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Lecture 2 Derivatives dY Y lim dX X 0 X Derivative of a constant Y dY Y Y 1 Y 1 lim lim X 0 dX X X 0 X 2 X 1 Y=3 Y1 dY 0 lim 0 dX X 0 X 2 X 1 X1 X2 X MGMT 7730 - © 2011 Houman Younessi Lecture 2 Derivative of a line dY Y Y 2 Y 1 lim lim dX X 0 X X 0 X 2 X 1 Y dY 15 10 5 lim 5 X 0 dX 32 1 Y=5X Y2=15 Y1=10 X1=2 X2=3 X MGMT 7730 - © 2011 Houman Younessi Lecture 2 Derivative of a polynomial function Y ab X n ab 1 X ( n 1 ) ab 2 X ( n 2 ) ...... a0 dY nab X ( n 1 ) ( n 1 )ab 1 X ( n 2 ) ( n 2 )ab 2 X ( n 3 ) ...... a1 dX Examples: Y 3X dY 3 2 X 2 1 6 X dX 2 Y KX CX 3 X 1 3 2 dY 3KX 2 2CX 3 dX MGMT 7730 - © 2011 Houman Younessi Lecture 2 Derivatives of sums and differences In general: For Y ( x ) W ( x ) Z( x ) dY dW dZ dX dX dX or Y ( x ) W ( x ) Z( x ) dY dW dZ dX dX dX MGMT 7730 - © 2011 Houman Younessi Lecture 2 Derivatives of products and quotients In general: Y ( x ) W ( x )Z( x ) For dY dZ dW W Z dX dX dX Examples: Y( x ) 6 X( 3 X 2 ) W( x ) 6X Z( x ) 3 X 2 dY dZ dW 6X (3 X 2 ) dX dX dX dY 6 X ( 2 X ) ( 3 X 2 )6 dX 12 X 2 18 6 X 2 18 18 X 2 18( 1 X 2 ) or Y ( x ) W ( x ) / Z( x ) dY Z( dX dW dX 5X 3 Y( x ) 34X W( x ) 5X 3 Z( x ) 3 4 X ) W ( dZ dX Z2 dW 15 X 2 dX dZ 4 dX dY ( 3 4 X )( 15 X 2 ) 5 X 3 ( 4 ) dX ( 3 4 X )2 MGMT 7730 - © 2011 Houman Younessi 45 X 2 40 X 3 X 2 ( 45 40 X ) ( 3 4 X )2 ( 3 4 X )2 ) Lecture 2 Derivative of a derivative Profit 0 d 2Y d dY ( ) 2 dX dx dX A Q 1 Number of units of output Y 3X 2 4 X 1 dY 6X 4 dX d 2Y 6 dX 2 Profit A 0 Q 1 Y 4 X 2 7 X 3 dY 8 X 7 dX d 2Y 8 dX 2 Number of units of output MGMT 7730 - © 2011 Houman Younessi Lecture 2 Partial Derivative A derivate with respect to only one variable when the function is the function of more than just that variable A single variable function: A multi-variable function: Y( x ) 3X 3 4 X 1 dY 9X 2 4 dX Y ( x ,w ) 3 X 3 4 XW 2 2 X 2 W 1 Y 9 X 2 4W 2 4 X X Y 0 8 XW 0 1 8 XW 1 W MGMT 7730 - © 2011 Houman Younessi Lecture 2 Optimization Theory Unconstrained Optimization Unconstrained optimization applies when we wish to find the maximum or minimum point of a curve. In other words we wish to find the value of the independent variable at which the dependent variable is maximized or minimized without any other external conditions restricting it. Let us assume that there is an activity x which generates both value V(x) and cost C(x). Net value would therefore be: The necessary condition to find the optimal level is: Or: NV ( x ) V ( x ) C( x ) dNV ( x ) dV ( x ) dC( x ) 0 dx dx dx dV ( x ) dC( x ) dx dx MGMT 7730 - © 2011 Houman Younessi Lecture 2 Unconstrained Optimization: Multiple variables In the case where there are more than one activity, say when the value function is a function of x and y, we take the derivative of the function with respect to each variable and set them all to zero. NV ( x , y ) V ( x , y ) C( x , y ) NV ( x , y ) V ( x , y ) C( x , y ) 0 x x x NV ( x , y ) V ( x , y ) C( x , y ) 0 y y y V ( x , y ) C( x , y ) x x and As such we have: V ( x , y ) C( x , y ) y y MGMT 7730 - © 2011 Houman Younessi Lecture 2 Example: ABCO LLC has two product lines: gadgets and widgets. ABCO produces G of gadgets and W of widgets annually The profit made by ABCO is of course related to their quantity of widgets and gadgets sold. The following equation shows this relationship: P( G ,W ) 5GW 10G 2 10W 2 80W 113.75G 20 Find the derivative (partial derivative) of profit with respect to G. P 5W 20G 113.75 G MGMT 7730 - © 2011 Houman Younessi Lecture 2 Find the derivative (partial derivative) of profit with respect to W. P 5G 20W 80 W Now, using this information find the quantities of G and W that ABCO must manufacture to maximize profit. To answer this question, we remember that a point is either a maximum or minimum when the derivative for that point is zero. For P to be maximized both derivatives with respect to G and W must be zero. 5W 20G 113.75 0 5G 20W 80 0 G 5.0 W 2.75 MGMT 7730 - © 2011 Houman Younessi Lecture 2 Constrained Optimization Constrained optimization applies when we wish to find the maximum or minimum point of a curve but there are also other limiting factors. In other words we wish to find the value of the independent variable at which the dependent variable is maximized or minimized with other external conditions restricting it. Let us start – without loss of generality -with the marginal value for a two variable case: The constraint is that the total cost must equal a specified level of cost relating to the price and quantities of the two components x, and y: V ( x , y ) V ( x , y ) and x y C( x , y ) Px x Py y C MGMT 7730 - © 2011 Houman Younessi Lecture 2 There are two equivalent ways of solving such problems: 1. Simple simultaneous equations: In this approach we solve the set of equations: V ( x , y ) 0 x V ( x , y ) 0 y Px x Py y C 0 2. Lagrangian method: The Lagrangian method works on the basis of adding “meaningful zeros” to the original equation and then assess their impact. MGMT 7730 - © 2011 Houman Younessi Lecture 2 The first thing we do is to form the Lagrangian function. To do so, we first rearrange our constraint formula or formulas so that they all evaluate to zero: Px x Py y C Px x Py y C 0 Then, we add “zero” to the original value function: L V ( x , y ) ( C Px x Py y ) Now we take partial derivatives of the value function wrt x, y and λ, set these to zero and solve. MGMT 7730 - © 2011 Houman Younessi Lecture 2 Example: Cando Co wishes to minimize the cost of their production governed by: TC 4Q12 5Q22 Q1Q2 The constraint is that the company can only make 30 units of product in total Q1 Q2 30 or The Lagrangian becomes: As such we have: 30 Q1 Q2 0 L 4Q12 5Q22 Q1Q2 ( 30 Q1 Q2 ) L( Q1 ,Q2 , ) 8Q1 Q2 0 Q1 L( Q1 ,Q2 , ) Q1 10Q2 0 Q2 L( Q1 ,Q2 , ) Q1 Q2 30 0 MGMT 7730 - © 2011 Houman Younessi Lecture 2 Solving 8Q1 Q2 0 Q1 10Q2 0 For Q1, Q2 and λ Q1 Q2 30 0 We get: Q1=16.5 Q2=13.5 λ= 118.5 What does λ mean? It means that if the constraint were to be relaxed so that more than 30 units could be produced, the cost of producing the 31st is $118.5 MGMT 7730 - © 2011 Houman Younessi Lecture 2 Example 2: Imagine that you are running a manufacturing plant. This plant has the capacity of making 30 units of either widgets or gadgets. Furthermore, the total cost of the manufacturing operation is: 2 2 C 4G 5W GW How many widgets and how many gadgets should you manufacture to minimize cost? To minimize cost, we must find the minimum of the cost function above. We also must make sure that the total units manufactures equals 30. As such: G W 30 therefore G 30 W MGMT 7730 - © 2011 Houman Younessi Lecture 2 Substituting: G 30 W Into: C 4G 2 5W 2 GW C 4( 30 W )2 5W 2 ( 30 W )W 10W 2 270W 3600 Taking the derivative and setting it to zero, we get: dC 20W 270 0 dW or W 13.5 G 30 13.5 16.5 To make sure this is a minimum point: d dC ( ) 20 0 dW dW MGMT 7730 - © 2011 Houman Younessi Lecture 2 Market Demand and the Demand Function Market Demand Schedule for laptops Price per unit ($) Quantity demanded per year (‘000) 3000 800 2750 975 2500 1150 2250 1325 2000 1500 MGMT 7730 - © 2011 Houman Younessi Lecture 2 Demand Curve Price 3000 2500 2000 800 1000 1200 1400 1600 Quantity MGMT 7730 - © 2011 Houman Younessi Lecture 2 Influences on Demand Price Price 3000 2500 2000 3000 2500 Increase in customer preference for laptops 8 0 0 1 0 0 0 Price 1 2 0 0 2000 1 4 0 0 1 6 0 0 8 0 0 Quantity 3000 Increase in customer per capita income 1 0 0 0 1 2 0 0 1 4 0 0 1 6 0 0 1 2 0 0 1 4 0 0 1 6 0 0 Quantity Price 3000 2500 2000 Increase in advertising for laptops 8 0 0 1 0 0 0 1 2 0 0 2500 2000 1 4 0 0 1 6 0 0 Quantity reduction in cost of software 8 0 0 1 0 0 0 MGMT 7730 - © 2011 Houman Younessi Quantity Lecture 2 Demand Function Q=f( price of X, Income of consumer, taste of consumer, advertising expenditure, price of associated goods,….) Example: Demand for laptops in 2007 is estimated to be: Q= -700P+200I-500S+0.01A where P is the average price of laptops in 2007 I is the per capita disposable income in 2007 S is the average price of typical software packages in 2007 A is the average expenditure on advertising in 2007 MGMT 7730 - © 2011 Houman Younessi Lecture 2 Now let us assume that in 2007: I=$33,000 S=$400 and A=$50,000,000 What will be the relationship between price and quantity demanded? Given that: Q= -700P+200I-500S+0.01A We have: Q= -700P+200(33,000)-500(400)+0.01(50,000,000) Q= -700P+6,900,000 MGMT 7730 - © 2011 Houman Younessi Lecture 2 Price Elasticity of Demand By what percentage would the quantity demanded change as a result of one unit of change in price? The percentage change of quantity would be: Q Q The percentage change of price would be: P P Q P Dividing one by the other: Q P P Q E ( ) Q P Rearranging: MGMT 7730 - © 2011 Houman Younessi Lecture 2 At the limit: Therefore: Q dQ P dP E ( P Q ) Q P becomes ( P dQ ) Q dP Example: Determine the price elasticity of demand for laptops in 2007 when price is $3000. We know that: Q= -700P+6,900,000 dQ 700 dP Q=-700(3000)+6900,000=4,800,000 P 3000 0.000625 Q 4800000 700 0.000625 0.4375 MGMT 7730 - © 2011 Houman Younessi Lecture 2 P P = -aQ+b as Q 0 b 1 Demand is price elastic 1 1 Demand is price inelastic 0 b/a MGMT 7730 - © 2011 Houman Younessi as Q P 0 Lecture 2 Exercise: Show that the price elasticity of demand on a demand curve k given by the equation Q P is always 1 dQ P ( ) dP Q 1 Qk p dQ 1 k 2 dP P P P P2 P Q k k P Demand Curve 1 P 2 kP 2 k 2 P k kP 2 1 MGMT 7730 - © 2011 Houman Younessi Q Lecture 2 Exercise: Given the price elasticity of demand and the price, find marginal revenue TR PQ dTR dPQ dQ dQ dQ dP dP P Q P Q dQ dQ dQ Q dP P 1 ( )( ) P dQ MR 1 MR P 1 MGMT 7730 - © 2011 Houman Younessi Lecture 2 Exercise: Given price elasticity of demand and marginal cost, what is the maximum price we should charge? We said that: 1 MR P 1 We also know that in order for price to be maximum, MR=MC, so 1 MR P 1 MC 1 MC P 1 or for P P max MC MC 1 1 1 is the maximum price you should charge MGMT 7730 - © 2011 Houman Younessi Lecture 2 Income Elasticity of Demand By what percentage would the quantity demanded change as a result of one unit of change in consumer income? Q The percentage change of quantity would be: Q I I The percentage change of income would be: Q I Dividing one by the other: Q Rearranging: At the limit: Therefore: E ( Q dQ I dI E ( I Q ) Q I becomes I I Q ) Q I I ( MGMT 7730 - © 2011 Houman Younessi I dQ ) Q dI Lecture 2 Example: Given that: Q= -700P+200I-500S+0.01A Determine the income elasticity of demand for laptops in 2007 when Income is $33000 S=$400 P=$3000 and A=$50,000,000 dQ 200 dI I ( I dQ ) Q dI I ( I dQ 33000 ) 200 1.375 Q dI 4800000 Therefore one percent increase in income leads to 1.375 percent increase in demand for laptops. MGMT 7730 - © 2011 Houman Younessi Lecture 2 Cross Elasticity of Demand By what percentage would the quantity demanded change as a result of one unit of change in the price of an associated product? Example: XY PY dQX ( ) Q X dPY Determine the cross elasticity of demand for laptops in 2007 when price of software is $400 XY ( PY dQ X 400 ) 500 0.042 Q X dPY 4800000 Therefore one percent increase in price of software leads to 0.042 percent decrease in demand for laptops. MGMT 7730 - © 2011 Houman Younessi