topic 6: optimization i · 2019-01-03 · if d2y/dx2 = f ′′ (x) = 0 ⇒ indeterminate the...
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Topic 6: Optimization I
Maximisation and MinimisationJacques (4th Edition):Chapter 4.6 & 4.7
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For a straight lineY=a+bXY= f (X) = a + bX
First DerivativedY/dX = f ′ = b
constant slope b
Second Derivatived2Y/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
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Y
X
Y=Xα α>1
X
Y Y=Xα 0< α < 1
For non-linear functions
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Y= f (X) = Xα
First Derivative dY/dX = f ′ = α Xα-1 > 0 Positive Slope: change in Y due to change X is Positive Second Derivative
d2Y/dX2 = f ′′= (α-1)α X(α-1)-1 or
d2Y/dX2 = f ′′= (α-1)α(Y/X2)
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d2Y/dX2 = f ′′= (α-1)α(Y/X2) Sign of Second Derivative?d2Y/dX2 = f ′′= 0 if α = 1 constant rate of changed2Y/dX2 = f ′′ > 0 if α > 1increasing 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 α < 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 functionSlope of tangent to curve is zero at stationary pointsStationary 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 isincreasing, decreasing, or does not change afterthe 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 =+= xdxdy
122
−=−=
xx
S.O.C. : check sign of second derivative at x=-1
022
2>=
dxyd
(slope increases after the stationary point, so mustbe a minimum at x= -1)
0
5
10
15
20
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-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 dem and function TR (Q ) = P.Q = 8Q - ½ Q 2 TC (Q ) = 1/3Q 3 - 3Q 2 + 12Q N ow write out the profit function M A X Π = TR - TC Π (Q ) = -4Q + 2 ½ Q 2 – 1/3Q 3
The function we want to Maximise is PROFIT….
And Profit = Total Revenue – Total Cost
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MAX Π (Q) = -4Q + 2 ½ Q2 – 1/3Q3
First Order Condition:dΠ/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:d2Π/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
242 −±−
=
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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 + 12QMC = dTC/dQ = Q2 – 6Q +12 Evaluate at Q = 4 ….then MC = 16 – 24 +12 = +4Thus 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, isbeing considered. At what value of t does the governmentmaximise tax revenue in marketequilibrium?
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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 – 3QNow Solve for QQe = 18 – ¼ tNow 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 t2t*
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 xThus, 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* = 36Qe = 18 – ¼ t* = 9 Tax Revenue T = t*.Qe = 18t* – ¼ t*2 = 324Pe = 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 = 26Ex-post (t* =36) Pe = 53The 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: 228 Q
KKC +=
(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 + 0.1Qand MC = dC/dQ = 0.2Q
First Order Condition: set first derivative (slope)=0AC is at min when dAC/dQ = 0So dAC/dQ = - 160/Q2 + 0.1 = 0And 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 + 0.1Then 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 = 0.2QSo 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
functionCtoQSubstitute2100028228
1000)(
+=+=
=
dC/dK = 8 – (2(10002 )/ K2 )= 0Solving ⇒ 8K2 = 2.(1000)2
⇒ K2 = ¼ .(1000)2
⇒ optimal K* = √¼.(1000)=½(1000)=500
∴if Q = 1000, optimal K* = 500more 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 = 8000or 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 functionsIdentifying the max and min of various functions –sketch graphsFinding 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 derivativesStationary PointsOptimisation IApplications