Y Y=X 0< < 1

For non-linear functions

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Y= f (X) = X

First Derivative

dY/dX = f = X-1 > 0

Positive Slope: change in Y due to change X is Positive

Second Derivative

d2Y/dX2 = f = (-1) X(-1)-1 or

d2Y/dX2 = f = (-1)(Y/X2)

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d2Y/dX2 = f = (-1)(Y/X2)

Sign of Second Derivative? d2Y/dX2 = f = 0 if = 1 constant rate of

change d2Y/dX2 = f > 0 if > 1

increasing rate of change (change Y due to change X is bigger at higher X – the change in the slope is positive)

d2Y/dX2 = f < 0 if < 1

decreasing rate of change (change Y due to change X is smaller at higher X – the change in the slope is negative)

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Maximisation and Minimisation

Stationary Points

•Second-order derivatives

•Applications

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Y

If f (X) < 0 as X will Y

If f (X) > 0 as X will Y

If f (X) = 0 slope 0…as X no change Y

X*=1

A

B

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Definition

Stationary points are the turning points or critical points of a function

Slope of tangent to curve is zero at stationary points

Stationary point(s) at A & B: where f (X) = 0

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Are these a Max or Min point of the function?

1) examine slope in region near the stationary point

Sign of first derivative around a turning point: Before At After Maximum plus zero minus Minimum minus zero plus

dY/dX = f (X) is (–, 0, +) min dY/dX = f (X) is (+, 0, -) max

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if d2Y/dX2 = f (X) > 0 a minimum the change in the slope is positive beyond the stationary point, so the point is a local minimum

if d2Y/dX2 = f (X) < 0 a maximum the change in the slope is negative beyond the stationary point, so the point is a local maximum

if d2Y/dX2 = f (X) = 0 indeterminate the change in the slope is zero beyond the stationary point - could be a max, or a min, or an inflection point

2) or calculate the second derivative……

look at the change in the slope beyond the stationary point

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e.g. inflection point

f =0 & f =0

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To find the Max or Min of a function Y= f(X)

1) First Order Condition (F.O.C.):

set slope dY/dX = f (X) = 0

this identifies the stationary point(s)

2) Second Order Condition (S.O.C.):

check the sign of the second derivative (gives the change in the slope)

d2Y/dX2 = f (X) > 0 a minimum

d2Y/dX2 = f (X) < 0 a maximum d2Y/dX2 = f (X) = 0 indeterminate

this identifies whether the slope of the function is increasing, decreasing, or does not change after the stationary point(s)

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Find the Maxima and Minima of the following functions:

122 xxy F.O.C. : slope=0 at stationary point

022 xdx

dy

1

22

x

x

S.O.C. : check sign of second derivative at x=-1

022

2

dx

yd (slope increases after the stationary point, so must

be a minimum at x= -1)

0

5

10

15

20

25

30

-4 -3 -2 -1 0 1 2 3 4

X

Y

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Example 1: Profit Maximisation

Question.

A firm faces the demand curveP=8-0.5Q

and total cost function TC=1/3Q3-3Q2+12Q.

Find the level of Q that maximises total profit and verify that this value of Q is where MC=MR

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Answer….going to take a few slides!

Find Total Revenue….

P = 8 - 0.5Q inverse demand function

TR (Q) = P.Q = 8Q - ½Q2

TC (Q) = 1/3Q3 - 3Q2 + 12Q

Now write out the profit function

MAX = TR - TC (Q) = -4Q + 2 ½ Q2 – 1/3Q

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The function we want to Maximise is PROFIT….

And Profit = Total Revenue – Total Cost

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MAX (Q) = -4Q + 2 ½ Q2 – 1/3Q3

First Order Condition:d/dQ = f (Q)= - 4 + 5Q – Q2 = 0 (solve quadratic – Q2 + 5Q – 4 by applying formula: )

Optimal Q solves as: Q*=1 and Q*= 4

Second Order Condition:d2/dQ2 = f (Q) = 5 – 2Q Sign ? f = 3 > 0 if Q*= 1 (Min)f = - 3 < 0 if Q*= 4 (Max)

So profit is max at output Q = 4

a

acbbQ

2

42

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Continued….. Verify that MR = MC at Q = 4:

TR (Q) = 8Q - ½Q2

MR = dTR/dQ = 8 – Q

Evaluate at Q = 4 …..then MR = 4

TC (Q) = 1/3Q3 - 3Q2 + 12Q

MC = dTC/dQ = Q2 – 6Q +12

Evaluate at Q = 4 ….

then MC = 16 – 24 +12 = +4

Thus At Q = 4, we have MR = MC

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Maximisation and MinimisationTax Example

The (inverse) Supply and Demand Equations of a good are given, respectively, as P- t = 8 + QS P = 80 – 3QD A tax t per unit, imposed on suppliers, is being considered. At what value of t does the government maximise tax revenue in market equilibrium?

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What do we want to maximise?Tax revenue in equilibrium…..

This will be equal to the tax rate t multiplied by the equilibrium quantity

So first we need to find the equilibrium quantity

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Solution

To find equilibrium Q, Set Supply equal to Demand….

In equilibrium, QD = QS so

Q + 8 + t = 80 – 3Q

Now Solve for Q

Qe = 18 – ¼ t

 Now we can write out our objective function…

Tax Revenue T = t.Qe = t(18 – ¼ t)

MAX T(t) = 18t – ¼ t2

t*

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MAX T(t) = 18t – ¼ t2

t*

First Order Condition for max: set the slope (or first derivative) = 0

dT/dt = 18 – ½ t = 0

t* = 36

Second Order Condition for max: check sign of second derivative

d2T/dt2 = -½ < 0 at all values of x

Thus, tax rate of 36 will Maximise tax revenue in equilibrium

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Now we can compute out the equilibrium P and Q and the total tax revenue when t = 36

At t* = 36

Qe = 18 – ¼ t* = 9

Tax Revenue T = t*.Qe = 18t* – ¼ t*2 = 324

Pe = Qe + 8 + t*= 53

 

If t = 0, then tax revenue = 0,

Qe = 18 ,

Pe = Qe + 8 = 26

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Is the full burden of the tax passed on to consumers?

Ex-ante (no tax) Pe = 26

Ex-post (t* =36) Pe = 53

The tax is t*= 36, but the price increase is only 27 (75% paid by consumer)

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Another example

Cost Producing Q output given capital K is: 22

8 QK

KC

(a) if K=20 in Short Run, find the level of Q at which AC is minimised.

(b) Show that MC and AC are equal at this point.

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Solution Substituting in K = 20 to our C function:

C = (8*20) + (2/20)Q2 = 160 + 0.1Q2

AC = C/Q = 160/Q + 0.1Q

and MC = dC/dQ = 0.2Q

 

First Order Condition: set first derivative (slope)=0

AC is at min when dAC/dQ = 0

So dAC/dQ = - 160/Q2 + 0.1 = 0

And this solves as Q2 = 1600 Q = 40

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Second Order Condition

Second Order Condition: check sign at Q = 40

If d2AC/dQ2 >0 min.

Since dAC/dQ = - 160/Q2 + 0.1

Then d2AC/dQ2 = + 320/Q3

Evaluate at Q = 40,

d2AC/dQ2 = 320 / 403 >0

min AC at Q = 40

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b) Now show MC = AC when Q = 40:

AC = C/Q = 160/Q + 0.1Q

AC at Q=40: 160/40 + (0.1*40) = 8

 

and MC = dC/dQ = 0.2Q

So MC at Q = 40: 0.2*40 = 8

MC = AC at min AC when Q=40

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c) What level of K minimises C when Q = 1000?

KKQ

KKC

functionCtoQSubstitute210002

8228

1000

)(

dC/dK = 8 – (2(10002 )/ K2 )= 0

Solving 8K2 = 2.(1000)2

K2 = ¼ .(1000)2

optimal K* = ¼.(1000)

=½(1000)=500

if Q = 1000, optimal K* = 500

more generally, if Q = Q0, optimal K = ½ Q0

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Second Order Condition: check sign of second derivative at K = 500

If d2C/dK2 >0 min.

Since dC/dK = 8 – (2(10002 )/ K2 )

d2C/dK2 = + (2.(10002).2K )/ K4 >0 for all values of K>0 and so C are at a min when K = 500

The min cost producing Q =1000 occurs when K = 500

Subbing in value k = 500 we get C = 8000

 or more generally, min cost producing Q0 occurs when K =

Q0/2 and so C = 4Q0 + 4Q0 = 8Q0

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Topic 6: Maximisation and Minimisation Second DeIdentifying the max and min of various

functions Identifying the max and min of various functions –

sketch graphs Finding value of t that maximises tax revenues, given

D and S functions Identifying all local max and min of various

functions. Identifying profit max output level. Differentiate various functions.

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Maximisation and Minimisation

Second-order derivatives Stationary Points Optimisation I Applications