Currently, science is growing, with this development strength, the application of science and scientific patents in schools is very feasible and significant. Since the first year, Ho Chi Minh City University of Technology lecturers have helped technical students get acquainted with programming applications such as MATLAB. MATLAB is a rising and programming environment that allows calculating numbers with matrices, graphing functions or charting information, implementing algorithms, creating user interfaces, and linking to computer programs written in many other programming languages. MATLAB enables simulation of calculations with the Toolbox library, experimenting with many models in practice and techniques. With more than 40 years of establishment and development, MATLAB is an effective calculation tool to solve technical problems today with a relatively simple and universal design. Therefore, for the problems in Algebra, especially matrix problems, we can use MATLABs computing applications to solve most simply and understandably, helping us get acquainted and add more skills to use programs application for students.
HCMC University of Technology Department of Applied Mathematics MATLAB Project Subject: Calculus Introduction Currently, science is growing, with this development strength, the application of science and scientific patents in schools is very feasible and significant Since the first year, Ho Chi Minh City University of Technology lecturers have helped technical students get acquainted with programming applications such as MATLAB MATLAB is a rising and programming environment that allows calculating numbers with matrices, graphing functions or charting information, implementing algorithms, creating user interfaces, and linking to computer programs written in many other programming languages MATLAB enables simulation of calculations with the Toolbox library, experimenting with many models in practice and techniques With more than 40 years of establishment and development, MATLAB is an effective calculation tool to solve technical problems today with a relatively simple and universal design Therefore, for the problems in Algebra, especially matrix problems, we can use MATLAB's computing applications to solve most simply and understandably, helping us get acquainted and add more skills to use programs application for students Solutions: Question 1: Find the extreme values of the function f(x; y) = x + 2y2 on the circle x2 + y2 = Sketch the given surface and show the extreme values fx = 2x ; fy = 4y, the only critical point is (0;0) and f(0;0) =0 To find the critical points on the boundary x 2+ y 2=1 we follow the method of Lagrange Multiplies with f( x,y) = x 2+ y and g(x;y) = x 2+ y We have the system 2x = 2x, 4y= lamda.2y, x 2+ y 2=1 Solving this system give us four extreme points: (0;1) (0;-1) (1;0) and (1;0) The values for f at these point are 2; 2; 1; Therefore the maximum value is f ( ; ± )=2and the minimun point is f ( ; )=0 Question 2: Find the extreme values of the function f(x; y) = x 2y on the curve x2 + 2y2 = 6: Sketch the given surface and show the extreme values L(x , y )=f (x , y )+φ (x , y )=x y +( x2 +2 y 2−6) L' x =2 xy +2 x =0 α =0 ; ± ¿0;±2 L ' y =x2 + y=0 2 L ' α =x +2 y =6 y =± √ , ±1 We have stationary points P1 ¿ d L=L''xx ( x o , y o ) d x +2 L'xy' ( x o , y o ) dxdy + L'yy' ( x o , y o ) d y −x φ (x , y )=x 2+ y 2−6=0 dy = dx 2y When ¿ d L (P ¿¿ 1)0 ¿ so P2 is minimum point ¿ d L(P¿¿ 3) , d L(P¿¿ )< so P3 ; P4 ¿ ¿is local maximum ¿−1 d2 L(P¿¿ 5),d L( P¿¿ 6)>0 so P5 ; P6 islocal minimum ¿ ¿ { 1|56 { Question 3: Find the extreme values of the function f(x; y) = - 5x - 4y on the curve x2 + y2 = Sketch the given surface and show the extreme values f x =−5 f y =−4 g ( x , y )=x + y 2=9 → ∇ f =λ ∇ g { { { 15 √ 41 41 f x =λ g x −5= λ x 12 f y =λ g y → −4=λ y → y=± √ 41 41 2 2 x + y =9 x + y =9 √ 41 λ=± x=± f has possible extreme values at the points ( 1541√ 41 , 1241√ 41 ) ;( −1541√ 41 , −1241√ 41 ) ; 15 √ 41 12 √ 41 f( , =6−3 √ 41 41 41 ) −15 √ 41 −12 √ 41 f( , =6+3 √ 41 41 41 ) The maximum value of f(x,y) is 6+3 √ 41 and the minimum is 6−3 √ 41 Question 4: Find the extreme values of the function f(x; y) = - 4x - 8y on the curve x2 + 8y2 = Sketch the given surface and show the extreme values Solve equations ∇f= λ∇g and g(x,y)=8 using Lagrange multipliers Constraint g(x,y) = x2 + 8y2 = fx = λgx fy = λgy g(x,y) = which become −4 -4 = 2xλ => x = (1) 2λ -8 = 16yλ x2 + 8y2 = => y = −8 16 λ (2) (3) subtitude (1) and (2) into (3), we have: √3 ¿)2 + 8( )2 = => λ = ± 2λ These values of λ then give the corresponding points (x,y) 3 3 (- √ ,- √ ) and ( √ , √ ) 3 3 3 f(- √ ,- √ ) = + 8√ 3 3 f( √ , √ ) = – 8√ 3 2|56 Therefore the maximum value of f(x,y) on the curve x2 + 8y2 = is f(,- √3 √ ) = + and the minimum value is f( √3 , √3 ) = - √ √ 3 Question 5: Find the extreme values of the function f(x; y) = x + y2 + xy on the curve x2 + 2y2 = Sketch the given surface and show the extreme values Consider Lagrange function: L ( x , y , λ )=x + y 2+ xy−λ (1−x 2−2 y 2) Find stationary points: { L'x =0 L'y =0 L'λ=0 → x+ y+ λx =0 y + x + λy=0 2 x + y =1 { ↔ { −2 x + y 2x −2 y + x λ= 4y 2 x +2 y =1 λ= −2 x + y −2 y + x ↔ y ( x + y )=2 x (2 y + x ) ↔ x=( 1+ √ ) y , x=( 1−√ ) y = 2x 4y x 2+ y 2=1 ↔ [ ( 1+ √ ) y ] +2 y =1 ↔ y=± x=( 1−√ ) y : x 2+ y 2=1 ↔ [ ( 1− √ ) y ] +2 y 2=1 ↔ y=± 3−√ 12 √ √ x=( 1+ √ ) y : 3+ √ 13 12 There are four stationary points L'xx' L''xy −φ'x 2+ λ −2 x '' '' ' ∆=−det ( ) ∆=−det ( L xy L yy −φ y ↔ 2+ λ −2 y ) Compute: −2 x −2 y −φ'x −φ'y Stationary points 3−√3 3−√ ((1+ √3) ; ) 12 12 3− √3 3−√ ((1+ √3)(− );− ) 12 12 3+ √13 3+ √ 13 ((1−√ 3) ; ) 12 12 3+ √13 3+ √13 ((1−√ 3)(− );− ) 12 12 √ √ √ √ √ √ √ √ ∆ 0 Minimum point Question 6: Find the extreme values of the function f(x; y) = 2x + 12xy + y2 on the curve x2 + 4y2 = 25 Sketch the given surface and show the extreme values f ( x , y )=2 x2 +12 xy+ y x 2+ y 2=25 fx = 4x+12y fy = 12x+2y gx = 2x gy = 2y fx = lamda.gx => 4x +12y = lamda.2x (1) fy = lamda.gy => 12x +2y = lamda.2y (2) 3|56 1, => x +12 y 12 x+2 y = 2x 2y xy+ 12 y 2=12 x 2+ xy y 2−6 x 2−xy =0 We have ( 25−x 2) −6 x 2−x √25−x 2=0 x= 3.39 and x=-3.68 Question 7: Find the extreme values of the function f ( x , y )=x 2+ y on the x y + =1 Sketch the given surface and show the extreme values plane x y L(x , y )=f ( x , y )+φ ( x , y )=x 2+ y2 − + −1 ( { ) x 18 x− =0 x = 13 L ' x =0 y 12 L' y =0 y− =0 y= 13 L' α =0 x y 42 + −1=0 ¿ 13 { { ( 1813 , 1213 ), is a stationary point M d L=L''xx ( x o , y o ) d x +2 L'xy' ( x o , y o ) dxdy + L'yy' ( x o , y o ) d y x y −3 φ (x , y )= + −1=0 dy= dx −3 22 d L=L''xx ( x o , y o ) d x +2 L'xy' ( x o , y o ) d x+ L'yy' ( x o , y o ) d x= + d x 13 13 18 12 , is the minimum point Then d L> so M 3 ( ( ) ) Question 8: Find the local maximum and minimum values and saddle points of f ( x , y )=x + y −2 x2 + xy−2 y Sketch the given surface and show the extreme values f x =4 x 3−4 x+ y f y =4 y +4 x−4 y f x =0 3 2 → x 3−4 x +4 y =0 → x + y =0 →(x+ y )(x −xy + y )=0 f y =0 y + x−4 y=0 { { x + y=0∨x 2−xy + y 2=0 (0,0) x=0 y=0 → x=− y → x 3−x−x=0→ x=√ → y =−√ → ( √ ,−√ 2) x=−√ y =√ (−√ , √ 2) { l= ∂2 f =12 x −4 ∂x ∂2 f m= =4 ∂ x∂ y n= ∂2 f =12 y−4 ∂ y2 At point (0,0) l=12 ×0−4=−4< m=4 4|56 { { n=12× 0−4=−4 m=4 n=12× 2−4=20>0 ln −m 2=20.20−4 2=384> 0∧l>0 → f has a minimum value=−8 At point(−√ , √ 2) l=12 ×2−4=20> m=4 n=12× 2−4=20>0 ln −m2=20.20−4 2=384> 0∧l>0 → f has a minimumvalue=−8 Question 9: Find the local maximum and minimum values and saddle points of f ( x , y )=2 x + y −x2 −2 y Sketch the given surface and show the extreme values fx = 8x3 – 2x = => x = fy = 4y3 – 4y = x= x= −1 => y = y=1 y = -1 fxx = 24x2 – fyy = 12y2 – fxy = ∆ = fxx.fyy – fxy2 = 288x2y2 – 96x2 – 24y2 + (0,0) (0,1) (0,-1) ( ,0) ( ,1) ( ,-1) −1 ( ,0) −1 ( ,1) −1 ( ,-1) 5|56 fxx -2 -2 -2 fyy -4 8 -4 ∆ = fxx.fyy – fxy2 -16 -16 -16 Local maximum Saddle point Saddle point Saddle point 32 Local minimum 32 Local minimum -4 -16 Saddle point 32 Local minimum 32 Local minimum Therefore, the maximum value of f is f(0,0) = 0, and f( ,1) = - , f( ,-1) = −1 −1 , f( ,1) = - , f( ,-1) = - are the minimum point 8 Question 10: Find the local maximum and minimum values and saddle points of f ( x , y )=(x 2−2 y )e x− y Sketch the given surface and show the extreme values Finding stationary points: f 'x =0 f 'y =0 { ( x ) e x− y + ( x 2−2 y ) e { ↔ (−4 y ) e x− y x− y =0 x− y + ( x −2 y ) (−1 ) e =0 2 → x=0, y=0 {x=−4, y=−2 The two stationary points are (0;0) and (-4;-2) A=f 'xx' =( 2+2 x ) e x− y +(2 x+ x 2−2 y 2) e x− y Compute: B=f 'xy' = (−4 y ) e x− y + ( x+ x 2−2 y ) (−1 ) e x− y C=f 'yy' =−( 4−4 y ) e x− y −(4 y+ x2 −2 y )e x− y { Stationar y point (0;0) Conclusio n -4 0 Maximum point Question 11: Find the local maximum and minimum values and saddle points of f ( x , y )=(x + y +2 y)e x Sketch the given surface and show the extreme values Finding stationary points: f 'x =0 f 'y =0 { A B e x (1+ x +2 y 2+ y )=0 x (2 y +2)e =0 { ↔ C → ∆=AC-B2 {x=0.5 y=−1 The two stationary points are (0;0) and (-1;-1) A=f 'xx' =2 e2 x (2+ x +2 y 2+ y ) B=f 'xy' =e x (4 y +4) Compute: C=f 'yy' =2e x { Stationar y point (0.5;-1) A B C ∆=AC-B2 >0 >0 >0 Conclusio n Maximum point Question 12: Find the local maximum and minimum values and saddle x points of f ( x , y )=( x + y )e Sketch the given surface and show the extreme values x f (x , y )=(x+ y ) e x x f x=e + ( x+ y 2) e =0 x=−2 x y =0 ' f y=2 y e =0 M(-2,0) is a stationary point { ' 6|56 { A=f 'xx' 0.184 B=f ''xy =0 ∆= AC −B 2> 0∧ A> We have C=f 'yy' = e So M(-2,0) is the minimum point Question 13: Find the local maximum and minimum values and saddle points of f ( x , y )=( x + y )(e−x − y −1) Sketch the given surface and show the extreme values f(x, y) = (x + y )(e −x 2−y − 1) { 2 2 f x =2 x e− x − y ( 1−x 2− y 2) −2 x 2 f y =2 y e− x − y ( 1− x2− y 2) −2 y f x =0 → { x=0 , y=0 f y =0 { Question 14: Find the local maximum and minimum values and saddle points of f ( x , y )=3 x 2−x 3+2 y +4 y Sketch the given surface and show the extreme values fx = 6x - 3x2 = => x = x=2 fy = 4y + = => y = -1 fxx = – 6x fyy = fxy = ∆ = fxx.fyy – fxy2 = 24 - 24x fxx 6>0 -6 (0,-1) (2,-1) fyy 4 ∆ = fxx.fyy – fxy2 24 > -24 < Local minimum Saddle point Question 15: Find the local maximum and minimum values and saddle points of f ( x , y )=x 3+ xy + y Sketch the given surface and show the extreme values Finding stationary points: f 'x =0 f 'y =0 { x +3 y=0 x +3 y 2=0 { ↔ → x =0, y =0 {x=−1, y=−1 The two stationary points are (0;0) and (-1;-1) A=f ''xx =6 x '' Compute: B=f xy =3 C=f 'yy' =6 y { Stationar y point (0;0) A B C ∆=AC-B2 0 7|56 Conclusio n Saddle point Local maximum point Question 16: Find the local maximum and minimum values and saddle points of f ( x , y )=4 xy−x 4− y Sketch the given surface and show the extreme values f ( x , y )=4 xy−x 4− y fx= y−4 x 3=0=¿y - x 3=0 fy= x−4 y 3=0 → x− y 3=¿0 x=0, y=0 x=-1, y=-1 x=1, y=1 Critical points (0,0); (-1;1); (1;1) fxy=-12x =A ∆= AC−B2 fxy=4=B fyy= -12y =C At (0;0) A=0, B=4, C=0 ∆ 0, A >0 → ( ; ) is a local maximum At (1;-1) A= -12, B=4, C=-12 ∆ >0, A So M is maximum point B=f 'xy' =0 ∆=AC −B ∆3 >0∧ A