## Full text

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WEN-CHING LIEN

Department of Mathematics National Cheng Kung University

2009

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## 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 .

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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.

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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.

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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).

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Example (11.54)

Find the maximum and minimum of

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

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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 .

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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 .

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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 .

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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 .

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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).

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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.

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Example (11.64)

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

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