Advanced Calculus (II)
WEN-CHING LIEN
Department of Mathematics National Cheng Kung University
2009
Ch11: Differentiability on R
n11.7: 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∈Bρ(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 kxkx ∈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 kxkx ∈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 kxkx ∈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 kxkx ∈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 kxkx ∈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 kxkx ∈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 kxkx ∈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 kxkx ∈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 kxkx ∈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 kxkx ∈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 kxkx ∈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=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.
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=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.
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=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.
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=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.
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.
