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Chapter 8 Calculus of Several Variables

Chapter 8 Calculus of Several Variables. 8.1 Functions of Several Independent Variables. Function of Several Variables. The notion for functions of two or more variables is similar to that used for functions of a single variable. and

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Chapter 8 Calculus of Several Variables

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  1. Chapter 8Calculus of Several Variables

  2. 8.1 Functions of Several Independent Variables

  3. Function of Several Variables • The notion for functions of two or more variables is similar to that used for functions of a single variable. • and are functions of two variables. • is a function of three variables.

  4. Example • For f(x,y) = 2x2 –y2, you can evaluate f(2,3) as follows. • For f(x, y, z) = ex (y + z), you can evaluate f(0, -1, 4) as follows.

  5. The Three Dimensional Coordinate System(三維直角坐標系)

  6. Graph of function of two variables • The graph of f (x, y), called a surface, is a collection of all points (x, y, z) such that z =f (x, y). • f (x, y)=2x2+y2+8x-6y+20

  7. Cobb-Douglass生產函數 • 在1928年，Cobb和Douglass建立一簡單的數學模型，來描述美國在1899年初到1922年間總體的生產模式。令 P為某一年的總年產量， K及 L分別表示該年的投資總額及所有勞動力，他們發現 P與 L、K的關係為

  8. Cobb-Douglass生產函數 • 一般而言，分析一個工廠的生產量 P時，也可運用Cobb-Douglass的概念來描述。設有 K單位的資金及L單位的勞動力，則生產量P(L, K)與 K、L的關係式則為 其中 a, b和c為常數。我們稱此等式為古柏道格拉斯生產函數(Cobb-Douglass production function)。

  9. Example 假設有一家公司的Cobb-Douglass生產函數為 P(L, K)=1.2L2/3K1/3。 (a)設每單位資金為300元，每單位勞動力為10小時； 在資金為30,000元，工作時數為500小時下，求該公司的生產量。 • 每單位資金為300元，故共有30000/300=100單位資金。 • 每單位為10小時，故共有500/10=50單位勞動力。 • 所以生產量為P(L, K)=1.2L2/3K1/3 =1.2502/31001/3  95.24

  10. Example 假設有一家公司的Cobb-Douglass生產函數為 P(L, K)=1.2L2/3K1/3。 (b)試問資金及勞動力單位數都加倍時，公司的生產 量是否也加倍？ • 資金及勞動力單位都加倍時，即 K變為2K、L變為2L • P(2L,2 K)=1.2(2L) 2/3(2K)1/3 =1.2(2) 2/3L2/3(2)1/3K1/3 = 2  1.2 L2/3K1/3 • 所以在資金及工作時數都加倍時，其生產量也是加倍的。

  11. 定義域 • 如同單變數函數，二變數函數若未指明定義域，則內定其定義域(domain)即為使函數值有意義的xy-平面上的點(x,y)所成的集合。 • 即使得f (x,y)有意義 ( 或存在 ) 的所有有序數對(x,y)所成的集合。

  12. Example • f (x, y)=2x2+y2+8x-6y+20的定義域為整個xy-平面 2=={(x,y)|x,y} • f (x, y)=x+y - 5的定義域為xy-平面的上半平面,即為{(x,y)|x , y0}

  13. Exercise 8.1 • 11,12,13,14,25,28,74,75

  14. 8.2 Level Curves, Contour Maps, and Cross-Sectional Analysis

  15. 等高線(level curve) • 相同高度的曲線，就是所謂等高線(level curve)，它是指在 xy平面上，方程式滿足f(x, y) = C的所有點(x, y)所成之集合，其中 C為一個常數。

  16. Example • 左圖3表示函數f(x, y) = x2 + y2的圖形。而右圖是C分別以1、2、3、4代入f(x, y) = C 所得到不同高度的曲線。

  17. 8.3 Partial Derivatives and Second-Order Partial Derivatives

  18. Partial Derivatives of a Function of Two Variables • If z = f( x, y), then the first partial derivatives(偏微分) of f with respect to x and y are the functions and , defined as follows.

  19. The partial derivatives of a function of two variables are determined by temporarily considering one variable to be fixed. Hold y constant and differentiate with respect to x. Hold x constant and differentiate with respect to y.

  20. Notation for First Partial Derivatives • The first partial derivatives of z=f( x, y) are denoted by • The values of the first partial derivatives at the point (a, b) are denoted by and

  21. Example • Find the first partial derivatives of f (x, y)= exyln(x+y). • Solution:

  22. 偏導數的幾何意義 • 對x的偏導數fx(x, y0)為曲面z=f(x, y)在平面y=y0的(切線)斜率。(或slope in x-direction). • 對y的偏導數fy(x0, y)為曲面z=f(x,y)在平面x=x0的(切線)斜率。(或slope in y-direction).

  23. Cobb-Douglass生產函數 • 設有 K單位的資金及L單位的勞動力，則生產量P(L, K)與 K、L的關係式則為 其中 a, b和c為常數。 • 在 K變化時，生產量之變化率PK(L, K)稱之為「資金邊際生產效率」。 • 在 L變化時，生產量之變化率PL(L, K)稱之為「工時邊際生產效率」。

  24. Example • 假設Cobb-Douglass生產函數是 P(L, K)=20L1/4K3/4。 求當K = 16單位、L = 81單位時的資金邊際生產效率與工時邊際生產效率。 • 解: • 資金邊際生產效率PK(81,16)=20(81)1/4 (16)3/4

  25. Second-order Partial Derivatives

  26. Example • Find the second partial derivatives of and determine the value of fxy(-1,2). • Solution:

  27. Exercise 8.3 • 9,19,22,28,31,37,43,47,57,60

  28. 8.4 Maxima and Minima

  29. Extrema of Functions of Two Variables • Let f be a function defined on a region containing (a, b). • f(a, b) is a relative maximum of f if there is a cricular region R centered at (a, b) such that f(x, y)f(a, b) for all (x, y) in R. • f(a, b) is a relative minimum of f if there is a cricular region R centered at (a, b) such that f(x, y)f(a, b) for all (x, y) in R.

  30. Critical point • A point (x0, y0) is a critical point of f if fx(x0, y0) or fy(x0, y0) is undifined of if fx(x0, y0)=0 and fy(x0, y0)=0. • A point (x0, y0) is a saddle point of f if (x0, y0) is a critial point and f(x0, y0) is not a relative extremum.

  31. Example • Find the critical points for f(x,y)= 2x2 + y2 + 8x - 6y + 20. • Solution: • fx(x, y) = 4x+ 8and fy(x, y) = 2y- 6 • Solvefx(x, y) = 0and fy(x, y) = 0 • We have x = -2and y = 3 • So point (-2,3) is the only critical point of f.

  32. Second-Derivative Test • If f has continuous first and second partial derivatives in an open region and there exists a point (a,b) in the region such that then fx(a,b)=0 and fy(a,b)=0, then the quantity D = fxx(a,b)fyy(a,b)-[fxy(a,b)]2can be used as shown. • f(a,b) is a relative minimum if D > 0 and fxx(a,b)>0. • f(a,b) is a relative maximum if D > 0 and fxx(a,b)<0. • (a,b,f(a,b)) is a saddle point if D < 0 . • The test gives no information if D = 0.

  33. Example • Find the relative extreme of f(x,y)= 2x2 + y2 + 8x - 6y + 20. • Solution: • fx(x, y) = 4x+ 8and fy(x, y) = 2y- 6 • Solvefx(x, y) = 0and fy(x, y) = 0 • We have x = -2and y = 3 • So point (-2,3) is the only critical point of f. • D = fxx(-2,3)fyy(-2,3)-[fxy(-2,3)]2 = 42-[0]2 = 8 > 0and fxx(-2,3)=4>0. • We have the relative minimum f(-2,3) = 3.

  34. Example • Find the relative extreme and saddle points of f(x,y)= xy – 1/4x4 – 1/4y4 . • Solution: • fx(x, y) = y – x3and fy(x, y) = x – y3 • fxx(x, y) = -3x2, fxy(x, y) = 1,fyy(x, y) = -3y2. • Solvefx(x, y) = 0and fy(x, y) = 0 • We have critical point (0,0),(1,1),and (-1,-1). • D(0,0)= 00-[1]2 = -1 < 0. • D(1,1)= (-3)(-3)-[1]2 = 8 > 0and fxx(1,1)=-3<0. • D(-1,-1)= (-3)(-3)-[1]2 = 8 > 0and fxx(-1,-1)=-3<0. • We have the relative maximum f(1,1) = 1/2 and f(-1,-1) = 1/2. • And (0,0)is a saddle point.

  35. Example • 某兩產品的需求函數分別為x1 = 200(p2 - p1)與x2 = 500+100p1 - 180p2。其中x1與x2分別表示這兩個產品每單位價格為p1與p2時的銷售量。若這兩個產品每單位的生產成本分別為0.5與0.7。求產生最大利潤的售價。 • Solution: • C = 0.5x1+ 0.7x2 = 375 - 25p1 - 35p2 • R = p1x1+ p2x2 = -200p12 - 180p22 + 300p1p2+ 500p2 • 所以總利潤 P=R-C = -200p12 - 180p22 + 300p1p2+ 25p1+535p2 - 375

  36. Example P= -200p12 - 180p22 + 300p1p2+ 25p1+535p2 - 375 • Solve • We have p1=3.14 and p2=4.10 • D = (-400)(-360)-[300]2 = 54000 > 0and • 所以在p1=3.14與p2=4.10時有最大利潤P(3.14,4.10)=761.48

  37. Exercise 8.4 • 7,11,17,20,25,26,33,34,35

  38. 8.5 Lagrange Multipliers

  39. Lagrange Multipliers • If f(x,y) has a maximum or minimum subject to the constraint g(x,y)=0, then it will occur at one of the critical points of the function F defined by F(x,y,)= f(x,y)- g(x,y). • The value  is called a Lagrange Multiplier.

  40. How to Use the Method of Lagrange • Write the situation in the form • Define • Determine the partial derivatives • Solve the system of equations • The maximum (or minimum) of f (x, y) is among the values in step 4. Simply evaluate z = f (x, y) at each.

  41. Example • Find the maximum of A=xy subject to the constraint 6x+4y- 24 = 0. • Solution: • Let f(x, y) = xyand g(x, y) = 6x+4y-24 • Then F( x, y, ) = f(x,y) - g(x,y). = xy - (6x+4y- 24 ) • Solve Fx(x, y, ) = y - 6 = 0, Fy(x, y, ) = x - 4 = 0, F(x, y ) = -6x -4y+ 24 = 0, • We have x = 2, y = 3. • So the maximum is A = xy= (2)(3) = 6.

  42. Example • Find the minimum of f(x,y) = x2 + y2 subject to the constraint 3x+ 2y- 13 = 0. • Solution: • F( x,y,) =x2+y2 - (3x+2y- 13) • Solve Fx(x, y, ) = 2x - 3 = 0, Fy(x, y, ) = 2y - 2 = 0, F(x, y, ) = -3x-2y + 13= 0, • We have x = 3, y = 2. • So the minimum is f(3, 2) = 13.

  43. Example • The product of two positive numbers x and y is 54. Minimize 3x + 2y. • Solution: • Let f(x, y) = 3x + 2yand g(x, y) = xy-54 • F( x,y,) = 3x+ 2y- (xy- 54) • Solve Fx(x, y, ) = 6x - y = 0, Fy(x, y, ) = 4y - x = 0, F(x, y, ) = -xy + 54= 0, • We have x = 6, y = 9. • So the minimum is f(6, 9) = 36.

  44. Example • 某產品的生產函數為f(x,y)= 100x3/4y1/4。其中x與表示勞動力單位、y表示資本單位。勞動力與資本的每單位成本分別為150元與250元。若勞動力與資本的總預算為50000元。求最大的生產水準。 • Solution: • 150x +250y= 50000 • The constraintg(x, y) = 150x +250y- 50000 • Then F( x, y, ) = f(x,y)- g(x,y). = 100x3/4y1/4 - (150x+250y- 50000)

  45. Example • F( x, y, ) = 100x3/4y1/4 - (150x+250y- 50000) • Solve Fx(x, y,) = 75x-1/4y1/4 - 150 = 0, Fy(x, y,) = 25x3/4y-3/4 - 250 = 0, F(x, y, ) = 150x+250y- 50000 = 0, • We have x = 250, y = 50. • 所以最大的生產水準為 f(250,50)= 100(250)3/4(50)1/4  16719.

  46. marginal productivity of money • 在前述例子我們可得在x = 250, y = 50時  0.334 • 在經濟學上，我們稱此Lagrange Multiplier為marginal productivity of money(金錢的邊際生產力)。此意味著每增減一單位的金錢會產生單位的生產力變化。

  47. Example • 在前一個例子中，若勞動力與資本的總預算增為為70000元。求最大的生產水準。 • Solution: • 150x +250y= 50000 • 此時大的生產水準為 16719 + (70000-50000) 16719 +(0.334)(20000) = 23399

  48. Exercise 8.5 • 7,13,17,21,23,24,30

  49. 8.6 Double Integrals

  50. If you are given the partial derivative • Then by holding y constant, you can integrate with respect to x to obtain • This procedure is called partial integration with respect to x. • Note that the “constant of integration” C(y) is assumed to be a function y, because y is fixed during integration with respect to x.

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