Answer ”True” if it is correct

1082 ®M4-(“ßF¡„É( sãÊU–

1. (10%) Read the following statements. Answer ”True” if it is correct. Answer ”False” if it is incorrect.

(a) (2%) True A is an n× m matrix. The dimension of A’s column space equals the dimension of A’s row space.

(b) (2%) True A is an n× n matrix. If detA=0, then 0 is an eigenvalue of A.

(c) (2%) True A is an n× m matrix. The rank of A is less than or equal to n. The rank of A is less than or equal to m.

(d) (2%) False The set S=⎧⎪⎪⎪

⎨⎪⎪⎪⎩

⎡⎢⎢⎢

⎢⎢⎣ 1 2 3

⎤⎥⎥⎥

⎥⎥⎦ ,

⎡⎢⎢⎢

⎢⎢⎣

−1 3

−2

⎤⎥⎥⎥

⎥⎥⎦ ,

⎡⎢⎢⎢

⎢⎢⎣ 1 7 4

⎤⎥⎥⎥

⎥⎥⎦

⎫⎪⎪⎪⎬⎪⎪

⎪⎭ is a basis for R³.

(e) (2%) False B is a k× n matrix. Suppose that there is a x0≠ 0, x0∈ Rⁿ such that Bx₀= 0. Then B is not full rank.

2. (18%) Consider the quadratic form f(x, y, z) = −2x²+ 2xy − 2y²− 5z². (a) (3%) Express f in the form of

f(x, y, z) = (x, y, z) M ⎛

⎜⎝ x y z

⎞⎟

⎠, where M is a real-valued symmetric 3× 3 matrix. Find M.

(b) (6%) Find the eigenvalues of M and their corresponding eigenvectors.

(c) (4%) Diagonalize M . That is, find a nonsingular matrix P and a diagonal matrix D such that P⁻¹M P= D.

(d) (5%) Use Sylvester’s criterion to determine the definiteness of M , and thus f .

Solution:

(a) M =⎛

⎜⎝

−2 1 0

1 −2 0

0 0 −5

⎞⎟

⎠. (3pts) (b)

0=det ⎛

⎜⎝

−2 − λ 1 0

1 −2 − λ 0

0 0 −5 − λ

⎞⎟

⎠ characteristic polynomial

=(−5 − λ) ((−2 − λ)²− 1) expand

=(−5 − λ) (λ²+ 4 λ + 3) expand

= − (λ + 5) (λ + 1) (λ + 3) factor

λ= − 5, −1, −3 solve (3pts)

(1) λ= −5 Ô⇒ ⎛

⎜⎝

−2 + 5 1 0

1 −2 + 5 0

0 0 −5 + 5

⎞⎟

⎠

⎛⎜

⎝ x y z

⎞⎟

⎠=⎛

⎜⎝ 0 0 0

⎞⎟

⎠

Ô⇒ eigenvector is⎛

⎜⎝ x y z

⎞⎟

⎠= t⎛

⎜⎝ 0 0 1

⎞⎟

⎠for any t∈ R with t ≠ 0. (1pt)

(2) λ= −3 Ô⇒ ⎛

⎜⎝

−2 + 3 1 0

1 −2 + 3 0

0 0 −5 + 3

⎞⎟

⎠

⎛⎜

⎝ x y z

⎞⎟

⎠=⎛

⎜⎝ 0 0 0

⎞⎟

⎠

Ô⇒ eigenvector is⎛

⎜⎝ x y z

⎞⎟

⎠= s⎛

⎜⎝ 1

−1 0

⎞⎟

⎠for any s∈ R with s ≠ 0. (1pt)

(3) λ= −1 Ô⇒ ⎛

⎜⎝

−2 + 1 1 0

1 −2 + 1 0

0 0 −5 + 1

⎞⎟

⎠

⎛⎜

⎝ x y z

⎞⎟

⎠=⎛

⎜⎝ 0 0 0

⎞⎟

⎠

Ô⇒ eigenvector is⎛

⎜⎝ x y z

⎞⎟

⎠= r⎛

⎜⎝ 1 1 0

⎞⎟

⎠for any r∈ R with r ≠ 0. (1pt)

(c) P =⎛

⎜⎝

0 1 1

0 −1 1

1 0 0

⎞⎟

⎠(2pts) Ô⇒ P⁻¹M P =⎛

⎜⎝

−5 0 0

0 −3 0

0 0 −1

⎞⎟

⎠= D. (2pts) (d) M is negative definite (2pts), and thus f , since (3pts)

−2 < 0, det ( −2 1

1 −2 ) =3> 0, det M = −15 < 0.

3. (20%) Find the maximum value of the function f(x, y) = x²+ 2y² subject to the constraints x+ 2y ≤ 9, x²+ y²≥ 16, x ≥ 0 and y≥ 0.

(a) (2%) Could both constraints x+ 2y ≤ 9 and x²+ y²≥ 16 be binding?

(b) (2%) Check whether the Kuhn - Tucker version NDCQ is satisfied.

(c) (2%) Write down the Kuhn - Tucker version Lagrangian function ˜L.

(d) (3%) Write down the Kuhn - Tucker version first order conditions.

(e) (4%) Is there a solution such that the constraint x²+ y²≥ 16 is binding?

(f) (5%) Find solution(s) such that the constraint x+ 2y ≤ 9 is binding.

(g) (2%) Find the maximum value.

Solution:

Let g₁(x, y) = x + 2y and g2(x, y) = −x²− y². Suppose that(x^∗, y^∗) is the maximizer of the problem.

(a) We want to find(x, y) satisfies x + 2y = 9 and x²+ y²= 16. Since x = 9 − 2y,

(9 − 2y)²+ y²= 16 ⇒ 65 − 36y + 5y²= 0 ⇒ it has no real solution becase 36²− 65 × 20 < 0.

(2 points).

(b) We have that∇g1(x, y) = ⟨1, 2⟩ and ∇g2(x, y) = ⟨−2x, −2y⟩.

From (a), we know that g1≤ 9 and g2≤ 16 can not both binding.

Suppose tht g1(x^∗, y^∗) = 9. Since ∇g1(x, y) = ⟨1, 2⟩, (^∂g_∂x¹), (^∂g_∂y¹), (^∂g_∂x^{1 ∂g}_∂y¹) all have rank 1. (1 point).

Suppose that g2(x^∗, y^∗) = 16, then (x^∗)²+ (y^∗)²= 16 ⇒ x^∗ and y^∗ can not be both 0. If x^∗> 0 but y^∗= 0, then (^∂g_∂x² = (2x^∗) has rank 1.

If x^∗= 0 but y^∗> 0, then (^∂g_∂y² = (2y^∗) has rank 1.

If x^∗> 0 and y^∗> 0, then (^∂g_∂x^{1 ∂g}_∂y¹) has rank 1. (1 point)Therefore it satisfies Kuhn-Tucker version NDCQ.

(c) ˜L(x, y, l1, l2) = x²+ 2y²− l1(x + 2y − 9) − l2(−x²− y²+ 16).(2 points).

(d) At the maximizer (x^∗, y^∗), there exist l1^∗≥ 0, l^∗2≥ 0 satisfy

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎨⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎩

∂ ˜L

∂x = 2x^∗− l^∗1+ 2x^∗l^∗₂≤ 0...(1)

∂ ˜L

∂y = 4y^∗− 2l1^∗+ 2y^∗l^∗₂≤ 0...(2) x^{∗ ∂ ˜}_∂x^L = 0 = x^∗⋅ (2x^∗− l1^∗+ 2x^∗l₂^∗)...(3) y^{∗ ∂ ˜}_∂y^L = 0 = y^∗⋅ (4y^∗− 2l1^∗+ 2y^∗l^∗₂)...(4)

∂ ˜L

∂λ₁ = −(x^∗+ 2y^∗− 9) ≥ 0...(5)

∂ ˜L

∂λ2 = (x^∗)²+ (y^∗)²− 16 ≥ 0...(6) l^∗_1∂λ^{∂ ˜L}

1 = 0 = l^∗1⋅ (x^∗+ 2y^∗− 9) = 0...(7) l^∗_2∂λ^{∂ ˜L}

2 = 0 = l^∗2⋅ [(x^∗)²+ (y^∗)²− 16] = 0...(8)

If you write down one correct equation, you get 0.5 point. If you write down all correct equations, you get 3 points.

(e) If(x^∗)²+ (y^∗)²= 16, then x^∗+ 2y^∗≠ 9.

(7) ⇒ l^∗1= 0.

(2) ⇒ 2y^∗(2 + l2^∗) ≤ 0. Since l^∗2≥ 0 and y^∗≥ 0, y^∗= 0. It implies that x^∗= 4.

(3) ⇒ l2^∗= −1. It is a contradiction. Therefore (x^∗)²+ (y^∗)²≥ 16 can not be binding. If you get one of correct x^∗, y^∗, l^∗₁, l^∗₂, you get one point.

(f) If x^∗+ 2y^∗= 9, then (x^∗)²+ (y^∗)²> 16.

(8) ⇒ l^∗2= 0.(1 point)

(3), (4) ⇒ { x^∗(2x^∗− l^∗1) = 0 y^∗(4y^∗− 2l^∗1) = 0.

If x^∗= 0, then y^∗= ⁹₂⇒ l^∗1= 9. It implies that (x^∗, y^∗, l^∗₁, l₂^∗) = (0,⁹₂, 9, 0).(1 point) If y^∗= 0, then x^∗= 9 ⇒ l^∗1= 18. It implies that ((x^∗, y^∗, l^∗₁, l₂^∗) = (9, 0, 18, 0).(1 point)

If x^∗ > 0, y^∗ > 0, then l1 = 2x^∗ = 2y^∗ ⇒ x^∗ = y^∗. Thus 3x^∗ = 9 ⇒ x^∗ = 3 ⇒ l^∗1 = 6. Thus ((x^∗, y^∗, l^∗₁, l^∗₂) = (3, 3, 6, 0).(2 points)

4. (22%) Consider a C² utility function U(x1, x2) such that _∂x^∂U₁ > 0 and _∂x^∂U₂ > 0. We want to maximize U(x1, x2) under constraints P1x1+ P2x2≤ I, x1≥ 0, and x2≥ 0, where P1> 0, P2> 0 are unit prices and I > 0 is the budget.

(a) (2%) Write down the usual version Lagrangian function.

(b) (2%) Show that the usual version NDCQ is satisfied.

(c) (3%) List the usual version first order conditions.

(d) (3%) From the first order conditions show that at the maximizer the constraint P1x1+ P2x2≤ I must be binding.

(e) (6%) Suppose that the first order conditions has a solution (x^∗1, x^∗₂) with x^∗1 > 0, x^∗2 > 0. Write down the bordered Hessian matrix at(x^∗1, x^∗₂). List the condition which guarantees that U(x^∗1, x^∗₂) is a local maximum value.

(f) (6%) Now P1, P2and I are parameters. Then the maximum value of U depends on P1, P2, and I, which is denoted by ˜U(P1, P2, I). Compute _∂P^{∂ ˜U}₁, _∂P^{∂ ˜U}

2, and ^{∂ ˜}_∂I^U. Determine the signs of these partial derivatives.

Solution:

(a) L(x1, x2, λ1, λ2, λ3) = U(x1, x2) − λ1(P1x1+ P2x2− I) + λ2x1+ λ3x2. (: λ2x1+ λ3x2 c λ2x1+ λ3x2 ë −λ2x1− λ3x2 c ) (b) Let g1(x1, x2) = P1x1+ P2x2, g2(x1, x2) = −x1, g3(x1, x2) = −x2.

Ð⇀∇g1= (P1, P₂) ≠ Ð⇀0 , Ð⇀∇g2= (−1, 0) ≠ Ð⇀0 , Ð⇀∇g3= (0, −1) ≠ Ð⇀0 . (1pt: —ú Ð∇g⇀ 1, Ð⇀∇g2, Ð∇g⇀ 3)

At the maximizer at most two of the constraints are binding and any two of {Ð⇀∇g1, Ð⇀∇g2, Ð∇g⇀ 3} are linearly independent. Hence the NDCQ is satisfied. (1pt)

(c) First Order conditions : At the maximizer(x^∗1, x^∗₂) there are λ^∗1, λ^∗₂, λ^∗₃ s.t

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎨⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎩

∂L

∂x₁ = _∂x^∂U₁− λ^∗1P1+ λ^∗2= 0⋯ 1O

∂L

∂x₂ = _∂x^∂U₂− λ^∗1P2+ λ^∗3= 0⋯ 2O λ^∗₁(P1x^∗₁+ P2x^∗₂− I) = 0⋯ 3O λ^∗₂x^∗₁= 0⋯ 4O

λ^∗₃x^∗₂= 0⋯ 5O P₁x^∗₁+ P2x^∗₂≤ I⋯ 6O

x^∗₁≥ 0, x^∗2≥ 0, λ^∗1≥ 0, λ^∗2≥ 0, λ^∗3≥ 0⋯ 7O 1pt 1O+ 2O

1pt 3O+ 4O+ 5O 1pt 6O+ 7O

(d) If P₁x^∗₁+P2x^∗₂< I, then 3O ⇒ λ^∗1= 0. (1pt) Thus 1O ⇒ _∂x^∂U_i(x^∗1, x^∗₂)+λ^∗2= 0 ⇒ _∂x^∂U_i(x^∗1, x^∗₂) = −λ^∗2≤ 0 contradiciton!

(2pts)

Hence P1x^∗₁+ P2x^∗₂= I.

(e) Since only the constraint P1x1+P2x2≤ I is binding at (x^∗1, x^∗₂), the bordered Hessian matrix is H =⎛

⎜⎝

0 P1 P2

P1 Lx₁x₁ Lx₁x₂

P2 Lx₂x₁ Lx₂x₂

⎞⎟

⎠ (1pt)

Because Lx₁x₁= Ux₁x₁, Lx₁x₂= Ux₁x₂, Lx₂x₂= Ux₂x₂, the bordered Hessian matrix is⎛

⎜⎝

0 P₁ P₂

P₁ U_x1x1 U_x1x2 P₂ U_x2x1 U_x2x2

⎞⎟

⎠(1pt) There are 2 variables(x1, x2) and one binding constraint.

We need to check the last(2 − 1) leading principal minor which is det⎛

⎜⎝

0 P₁ P₂

P₁ U_x1x1 U_x1x2 P2 Ux₂x₁ Ux₂x₂

⎞⎟

⎠. (1pt) det H= −P1(P1U_x2x2− P2U_x2x1) + P2(P1U_x2x1− P2U_x1x1)

= −P1²Ux₂x₂+ 2P1P2Ux₁x₂− P2²Ux₁x₁∣

(x^∗₁,x^∗₂). (2pts) If det⎛

⎜⎝

0 P1 P2

P1 Ux₁x₁ Ux₁x₂

P2 Ux2x1 Ux2x2

⎞⎟

⎠ RRRRR RRRRR RRR(x^∗₁,x^∗₂)

= −P1²Ux₂x₂+ 2P1P2Ux₁x₂− P2²Ux₁x₁ > 0, then U(x^∗1, x^∗₂) is a local maximum

value. (1pt)

(f) By the envelope theorem

∂ ˜U

∂P₁ = ∂L

∂P₁ = −λ^∗1x^∗₁< 0

∂ ˜U

∂P₂ = ∂L

∂P₂ = −λ^∗1x^∗₂< 0

∂ ˜U

∂I = λ^∗1> 0 (1pt for _∂P^{∂ ˜U}

1 = −λ^∗1x^∗₁, 1pt for the sign< 0, 1pt for _∂P^{∂ ˜U}₂ = −λ^∗1x^∗₂, 1pt for the sign< 0, 1pt for ^{∂ ˜}_∂I^U = λ^∗1, 1pt for the sign> 0.)

÷× (f) „TH. _∂P^{∂ ˜U}₁ ë −λ1x₁ (’  *)_∂P^{∂ ˜U}₂ = −λ1x₂(’  *)^{∂ ˜}_∂I^U = λ1(’  *) c

5. (18%) Suppose that in the following week you have 12 hours each day to study for the final exams of Calculus 4 and English. Let C be the number of hours per day spent studying for Calculsu 4 and E be the number of hours per day spent studying for English. Let your grade point average from these two courses be GP A= f(C, E) =²₃(√

C+√ 2E).

(a) (6%) Solve the optimization problem : Maximize f(C, E) under the constraint C + E = 12.

Use the bordered Hessian matrix to verify that the solution is indeed a local maximum.

(b) (6%) To assure that you obtain certain grades for Calculus 4 and English individually, you impose inequality con- straints C≥ 5 and E ≥ 4.

Solve the optimization problem : Maximize f(C, E) under constraints C + E = 12, C ≥ 5, and E ≥ 4.

(c) (6%) By the meaning of multipliers, estimate the maximum value of GP A when the constraints are C+ E = 12.5, C≥ 5.5 and E ≥ 4.1.

Solution:

(a) Define L(x, y, µ) = ²₃(√ C+√

2E) − µ(C + E − 12)

⎧⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎩

∂L

∂C =¹₃^√1_C− µ = 0

∂L

∂E =^√₃^2√1_E− µ = 0

∂L

∂µ = C + E − 12 = 0

⇒ C= _9µ¹2

E= _9µ²2

and C+ E = 12 ⇒ µ =¹₆, C= 4, E = 8.

(4pts for correct answers C= 4, E = 8, µ = ¹₆) The bordered Hessian matrix is ˆH=⎛

⎜⎜⎝

0 1 1

1 _∂C^∂2₂L _∂E∂C^∂2 L 1 _∂C∂E^∂2 L _∂E^∂2₂L

⎞⎟⎟

⎠ (1pt) At(C, E) = (4, 8)

Hˆ =⎛

⎜⎝

0 1 1

1 −₄₈¹ 0 1 0 −₉₆¹

⎞⎟

⎠.

det ˆH= ₉₆¹ +₄₈¹ > 0. (1pt) Hence f(C, E) is a local maximum.

(b) Define L(C, E, µ, λ1, λ2) = ²₃(√ C+√

2E) − µ(C + E − 12) − λ¹(−C + 5) − λ2(−E + 4) (1pt) constraints are

h(C, E) = C + E = 12 g1(C, E) = −C ≤ −5 g₂(C, E) = −E ≤ −4 Ð⇀∇h = (1, 1), Ð∇g⇀ 1= (−1, 0), Ð⇀∇g2= (0, −1). NDCQ is satisfied.

(2pts)

⎧⎪⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎪⎪

⎪⎪⎪⎪⎪

⎪⎩

∂L

∂C = ₃^√1_C− µ + λ1= 0⋯ 1O

∂L

∂E = ₃^√√2_E− µ + λ2= 0⋯ 2O λ1(−C + 5) = 0⋯ 3O λ2(−E + 4) = 0⋯ 4O C+ E = 12⋯ 5O

λ1≥ 0, λ2≥ 0, C ≥ 5, E ≥ 4.

We disscuss cases{λ1> 0, λ2> 0}, {λ1= 0, λ2> 0}, {λ1> 0, λ2= 0}, {λ1= λ2= 0}

(å4 cases q3: BúcºãF/’ Öhè„ cases Lc 1 2) 1. {λ1> 0, λ2> 0} : 3O 4O 5O contradiciton.

2. {λ1= 0, λ2> 0} : 4O 5O ⇒ E = 4, C = 8 O ⇒ µ =1 ₆^√1₂

O ⇒2 ^√₆²−₆^√1₂+ λ2= 0 ⇒ λ2< 0 contradiction.

3. {λ1> 0, λ2= 0} : 3O 5O ⇒ C = 5, E = 7, 2O ⇒ µ =¹₃√

2 7

O ⇒ λ1 1= µ −₃^√1₅ =¹₃√

2

7−₃^√1₅ > 0.

Ans : (C^∗, E^∗, µ^∗, λ^∗₁, λ^∗₂) = (5, 7,¹₃√

2 7,¹₃

√2

7−₃^√1₅, 0).

4. {λ1= 0, λ2= 0} : 1O 2O 5O ⇒ C = 4, E = 8 contradiction to the constraint C ≥ 5.

Hence the maximum value is f(5, 7) = ²₃(√ 5+√

14).

(c) The maximum value for new constraints can be approximated by

f(5, 7) + µ^∗× (12.5 − 12) + (−λ^∗1) × (5.5 − 5) + (−λ^∗2) × (4.1 − 4) (4pts)

= f(5, 7) + 0.5(µ^∗− λ^∗1) − 0.1λ^∗2= 2 3(√

5+√

14) + 1 6√

5 (2pts)

6. (12%) Suppose that f(x, y, z) is a C²function and P = (1, 2, −3) is a critical point of f which means that fx= fy= fz= 0 at P . If at point P , fxx= 1, fyy = 1, fzz = −3, fxy = −1, fxz = 3, and fyz = 3, determine whether f(1, 2, −3) is a local extreme value.

(a) (1%) Write down the Hessian matrix of f at P which is denoted by H.

(b) (3%) Is H positive definite, negative definite, or indefinite? Is f(1, 2, −3) a local maximum, local minimum, or a saddle point?

(c) (3%) Now we want to find the extreme value of f under the constraint x²+xy+xz−3z = 9. Write down the Lagrangian functionL(x, y, z, µ). Find µ^∗ such that _∂x^∂ L = _∂y^∂ L = _∂z^∂ L = 0 when (x, y, z, µ) = (1, 2, −3, µ^∗).

(d) (5%) On the constraint set x²+ xy + xz − 3z = 9, is f(1, 2, −3) a local maximum, local minimum, or a saddle point?

Solution:

(a) H=⎛

⎜⎝

1 −1 3

−1 1 3

3 3 −3

⎞⎟

⎠(1pt)

(b) The leading principal minors of H are H1= 1, H2= ∣1 −1

−1 1 ∣ =0, H3=RRRRR RRRRR RRR

1 −1 3

−1 1 3

3 3 −3

RRRRR RRRRR

RRR= −36 (2pts)

∵H1⋅ H3= −36 < 0 ∴ H is indefinite. f(P ) is a saddle point. (1pt) (c) L(x, y, z, µ) = f(x, y, z) − µ(x²+ xy + xz − 3z − 9) (1pt)

At(1, 2, −3),

∂L

∂x =∂f

∂x(1, 2, −3) − µ × 1 = −µ

∂L

∂y =∂f

∂y(1, 2, −3) − µ × 1 = −µ

∂L

∂z =∂f

∂z(1, 2, −3) − µ × (−2) = 2µ when µ= 0, ^∂L_∂x =^∂L_∂y = ^∂L_∂z = 0. Hence µ^∗= 0. (2pts)

(d) S∶ g(x, y, z) = 9 where g(x, y, z) = x²+ xy + xz − 3z, Ð∇g(1, 2, −3) = (1, 1, −2).⇀

At (x, y, z, µ) = (1, 2, −3, 0), Lxx = fxx = 1, Lxy = fxy = −1, Lxz = fxz = 3, Lyy = fyy = 1, Lyz = fyz = 3, Lzz= fzz= −3. (1pt) The bordered Hessian is

Hˆ =

⎛⎜⎜

⎜⎝

0 1 1 −2

1 1 −1 3

1 −1 1 3

−2 3 3 −3

⎞⎟⎟

⎟⎠ (1pt)

There are 3 variables and 1 constraint. Hence we need to check the last 3− 1 leading principal minors of ˆH.

Hˆ3=RRRRR RRRRR RRR

0 1 1

1 1 −1

1 −1 1 RRRRR RRRRR

RRR= −4 < 0 (1pt) Hˆ4=RRRRR

RRRRR RRRRR RRR

0 1 1 −2

1 1 −1 3

1 −1 1 3

−2 3 3 −3

RRRRR RRRRR RRRRR RRR

= −36 < 0 (1pt)

Hˆ₃ and ˆH₄ both have the same sign as (−1)¹. Thus H is positive definite on the constraint set. f(1, 2, −3) is a local minimum on S. (1pt)