1. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Derivatives Derivative of a constant Y X Y=3 Y1 X1X2

2. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Derivative of a line Y X Y=5X Y1=10 X1=2X2=3 Y2=15

3. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Derivative of a polynomial function Examples:

4. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Derivatives of sums and differences In general: Foror

5. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Derivatives of products and quotients In general: Foror Examples:

6. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Derivative of a derivative 0Q1Q1 Profit Number of units of output A Q1Q1 Profit Number of units of output A 0

7. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 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:

8. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 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:

9. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 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. As such we have:

10. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 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: Find the derivative (partial derivative) of profit with respect to G.

11. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Find the derivative (partial derivative) of profit with respect to 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.

12. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 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:

13. Lecture 2 MGMT 7730 - © 2011 Houman Younessi There are two equivalent ways of solving such problems: 1. Simple simultaneous equations: In this approach we solve the set of equations: 2. Lagrangian method: The Lagrangian method works on the basis of adding “meaningful zeros” to the original equation and then assess their impact.

14. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 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: Then, we add “zero” to the original value function: Now we take partial derivatives of the value function wrt x, y and λ, set these to zero and solve.

15. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Example: Cando Co wishes to minimize the cost of their production governed by: The constraint is that the company can only make 30 units of product in total The Lagrangian becomes: As such we have:

16. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Solving For Q 1, Q 2 and λ We get: Q 1 =16.5 Q 2 =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 31 st is $118.5

17. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 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: 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: Example 2:

18. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Substituting:Into: Taking the derivative and setting it to zero, we get: To make sure this is a minimum point:

19. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Market Demand and the Demand Function Market Demand Schedule for laptops Price per unit ($)Quantity demanded per year (‘000) 3000 2750 2500 2250 2000 800 975 1150 1325 1500

20. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 2000 2500 3000 8001000120014001600 Demand Curve Price Quantity

21. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Influences on Demand 2000 2500 3000 800800 10001000 12001200 14001400 16001600 Price Quantity Increase in customer preference for laptops 2000 2500 3000 800800 10001000 12001200 14001400 16001600 Price Quantity Increase in customer per capita income 2000 2500 3000 800800 10001000 12001200 14001400 16001600 Price Quantity Increase in advertising for laptops 2000 2500 3000 800800 10001000 12001200 14001400 16001600 Price Quantity reduction in cost of software

22. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Demand Function Q=f( price of X, Income of consumer, taste of consumer, advertising expenditure, price of associated goods,….) Example: Q= -700P+200I-500S+0.01A where Demand for laptops in 2007 is estimated to be: 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

23. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Now let us assume that in 2007: I=$33,000S=$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

24. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 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: The percentage change of price would be: Dividing one by the other: Rearranging:

25. Lecture 2 MGMT 7730 - © 2011 Houman Younessi At the limit: Therefore:becomes Example: Determine the price elasticity of demand for laptops in 2007 when price is $3000. We know that:Q= -700P+6,900,000 Q=-700(3000)+6900,000=4,800,000

26. Lecture 2 MGMT 7730 - © 2011 Houman Younessi b P b/a P = -aQ+b Q Demand is price elastic Demand is price inelastic

27. Lecture 2 MGMT 7730 - © 2011 Houman Younessi P Demand Curve Q Exercise: Show that the price elasticity of demand on a demand curve given by the equation is always

28. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Exercise: Given the price elasticity of demand and the price, find marginal revenue

29. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Exercise: Given price elasticity of demand and marginal cost, what is the maximum price we should charge? We said that: We also know that in order for price to be maximum, MR=MC, so is the maximum price you should charge

30. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Income Elasticity of Demand By what percentage would the quantity demanded change as a result of one unit of change in consumer income? The percentage change of quantity would be: The percentage change of income would be: Dividing one by the other: Rearranging: Therefore:becomes At the limit:

31. Lecture 2 MGMT 7730 - © 2011 Houman Younessi Given that:Q= -700P+200I-500S+0.01A Example: Determine the income elasticity of demand for laptops in 2007 when Income is $33000 S=$400 P=$3000 and A=$50,000,000 Therefore one percent increase in income leads to 1.375 percent increase in demand for laptops.

32. Lecture 2 MGMT 7730 - © 2011 Houman Younessi 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: Determine the cross elasticity of demand for laptops in 2007 when price of software is $400 Therefore one percent increase in price of software leads to 0.042 percent decrease in demand for laptops.