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Derivatives

Derivatives. Derivative of a constant. Y. Y=3. Y1. X. X1. X2. Derivative of a line. Y=5X. Y. Y2=15. Y1=10. X1=2. X2=3. X. Derivative of a polynomial function. Examples:. Derivatives of sums and differences. In general:. For. or. Derivatives of products and quotients.

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Derivatives

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  1. Derivatives Derivative of a constant Y Y=3 Y1 X X1 X2

  2. Derivative of a line Y=5X Y Y2=15 Y1=10 X1=2 X2=3 X

  3. Derivative of a polynomial function Examples:

  4. Derivatives of sums and differences In general: For or

  5. Derivatives of products and quotients In general: For or Examples:

  6. A Profit Profit A 0 Q1 0 Number of units of output Q1 Number of units of output Derivative of a derivative

  7. 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. 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. 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. 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. 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. 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. 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. 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. 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. Solving For Q1, Q2 and λ Q1=16.5 Q2=13.5 λ= 118.5 We get: 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

  17. 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: 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:

  18. Substituting: Into: Taking the derivative and setting it to zero, we get: To make sure this is a minimum point:

  19. 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

  20. Demand Curve Price 3000 2500 2000 800 1000 1200 1400 1600 Quantity

  21. Price Price Price Price 3000 3000 3000 3000 2500 2500 2500 2500 Increase in customer preference for laptops 2000 2000 2000 2000 Increase in customer per capita income 800 800 800 800 1000 1000 1000 1000 1200 1200 1200 1200 1400 1400 1400 1400 1600 1600 1600 1600 Quantity Quantity Quantity Quantity Increase in advertising for laptops reduction in cost of software Influences on Demand

  22. 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

  23. 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

  24. 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. 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. P P = -aQ+b b Demand is price elastic Demand is price inelastic b/a Q

  27. P Demand Curve Q Exercise: Show that the price elasticity of demand on a demand curve given by the equation is always

  28. Exercise: Given the price elasticity of demand and the price, find marginal revenue

  29. 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. 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: At the limit: Therefore: becomes

  31. 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 Therefore one percent increase in income leads to 1.375 percent increase in demand for laptops.

  32. 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.

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