Should You Take the P/NP Option?
In this dilemma about choosing the P/NP option based on varying grades and required exam scores, the decision process and scenarios are broken down with calculations for different grade goals. The timeline guides students through review opportunities, exam availability, and grading specifics, ensuring they are well-prepared and informed.
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Lecturer: Dr. Monica Lambon-Quayefio Contact Information: mplambon-quayefio@ug.edu.gh College of Education School of Continuing and Distance Education 2014/2015 2016/2017
Session Overview The previous two sessions on differential calculus focused on independent variable functions. Many economic activities however involve functions of more than one independent variable. To be able to measure the effect of a change in one independent variable the dependent variable in a multivariable function, the partial derivative is needed. This session introduces students to the partial derivatives and also presents the rules of partial differentiation. Objectives At the end of the session, the student will Be able to determine the difference between the derivatives of a one variable function and a multivariate function. Be able to determine the total differentiation of a multivariate function using partial differentiation. Be able to perform unconstrained optimization Slide 2
Session Outline The key topics to be covered in the session are as follows: Derivatives of Multivariate Functions Total Differentiation of Multivariate Functions Unconstrained Optimization Slide 3
Reading List Sydsaeter, K. and P. Hammond, Essential Mathematics for Economic Analysis, 2nd Edition, Prentice Hall, 2006- Chapter 11 Dowling, E. T., Introduction to Mathematical Economics , 3rdEdition, Shaum s Outline Series, McGraw-Hill Inc., 2001.- Chapter 5 Chiang, A. C., Fundamental Methods of Mathematical Economics , McGraw Hill Book Co., New York, 1984.- Chapter 7 and Chapter 8 Slide 4
Topic One DERIVATIVES OF MULTIVARIATE FUNCTIONS Slide 5
Multivariate Functions: Definition Our study of derivatives so far has been limited to functions of single independent variables such as y=f(x). However, many economics activities involve functions with more than one independent variable Z= f(x,y) is defined as a function of two independent variables if there exists one and only one value of z in the range of f for each ordered pair of real numbers (x,y) in the domain of f Slide 6
PARTIAL DERIVATIVES To measure the effect of a change in a single independent variable (x or y) on the dependent variable (z) in a multivariate function, the partial derivative is required. The partial derivative of z with respect to x (y) measures the instantaneous rate of change of z with respect to x(y) while y(x) is held constant. The partial derivative of z or f with respect to x (y) is denoted by the following notations: ?? , ?? ?? ?? , ??(x,y),?? or ?? and ?? ?? ??, ?? , ??(x,y),?? or ?? Slide 7
Partial Derivatives: Mathematical Expression The partial derivative of a multivariate function can be expressed mathematically as: y h x f y x f h 0 y x f y x f h 0 + ( , ) ( , ) f x y = ( , ) lim x h + ( , ) ( , ) h f x y = ( , ) lim y h Rule for finding the partial differentiation of z = f(x,y): To find ??, regard y as a constant and differentiate f (x, y) with respect to x. To find ??, regard x as a constant and differentiate f (x, y) with respect to y.
Partial Derivatives: Examples Find the partial derivative of the multivariate function: Z= 3?2?3 Solution: Z= 3?2?3 When differentiating with respect to x, treat the y term as a constant by mentally bracketing it with the coefficient: Z= (3?3). ?2 Then take the derivative of the x term holding the y term constant: ?? ?? = ??= (3?3). 2x = 6x?3 When differentiation with respect to y, treat the x term as constant by mentally bracketing it with the coefficient: Z= (3?2).?3 Then take the derivative of the y term holding the x term constant: ?? ??= ??= (3?2).3?2= 9?2?2
Trial Question Take a few minutes to find the partial derivatives of the function Z= 5?3- 3?2?2+7?5, by following the steps in the immediate previous slide.
Partial Derivatives: Rules Partial derivatives follow the same basic patterns as the rules of differentiation discussed in the earlier sessions. The rules are as follows: The Generalized Power Function Rule: Given z= [?(?,?)]?: The partial derivatives with respect to x and y respectively are: ?? ??= n [?(?,?)]? 1. ?? ?? Example: Differentiate z= (?3+ 7?2)4 with respect to x and y. ?? ??= 4 (?3+ 7?2)3.(3?2)=13?2(?3+ 7?2)3 ?? ??=4(?3+ 7?2)3.(14y)=56y(?3+ 7?2)3
Partial Derivatives: Rules The Product Rule: Given z= g(x,y) x h (x,y).The partial derivatives with respect to x and y are: ?? ??=g(x,y). ? ?? ?? ??=g(x,y). ? ?? ?? + h(x,y). ?? ?? + h(x,y). ?? Example: Differentiate z= (3x+5)(2x+6y)with respect to x and y. ?? ??= (3x+5).(2)+(2x+6y)(3)=12x+10-18y ?? ??= (3x + 5).(6)+(2x+6y).(0)= 18x+30
Partial Derivatives: Rules The Quotient Rule: Given z= g(x,y) / h (x,y).The partial derivatives with respect to x and y are: ?? ?? ?(?,?) [ (?,?)]2 ?? ?? ?(?,?) [ (?,?)]2 ? ?,? . ?? ??= ?? ? ?,? . ?? ??= ?? Example: Differentiate z= (6x+7y)(5x+3y)with respect to x and y. ?? ??= [(5?+3?)]2 5?+3? . 6 6?+7? .(5) =30?+18? 30? 35? [5?+3?]2 17? (5?+3?)2 = ?? ??= 5?+3? . 7 6?+7? .(3) [(5?+3?)]2 = 35?+21? 18? 21? [5?+3?]2 17? = (5?+3?)2
Second Order Partial Derivatives Given a function z= f(x,y), the direct second order partial derivative signifies that the function has been differentiated with respect to one of the independent variables twice while the other independent variable has been held constant. This is denoted by ???=? ?? ?? ??=?2? ?? ??=?2? ??2 and ???= ? ??2 ?? ??? measures the rate of change of the first-order partial derivative ??with respect to x while y is held constant ??? measures the rate of change of the first order partial derivative ?? with respect to y while x is held constant
Second Order Partial Derivatives The cross or mixed partial derivatives ???and ??? indicate that the original function has been partially differentiated with respect to one of the independent variables and that first partial derivative has in turn been partially differentiated with respect to the other independent variable. ?? ??=?2? ?? ??=?2? This is denoted by ???=? ???? and ???= ? ?? ?? ???? ??? measures the rate of change of the first-order partial derivative ??with respect to y while x is held constant ??? measures the rate of change of the first order partial derivative ?? with respect to x while y is held constant Note that ???= ???according to Young s Theorem.
Examples Find the first, second and cross partial derivatives for z= 7?3+9xy+2?2 1. First order Partial Derivative: ??= 21?2+9y and ??=9x+4y 2. Second (direct) Partial Derivative: ???=42x and ???=4 3. Cross Partial Derivatives ???=? ?? ???=? ?? ? ??=? ? ??=? ??(21?2+9)=9 ??(9x+4y)=9 Slide 16
Topic Two TOTAL DIFFERENTIATION OF MULTIVARIATE FUNCTIONS Slide 17
Definitions For a function of two variables, the total differentiation measures the change in the dependent variable as a result of a small change in each of the independent variables. If z= f(x,y), the total differential of z, denoted by dz is expressed mathematically as : dz=??dx+??dy Where ?? and ?? are the partial derivatives with respect to xy and y respectively and dx and dy are small changes in x and y . The total differential can thus be found by taking the partial derivatives of the function with respect to each independent variable and substituting in the formula stated above To obtain the total derivative with respect to x, divide the formula through by dx to obtain: ?? ?? ??+?? ?? ??=??+?? ?? ?? ??= ?? 18
Examples 1.Find the total differential of z=?4+8xy+3?3 From the given function, ??= 4?3+8y and ??=8x+9?2 Substituting in the formula: dz=??dx+??dy, we have dz= (4?3+8y )dx + (8x+9?2)dy ?? given z=f(x,y)=6?3+7y where y= ?? 2. Find the total derivative g(x)=4?2+3x+8 Dividing the total derivative formula through by dx we have ??=?? +?? ?? From the question, ?? = 18?2 and ??= 7 and ?? Substituting, we have: ?? ?? ??= 8x+3 ?? ??= 18?2+7( 8x+3)= 18?2+56x+21 Slide 19
Topic Three UNCONSTRAINED OPTIMIZATION Slide 20
Unconstrained Optimization Unconstrained Optimization involves finding the optimum to some decision problem in which there are no constraints. For a multivariate function such as z=f(x,y) to have a minimum or maximum, the following three conditions have to be satisfied: The first order partial derivatives must equal zero simultaneously. This ensures that at a given point (a,b) called the critical point, the function is neither increasing nor decreasing. The second order direct partial derivatives, when evaluated at the critical point (a,b) must both be negative for a maximum and positive for a minimum. This ensures that at the critical point, the function is concave and moving downward in the case of a maximum and convex and moving upward in the case of a minimum. The product of the second-order direct partial derivatives evaluated at the critical point must exceed the product of the cross partial derivatives also evaluated at the critical point.
Unconstrained Optimization: Summary of Conditions For a Maximum: ??,?? =0 ???, ???<0 ???. ???> (???)2 Remember: ???= ??? For a Minimum ??,?? =0 ???,???>0 ???. ???>(???)2
Examples Optimize the function z=2?3-?3+147x-54y+12 by finding the critical points, and testing whether the function is at its minimum or maximum. For critical point, take the first order partial derivative, set them to zero and solve for x and y. ??=-3?2+147=0 ??=6?2-54=0 ?2=49 ?2=9 x= 7 y= 3 There are four distinct critical points: (7,3), (7,-3),(-7,3) and (-7,-3)
Examples Now take the second order direct partials and evaluate each of them at the critical points and check the signs. ???=-6x ???=12y a. ???7,3 = 42<0 ???(7,3)=36>0 b. ???7, 3 = 42<0 c. ??? 7,3 = 42 >0 d. ??? 7, 3 = 42 >0 ???(7,-3)=-36<0 ???(-7,3)=36>0 ???(-7,-3)=-36<0 Diagnosis: a) and d) have different signs for the second order partial differential. This means that the second condition has not been satisfied which also hints that the third condition cannot be satisfied. Therefore, the points (7,3) and (-7,-3) are neither maximum or minimum points. They are called saddle points.
Examples The signs of the second direct partials are negative in b) and positive in c) which indicates that the function may be at a maximum at (7,-3) and a minimum at (-7,3) We now test the third condition to be sure that this really holds true. At point (7,-3): ???7, 3 .???(7,-3) >(???(7, 3))2 (-42). (-36)> 02 At point (-7,3): ??? 7,3 .???(7,-3) >(???( 7,3))2 (42). (36)> 02 This means that the function is maximized at (7,-3) and minimized at (-7,3)
Example: Economic Application Sales (S) are a function of advertising in newspapers and magazines ( X, Y) Max S = 200X + 100Y -10X2 -20Y2 +20XY Solution: Differentiate with respect to X and Y and set equal to zero. S/ X = 200 - 20X + 20Y=0 S/ Y = 100 - 40Y + 20X = 0 solve for X & Y and Sales Slide 26
Example Contd. 200 - 20X + 20Y= 0 (1) 100 - 40Y + 20X = 0 (2) Adding them, the -20X and +20X cancel, so we get 300 - 20Y = 0 Y =15 Plug Y= 15 into (1): 200 - 20X + 300 = 0, hence X = 25 To find Sales, plug X=25 and Y=15 into equation: S = 200X + 100Y -10X2 -20Y2 +20XY = 3,250 Slide 27
Trial Questions For each of the multivariable functions find a. the critical points at which the function may be optimized b. determine the nature of the critical points: Z=3?2-xy+2?2-4x-7y+12 Z=60x+34y-4xy-6?2-3?2+5
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