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# Advanced Calculus (II)

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## Advanced Calculus (II)

WEN-CHING LIEN

Department of Mathematics National Cheng Kung University

2009

(2)

## Ch11: Differentiability on R

n

### 11.7: Optimization

Definition (11.50)

Let V be open in Rn, let aV , and suppose that f :VR.

(i) f(a)is called a local minimum of f if and only if there is an r >0 such that f(a) ≤f(x)for all xBr(a).

(ii) f(a)is called a local maximum of f if and only if there is an r >0 such that f(a) ≥f(x)for all xBr(a).

(iii) f(a)is called a local extremum of f if and only if f(a)is a local maximum or a local minimum of f .

(3)

Remark (11.51)

If the first-order partial derivatives of f exist at a, and f(a) is a local extremum of f , then∇f(a) =0.

(4)

Remark (11.52)

There exist continuously differentiable functions that satisfy∇f(a) =0 such that f(a)is neither a local maximum nor a local minimum.

(5)

Definition (11.53)

Let V be open in Rn, let aV , and let f :VR be differentiable at a. Then a is call a saddle point of f if

∇f(a) =0 and there is a r0>0 such that given any 0< ρ < r0there are points x,yBρ(a)that satisfy

f(x) <f(a) < f(y).

(6)

Example (11.54)

Find the maximum and minimum of

f(x,y) =x2x +y22y on H =B1(0,0).

(7)

Lemma (11.55)

Let V be open in Rn, aV , and f :VR. If all second- order partial derivatives of f exist at a and D(2)f(a;h) >0 for all h 6=0, then there is an m >0 such that

(33) D(2)f(a;x) ≥mkxk2 for all xRn.

Proof.

Set H = {x∈Rn : kxk =1}and consider the function g(x) : =D(2)f(a;x) : =

n

X

j=1 n

X

k=1

2f

∂xk∂xj

(a)xjxk, xRn.

(8)

Lemma (11.55)

Let V be open in Rn, aV , and f :VR. If all second- order partial derivatives of f exist at a and D(2)f(a;h) >0 for all h 6=0, then there is an m >0 such that

(33) D(2)f(a;x) ≥mkxk2 for all xRn.

Proof.

Set H = {x∈Rn : kxk =1}and consider the function

g(x) : =D(2)f(a;x) : =

n

X

j=1 n

X

k=1

2f

∂xk∂xj

(a)xjxk, xRn.

(9)

Lemma (11.55)

Let V be open in Rn, aV , and f :VR. If all second- order partial derivatives of f exist at a and D(2)f(a;h) >0 for all h 6=0, then there is an m >0 such that

(33) D(2)f(a;x) ≥mkxk2 for all xRn.

Proof.

Set H = {x∈Rn : kxk =1}and consider the function

g(x) : = D(2)f(a;x) : =

n

X

j=1 n

X

k=1

2f

∂xk∂xj

(a)xjxk, xRn.

(10)

Lemma (11.55)

Let V be open in Rn, aV , and f :VR. If all second- order partial derivatives of f exist at a and D(2)f(a;h) >0 for all h 6=0, then there is an m >0 such that

(33) D(2)f(a;x) ≥mkxk2 for all xRn.

Proof.

Set H = {x∈Rn : kxk =1}and consider the function

g(x) : = D(2)f(a;x) : =

n

X

j=1 n

X

k=1

2f

∂xk∂xj

(a)xjxk, xRn.

(11)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact,it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0. If x6=0, then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x) = g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(12)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact, it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0. If x6=0, then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x) = g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(13)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact,it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0.If x6=0, then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x) = g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(14)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact, it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0. If x6=0,then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x) = g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(15)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact, it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0.If x6=0, then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x)= g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(16)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact, it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0. If x6=0,then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x) = g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(17)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact, it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0. If x6=0, then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x)= g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(18)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact, it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0. If x6=0, then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x) = g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(19)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact, it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0. If x6=0, then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x) = g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(20)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact, it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0. If x6=0, then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x) = g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(21)

Proof.

By hypothesis, g is continuous and positive on Rn\{0}, hence on H. Since H is compact, it follows from the Extreme Value Theorem that g has a positive minimum m on H.

Clearly, (33) holds for x=0. If x6=0, then kxkxH, and it follows from the choice of g and m that

D(2)f(a;x) = g(x)

kxk2kxk2=g

 x kxk



kxk2mkxk2. We conclude that (33) holds for all xRn.

(22)

Theorem (11.56 Second Derivative Test)

Let V be open in Rn, aV , and suppose that f :VR satisfies∇f(a) =0. Suppose further that the second- order total differential of f exists on V and is continuous at a.

(i) If D(2)f(a;h) >0 for all h6=0, then f(a)is a local minimum of f .

(ii) If D(2)f(a;h) <0 for all h6=0, then f(a)is a local maximum of f .

(iii) If D(2)f(a;h)takes on both positive and negative values for hRn, then a is a saddle point of f .

(23)

Theorem (11.56 Second Derivative Test)

Let V be open in Rn, aV , and suppose that f :VR satisfies∇f(a) =0. Suppose further that the second- order total differential of f exists on V and is continuous at a.

(i) If D(2)f(a;h) >0 for all h6=0, then f(a)is a local minimum of f .

(ii) If D(2)f(a;h) <0 for all h6=0, then f(a)is a local maximum of f .

(iii) If D(2)f(a;h)takes on both positive and negative values for hRn, then a is a saddle point of f .

(24)

Theorem (11.56 Second Derivative Test)

Let V be open in Rn, aV , and suppose that f :VR satisfies∇f(a) =0. Suppose further that the second- order total differential of f exists on V and is continuous at a.

(i) If D(2)f(a;h) >0 for all h6=0, then f(a)is a local minimum of f .

(ii) If D(2)f(a;h) <0 for all h6=0, then f(a)is a local maximum of f .

(iii) If D(2)f(a;h)takes on both positive and negative values for hRn, then a is a saddle point of f .

(25)

Theorem (11.56 Second Derivative Test)

Let V be open in Rn, aV , and suppose that f :VR satisfies∇f(a) =0. Suppose further that the second- order total differential of f exists on V and is continuous at a.

(i) If D(2)f(a;h) >0 for all h6=0, then f(a)is a local minimum of f .

(ii) If D(2)f(a;h) <0 for all h6=0, then f(a)is a local maximum of f .

(iii) If D(2)f(a;h)takes on both positive and negative values for hRn, then a is a saddle point of f .

(26)

Remark (11.57)

If D(2)f(a;h) ≥0, then f(a)can be a local minimum or a can be a saddle point.

(27)

Lemma (11.58)

Let A,B,CR, D=B2AC, andφ(h,k) =Ah2+2Bhk +Ck2.

(i) If D <0, then A andφ(h,k)have the same sign for all (h,k) 6= (0,0).

(ii) If D >0, thenφ(h,k)takes on both positive and negative values as(h,k)varies over R2.

(28)

Lemma (11.58)

Let A,B,CR, D=B2AC, andφ(h,k) =Ah2+2Bhk +Ck2.

(i) If D <0, then A andφ(h,k)have the same sign for all (h,k) 6= (0,0).

(ii) If D >0, thenφ(h,k)takes on both positive and negative values as(h,k)varies over R2.

(29)

Lemma (11.58)

Let A,B,CR, D=B2AC, andφ(h,k) =Ah2+2Bhk +Ck2.

(i) If D <0, then A andφ(h,k)have the same sign for all (h,k) 6= (0,0).

(ii) If D >0, thenφ(h,k)takes on both positive and negative values as(h,k)varies over R2.

(30)

Theorem (11.59)

Let V be open in R2,(a,b) ∈V , and suppose that

f :VR satisfies∇f(a,b) =0. Suppose further that the second-order total differential of f exists on V and is continuous at(a,b), and set

D=fxy2(a,b) −fxx(a,b)fyy(a,b).

(i) If D <0 and fxx(a,b) >0, then f(a,b)is a local minimum.

(ii) If D <0 and fxx(a,b) <0, then f(a,b)is a local maximum.

(iii) If D >0, then(a,b)is a saddle point.

(31)

Theorem (11.59)

Let V be open in R2,(a,b) ∈V , and suppose that

f :VR satisfies∇f(a,b) =0. Suppose further that the second-order total differential of f exists on V and is continuous at(a,b), and set

D=fxy2(a,b) −fxx(a,b)fyy(a,b).

(i) If D <0 and fxx(a,b) >0, then f(a,b)is a local minimum.

(ii) If D <0 and fxx(a,b) <0, then f(a,b)is a local maximum.

(iii) If D >0, then(a,b)is a saddle point.

(32)

Theorem (11.59)

Let V be open in R2,(a,b) ∈V , and suppose that

f :VR satisfies∇f(a,b) =0. Suppose further that the second-order total differential of f exists on V and is continuous at(a,b), and set

D=fxy2(a,b) −fxx(a,b)fyy(a,b).

(i) If D <0 and fxx(a,b) >0, then f(a,b)is a local minimum.

(ii) If D <0 and fxx(a,b) <0, then f(a,b)is a local maximum.

(iii) If D >0, then(a,b)is a saddle point.

(33)

Theorem (11.59)

Let V be open in R2,(a,b) ∈V , and suppose that

f :VR satisfies∇f(a,b) =0. Suppose further that the second-order total differential of f exists on V and is continuous at(a,b), and set

D=fxy2(a,b) −fxx(a,b)fyy(a,b).

(i) If D <0 and fxx(a,b) >0, then f(a,b)is a local minimum.

(ii) If D <0 and fxx(a,b) <0, then f(a,b)is a local maximum.

(iii) If D >0, then(a,b)is a saddle point.

(34)

Remark (11.60)

If the discriminant D =0, f(a,b)may be a local maximum, a local minimum, or (a,b)may be a saddle point.

(35)

Definition (11.61)

Let V be open in Rn, aV , and f,gj :VR for j =1,2, . . . ,m.

(i) f(a)is called a local minimum of f subject to

constraints gj(a) =0, j =1, . . . ,m, if and only if there is a ρ >0 such that xBρ(a)and gj(x) =0 for all j =1, . . . ,m imply f(x) ≥f(a).

(ii) f(a)is called a local maximum of f subject to

constraints gj(a) =0, j =1, . . . ,m, if and only if there is a ρ >0 such that xBρ(a)and gj(x) =0 for all j =1, . . . ,m imply f(x) ≤f(a).

(36)

Theorem (11.63 Lagrange Multipliers)

Let m <n, V be open in Rnand f,gj :VR beC1 on V for j =1,2, . . . ,m. Suppose that there is an aV such that ∂(g1, . . . ,gm)

∂(x1, . . . ,xm)(a) 6=0.

If f(a)is a local extremum of f subject to constraints gk(a) =0, k =1, . . . ,m, then there exist scalars λ1, λ2, . . . , λm such that

(36) ∇f(a)+

m

X

k=1

λk∇gk(a) =0.

(37)

Example (11.64)

Find all extrema of x2+y2+z2subject to the constraints xy =1 and y2z2=1.

(38)

## Thank you.

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung

Department of Mathematics National Cheng Kung