Advanced Calculus (II)

Advanced Calculus (II)

W^EN-C^HING L^IEN

Department of Mathematics National Cheng Kung University

2009

Ch11: Differentiability on R

11.7: Optimization

Definition (11.50)

Let V be open in Rⁿ, let a∈V , and suppose that f :V →R.

(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 x∈Br(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 x∈Br(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 .

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.

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.

Definition (11.53)

Let V be open in Rⁿ, let a∈V , and let f :V →R 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,y∈B_ρ(a)that satisfy

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

Example (11.54)

Find the maximum and minimum of

f(x,y) =x²−x +y²−2y on H =B1(0,0).

Lemma (11.55)

Let V be open in Rⁿ, a∈V , and f :V →R. 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 x ∈Rⁿ.

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, x∈Rⁿ.

Lemma (11.55)

Let V be open in Rⁿ, a∈V , and f :V →R. 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 x ∈Rⁿ.

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, x∈Rⁿ.

Lemma (11.55)

Let V be open in Rⁿ, a∈V , and f :V →R. 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 x ∈Rⁿ.

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, x∈Rⁿ.

Lemma (11.55)

Let V be open in Rⁿ, a∈V , and f :V →R. 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 x ∈Rⁿ.

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, x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x) = g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x) = g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x) = g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x) = g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x)= g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x) = g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x)= g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x) = g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x) = g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x) = g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

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 _kxk^x ∈H, and it follows from the choice of g and m that

D⁽²⁾f(a;x) = g(x)

kxk²kxk²=g

 x kxk



kxk²≥mkxk². We conclude that (33) holds for all x∈Rⁿ.

Theorem (11.56 Second Derivative Test)

Let V be open in Rⁿ, a∈V , and suppose that f :V →R 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 h∈Rⁿ, then a is a saddle point of f .

Theorem (11.56 Second Derivative Test)

Let V be open in Rⁿ, a∈V , and suppose that f :V →R 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 h∈Rⁿ, then a is a saddle point of f .

Theorem (11.56 Second Derivative Test)

Let V be open in Rⁿ, a∈V , and suppose that f :V →R 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 h∈Rⁿ, then a is a saddle point of f .

Theorem (11.56 Second Derivative Test)

Let V be open in Rⁿ, a∈V , and suppose that f :V →R 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 h∈Rⁿ, then a is a saddle point of f .

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.

Lemma (11.58)

Let A,B,C ∈R, D=B²−AC, 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.

Lemma (11.58)

Let A,B,C ∈R, D=B²−AC, 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.

Lemma (11.58)

Let A,B,C ∈R, D=B²−AC, 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.

Theorem (11.59)

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

f :V →R 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=f_xy²(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.

Theorem (11.59)

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

f :V →R 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=f_xy²(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.

Theorem (11.59)

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

f :V →R 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=f_xy²(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.

Theorem (11.59)

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

f :V →R 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=f_xy²(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.

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.

Definition (11.61)

Let V be open in Rⁿ, a∈V , and f,gj :V →R 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 x∈Bρ(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 x∈Bρ(a)and gj(x) =0 for all j =1, . . . ,m imply f(x) ≤f(a).

Theorem (11.63 Lagrange Multipliers)

Let m <n, V be open in Rⁿand f,gj :V →R beC¹ on V for j =1,2, . . . ,m. Suppose that there is an a∈V 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.

Example (11.64)

Find all extrema of x²+y²+z²subject to the constraints x −y =1 and y²−z²=1.

Thank you.