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### Topic_6_-_Calculus_of_Several_Variables

Course: FOM bmt1024, Spring 2012
School: Multimedia University
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Word Count: 1290

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Partial CalculusofSeveralVariablesand ItsApplications Revision Derivatives Applied Maxima &amp; Minima. Lagrange Multipliers. FirstPartialDerivatives Suppose that z = f(x, y), The first partial derivative of f with respect to x at a z point (x, y) with y treated as a constant is x . The first partial derivative of f with repsect to y at a z point (x, y) with x treated as a constant is . y 05/03/12...

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Partial CalculusofSeveralVariablesand ItsApplications Revision Derivatives Applied Maxima & Minima. Lagrange Multipliers. FirstPartialDerivatives Suppose that z = f(x, y), The first partial derivative of f with respect to x at a z point (x, y) with y treated as a constant is x . The first partial derivative of f with repsect to y at a z point (x, y) with x treated as a constant is . y 05/03/12 BMT1024Tri32011/12 2 Problem1 Find the first partial derivatives for each functions: i) f ( x, y ) = 4 x 2 + 5 x 2 y 2 + 6 y 3 7 ii) z = x 2 y + e x ln( xy + 1) ( iii) R = st + t 3 ) 2 Solution: i) f x ( x, y ) = 4( 2 x ) + 5( 2 x ) y 2 = 8 x + 10 xy 2 f y ( x, y ) = 5 x 2 ( 2 y ) + 6( 3 x 2 ) = 10 x 2 y + 18 y 2 05/03/12 BMT1024Tri32011/12 3 Continued... ii) z = ( 2 x ) y + e x y = 2 xy + e x y x xy + 1 xy + 1 z x x 2 2 = x (1) + 0 =x y xy + 1 xy + 1 )[ ] iii) R = 2 st 3 + t (1) t 3 = 2t 3 st 3 + t s ( ( )[( ) ] ( ) R = 2 st 3 + t s 3t 2 + 1 = 2 st 3 + t 3st 2 + 1 t ( 05/03/12 BMT1024Tri32011/12 )( ) 4 SecondOrderPartialDerivatives The second partial derivatives of a function f with two variables, x and y, are fxx, fyy, fxy, and fyx 2 where f f xx = 2 = ( f x ) x x 2 f f yy = 2 = ( f y ) y y 2 f f xy = = ( fx ) yx y 2 f f yx = = ( fy ) xy x 05/03/12 BMT1024Tri32011/12 5 Problem2 zof x 2 y + e xy . Find each of the second partial derivatives = Solution: First partial derivatives: z = ( 2 x ) y + e xy ( y ) = 2 xy + ye xy x z = x 2 (1) + e xy ( x ) = x 2 + xe xy y 05/03/12 BMT1024Tri32011/12 6 Continued... Thus, the second partial derivatives: 2z = 2(1) y + ye xy ( y ) = 2 y + y 2 e xy x 2 2z = 0 + xe xy x = x 2 e xy y 2 [( ] 2z = 2 x(1) + y e xy x + e xy (1) = 2 x + xye xy + e xy yx 2z = 2 x + x e xy y + e xy (1) = 2 x + xye xy + e xy xy ) [( 05/03/12 ) BMT1024Tri32011/12 ] 7 ApplicationsofFirst PartialDerivatives An important problem in economics concerns how the factors necessary for a production determine the output of a product. For example, the output of a product depends on labor, land, capital, material and machines. If the amount of output z of a product depends on the amounts of two inputs x and y, then the quantity z is given by the production (function z = f x, y ). 05/03/12 BMT1024Tri32011/12 8 ApplicationsofFirst PartialDerivatives z x represents the rate of change in the output z with resepect to input x while input y remains constant. This partial derivative is called the z marginal productivity y x. The partial derivative of is the marginal prodcutivity of y and measures the rate of change of z with respect to input y. 05/03/12 BMT1024Tri32011/12 9 Example Suppose that the production function for a product is , where x represents the number z = x ln ( y + 1) of work-hours and y represents the available capital per week. Find the marginal productivity of x and the marginal productivity of y. z ln ( y + 1) = . Solution:The marginal productivity of x is x 2x z x . The marginal productivity of y is = y y + 1 05/03/12 BMT1024Tri32011/12 10 AppliedMaxima&Minima Partial derivatives are used to solve optimization problems involving non-linear functions. The process of solving such problem includes finding the relative maxima and minima for the function. The procedure: 1. Find fx and fy. 2. Find the points the satisfy both fx = 0 and fy = 0. These are the critical points. 3. Find all the second partial derivatives. 05/03/12 BMT1024Tri32011/12 11 AppliedMaxima&Minima 4. Evaluate D at each critical point. D( x, y ) = f xx f yy ( f xy ) 2 5. Use the test for maxima or minima to determine whether relative maxima or minima occurs. D(x, y) fxx (x,y) Interpretation + + Relative minimum at (x, y). + - Relative maximum at (x, y). 0 05/03/12 Neither maximum or minimum at (x, y). Test is inconclusive. BMT1024Tri32011/12 12 Example 3 2 Test z = x + y + 6 xy + 24 x minima. Solution: for maxima and First partial derivatives, z x = 3 x 2 + 6 y + 24 z y = 2 y + 6x Solve for x and y when zx = 0 and zy = 0, 3 x 2 + 6 y + 24 = 0 .........Equation(1) 2 y + 6x = 0 2 y = 6 x y = 3 x 05/03/12 BMT1024Tri32011/12 .........Equation(2) 13 Continued... Substitute equation (2) in (1): 3 x + 2 6( 3x ) + 24 = 0 3 x 2 18 x + 24 = 0 x2 6x + 8 = 0 ( x 4) ( x 2) = 0 x = 4 or 2 At x = 4, y = -3(4) = -12 and at x = 2, y = -3(2) = -6. Find all second partial derivatives: z xx = 6 x 05/03/12 z yy = 2 BMT1024Tri32011/12 z xy = 6 14 Continued... ( ) = ( 6 x ) ( 2) ( 6) Hence, D = z z z xx yy xy 2 2 = 12 x 36 D At x = 4 and y = -12, = 12( 4 ) 36 = 12 At x = 2 and y = -6, D = 12( 2 ) 36 = 12 Since D > 0 and zxx > 0 at the critical point (4, -12), z has a relative minimum point at x = 4 and y = -12. For the point (2, -16), D is negative thus the test is inconclusive. 05/03/12 BMT1024Tri32011/12 15 Problem Suppose that x units of one input and y units of a P = 40 x + 50 y x 2 y 2 xy second input result in units of product. Determine the inputs of x and y that will maximize P. What is the maximum pSolution: First partial derivatives, roduction? Px = 40 2 x y 05/03/12 Py = 50 2 y x BMT1024Tri32011/12 16 Continued... Solve for x and y when Px = 0 and Py = 40 2 x y = 0 .........Equation(1) 0, 50 2 y x = 0 x = 50 2 y .........Equation(2) Substitute equation (2) in (1): 40 2( 50 2 y ) y = 0 40 100 + 4 y y = 0 3 y = 60 y = 20 05/03/12 BMT1024Tri32011/12 x = 50 2( 20 ) = 10 17 Continued... Find all second partial derivatives: Pxx = 2 Pyy = 2 Pxy = 1 Hence 2 2 D = Pxx Pyy Pxy = ( 2 ) ( 2 ) ( 1) = 4 1 = 3 , Since D > 0 and Pxx < 0 at the critical point (10, 20), P has a relative maximum point. Therefore, P is maximized when 10 units of the first input and 20 units of the second input are produced. The maximum production is 2 2 () P = 40(10 ) + 50( 20) 10 20 10( 20) = 700 05/03/12 BMT1024Tri32011/12 18 LagrangeMultipliers We can obtain maxima and minima for a function z = f(x, y) subject to a constraint g(x,y) = 0 by using the method of Lagrange Multipliers. For example, a consumer may want to maximize utility derived from the consumption of commodities subject to budget constraint or a manufacturer may want to produce a box with a fixed volume so that the least of amount of material is used. 05/03/12 BMT1024Tri32011/12 19 MethodofLagrangeMultipliers Consider a function z = f(x, y) subject to a single constraint g(x, y) = 0. We introduce a new variable, , called the Lagrange multiplier and construct the new objective function, ) = f ( x, y ) + g ( x, y ) F ( x, y , Then, find the critical points of F that will satisfy Fx = 0, Fy = 0, F = 0 This critical points will satisfy the constraint g(x, y) and will be the critical points for f(x, y). 05/03/12 BMT1024Tri32011/12 20 MethodofLagrangeMultipliers However, this method will not tell us whether the critical points correspond to maxima or minima, but this can be determined by testing according to a similar procedure that is used for unconstrained maxima and minima for finding the relative maxima and minima of F. 05/03/12 BMT1024Tri32011/12 21 Example 2 z Find the maximum value of = x y x 0, y 0. subject + y = 9, x to Solution: The objective function:F ( x, y, ) = x 2 y + ( x + y 9 ) First partial derivatives, Fx = 2 xy + = 0 ..............Equation(1) 2 = x 2 ....Equation(2) Fy = x + = 0 F = x + y 9 = 0 ..............Equation(3) 05/03/12 BMT1024Tri32011/12 22 Continued... Substitute equation (2) in (1): ( ) 2 xy + x = 0 2 x 2 + 2 xy = 0 x( x 2 y ) = 0 x = 0 or x = 2 y Since x = 0 could not make z = x2y a maximum, substitute x = 2y in equation (3): ( 2 y) + y 9 = 0 Hence x = 2( 3) = 6 , 05/03/12 3y = 9 y =3 BMT1024Tri32011/12 23 Continued... To determine whether the critical points maximizes F, Fyy = 0 Fxx = 2 y Fxy = 2 x Hence D = F F ( F ) 2 = ( 6 ) ( 0 ) (12) 2 = 144 xx yy xy , Since D < 0, the test inconclusive. Test for several pair of points that satisfy the constraint x + y 9 =0 and see which point gives a maximum value to z. z = ( 52 )( 4 ) = 100 z = ( 6 2 )( 3) = 108 z = ( 7 2 )( 2 ) = 98 Therefore, the critical point x = 6 and y = 3 gives a maximum value to z = 108 while satisfying the constraint x + y = 9. 05/03/12 BMT1024Tri32011/12 24 LearningOutcomes Able to solve unconstrained optimization problem. Able to solve a single constrained optimization problem. 05/03/12 BMT1024Tri32011/12 25
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