1092 Calculus4 ME Final Exam

1092 Calculus4 ME Final Exam

June 19, 2021

1. Let V = span⎧⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎩

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 3

−4−5

−1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 4

−2−2 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1 3 2 3

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

−2 2

−4 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

⎫⎪⎪⎪⎪⎪

⎬⎪⎪⎪⎪⎪⎭and W = span⎧⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎩

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1 1

−1−1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1

−2−2 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1 1 1 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

−1 1 2 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

⎫⎪⎪⎪⎪⎪

⎬⎪⎪⎪⎪⎪⎭. (a) (5 pts) Find the dimension of the vector subspace V .

(b) (5 pts) Find a basis for W . (Hint: a basis is a linearly independent set of vectors that span W ) (c) (2 pts) Show that V and W are not equal.

Solution:

(a) The dimension is the rank of the matrix

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

3 −4 −5 −1

4 −2 −2 1

1 3 2 3

−2 2 −4 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

. We find the rank of the matrix via Gaussian elimination.

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

3 −4 −5 −1

4 −2 −2 1

1 3 2 3

−2 2 −4 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

1 3 2 3

−2 2 −4 1 3 −4 −5 −1

4 −2 −2 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

1 3 2 3

0 8 0 7

0 −13 −11 −10 0 −14 −10 −11

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎢⎢

⎢⎣

1 3 2 3

0 1 0 7

8 0 0 −11 11

8 0 0 −10 10

8

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎥⎥

⎥⎦

Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎢⎣

1 3 2 3

0 1 0 7

8 0 0 1 −1 0 0 0 80

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎥⎦

The dimension of V is 3.

Alternative methods for (a):

The determinant can be used to determine whether the matrix is full rank.

det

⎡⎢⎢⎢

⎢⎢⎢⎢

3 −4 −5 −1

4 −2 −2 1

1 3 2 3

⎤⎥⎥⎥

⎥⎥⎥⎥= 3 det⎡⎢

⎢⎢⎢⎢

−2 −2 1 3 2 3⎤⎥

⎥⎥⎥⎥+4 det⎡⎢

⎢⎢⎢⎢

4 −2 1 1 2 3⎤⎥

⎥⎥⎥⎥−5 det⎡⎢

⎢⎢⎢⎢

4 −2 1 1 3 3⎤⎥

⎥⎥⎥⎥+det⎡⎢

⎢⎢⎢⎢

4 −2 −2 1 3 2⎤⎥

⎥⎥⎥⎥

They can also find the number of non-zero eigenvalues via the characteristic polynomial, which is λ⁴− 4λ³+ 34λ²− 11λ. So the dimension of V is 3.

Here we also provide the Gaussian elimination without reordering if graders need to check for mistakes.

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

3 −4 −5 −1

4 −2 −2 1

1 3 2 3

−2 2 −4 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣ 1 −4

3

−5 3

−1 0 10 14 37

0 13 11 10

0 −2 −22 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣ 1 −4

3

−5 3

−1 0 1 1.4 0.73

0 0 −72 9

0 0 −96 12

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎢⎣

1 −4 3

−5 3

−1 3 0 1 1.4 0.7

0 0 1 −1

0 0 0 80

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎥⎦

And transposed.

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

3 4 1 −2

−4 −2 3 2

−5 −2 2 −4

−1 1 3 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣ 1 4

3 1 3

−2 0 10 13 −23 0 14 11 −22

0 7 10 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣ 1 4

3 1 3

−2 0 1 1.3 −0.23 0 0 −72 −192

0 0 9 24

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎢⎣

1 4 3

1 3

−2 0 1 1.3 −0.23

0 0 1 8

0 0 0 30

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎥⎦

(b) We find a basis of W via Gaussian elimination.

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

1 1 −1 −1

1 −2 −2 1

1 1 1 1

−1 1 2 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

1 1 −1 −1

0 −3 −1 2

0 0 2 2

0 2 1 −1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎢⎣

1 1 −1 −1

0 1 1

3

−2 3

0 0 2 2

0 0 1

3 1 3

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎥⎦

Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣

1 1 −1 −1

0 1 1

3

−2 0 0 1 31

0 0 0 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦ The vectors

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1 1

−1−1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 0 3 1

−2

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 0 0 1 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

form a basis of W . Because we didn’t switch the rows,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1 1

−1−1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1

−2−2 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1 1 1 1

⎤⎥⎥⎥

⎥⎥⎥⎥

also form a basis of W . ⎦

Alternative methods for (b):

If we start by checking whether the four vectors are linearly independent by definition, we would be solving a system of equations.

⎧⎪⎪⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎪⎪

⎩

x+ y + z − w = 0 x− 2y + z + w = 0

−x − 2y + z + 2w = 0

−x + y + z = 0

w=x+y+z

ÐÐÐÐ→⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

2x− y + 2z = 0 x+ 3z = 0

−x + y + z = 0

x=y+z

ÐÐÐ→⎧⎪⎪

⎨⎪⎪⎩

y+ 4z = 0 y+ 4z = 0

So x= 3, y = 4, z = −1, w = 6 is a nonzero solution. The four vectors are linearly dependent.

We can then check if three of them would be linearly independent.

⎧⎪⎪⎪⎪⎪⎪⎪

⎨⎪⎪⎪⎪⎪⎪⎪

⎩

x+ y + z = 0 x− 2y + z = 0

−x − 2y + z = 0

−x + y + z = 0

x=y+z

ÐÐÐ→⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

2y+ 2z = 0

−y + 2z = 0

−3y = 0

Page 2 of 13

Hence x = 0, y = 0, z = 0 is the only solution. The three vectors

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1 1

−1−1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1

−2−2 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ ,

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1 1 1 1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

are linearly

independent and form a basis of W .

(c) From the results of the Gaussian elimination we can see that the vector (0, 0, 1, 1) is in W but not in V .

Alternative methods for (c):

Since both vector subspaces are 3-dimensional, we can check whether they have the same

eigenvectors for λ = 0. The vector

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

−5−7 1 8

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

is an eigenvector for V and the vector

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣ 1

−1 1

−1

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ is an

eigenvector for W , not parallel.

We can also add a vector from V to W and show that the vector subspace would become 4-dimensional.

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

1 1 −1 −1

1 −2 −2 1

1 1 1 1

1 3 2 3

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

1 1 −1 −1

0 −3 −1 2

0 0 2 2

0 2 3 4

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

1 1 −1 −1

0 −3 −1 2

0 0 1 1

0 0 7 14

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦ Ð→

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

1 1 −1 −1

0 −3 −1 2

0 0 1 1

0 0 0 7

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

Grading:

(a) No points if the method is wrong. 2 points if they only showed that dim(V ) ≠ 4 (by determinant). 1 point off for each minor mistake (algebra, miscopy). 2 points off for each major mistake (inventing row operations). Do not take more points off for wrong answer if it is a result of mistakes.

(b) 3 points for finding a basis. 2 points for whether their work showed linearly independence.

Similar to (a), the work is more important than the answer.

(c) Depends on work shown in (a) and (b). Full credit or no credit.

Note:

For (a) and (b) if a minor mistake does not affect the answers, the grader may choose to take 0.5 off instead.

Because of the follow-through rule, they are allowed to say V and W are not equal due to dimension if they made a mistake in (a) or (b).

2. Let A=⎡⎢

⎢⎢⎢⎢

⎣

−1 3 1 3 −1 1

1 1 1

⎤⎥⎥⎥

⎥⎥⎦ .

(a) (8 pts) Find the eigenvalues of A and corresponding eigenvectors given that det(A − λI3) =

−λ³− λ²+ 12λ.

(b) (4 pts) Diagonalize A. That is, find an orthogonal matrix P (i.e. P^TP = I3) and a diagonal matrix D such that P^TAP = D.

(c) (4 pts) Determine whether A+ 5I3 is positive definite, negative definite, or indefinite.

(d) (4 pts) Determine whether A− 5I3 is positive definite, negative definite, or indefinite.

Solution:

(a) det(A − λI3) = −λ³− λ²+ 12λ = −λ(λ²+ λ − 12) = −λ(λ − 3)(λ + 4). Eigenvalues are 0, 3, −4.

The eigenvectors corresponding to each:

λ= 0 Ð→ A − λI3 =⎡⎢

⎢⎢⎢⎢

⎣

−1 3 1 3 −1 1

1 1 1

⎤⎥⎥⎥

⎥⎥⎦ Ð→⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

−x + 3y + z = 0 3x− y + z = 0 x+ y + z = 0

x=3y+z

ÐÐÐ→⎧⎪⎪

⎨⎪⎪⎩

8y+ 4z = 0

4y+ 2z = 0 Ð→⎡⎢

⎢⎢⎢⎢

⎣ x y z

⎤⎥⎥⎥

⎥⎥⎦

=⎡⎢

⎢⎢⎢⎢

⎣ 1 1

−2

⎤⎥⎥⎥

⎥⎥⎦ t

λ= 3 Ð→ A−λI3 =⎡⎢

⎢⎢⎢⎢

⎣

−4 3 1

3 −4 1

1 1 −2

⎤⎥⎥⎥

⎥⎥⎦ Ð→⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

−4x + 3y + z = 0 3x− 4y + z = 0 x+ y − 2z = 0

x=−y+2z

ÐÐÐÐ→⎧⎪⎪

⎨⎪⎪⎩

7y− 7z = 0

−7y + 7z = 0 Ð→⎡⎢

⎢⎢⎢⎢

⎣ x y z

⎤⎥⎥⎥

⎥⎥⎦

=⎡⎢

⎢⎢⎢⎢

⎣ 1 1 1

⎤⎥⎥⎥

⎥⎥⎦ t

λ= −4 Ð→ A − λI3=⎡⎢

⎢⎢⎢⎢

⎣

3 3 1 3 3 1 1 1 5

⎤⎥⎥⎥

⎥⎥⎦ Ð→⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

3x+ 3y + z = 0 3x+ 3y + z = 0 x+ y + 5z = 0

z=−3x−3y

ÐÐÐÐÐ→ {−2x − 2y = 0 Ð→⎡⎢

⎢⎢⎢⎢

⎣ x y z

⎤⎥⎥⎥

⎥⎥⎦

=⎡⎢

⎢⎢⎢⎢

⎣ 1

−1 0

⎤⎥⎥⎥

⎥⎥⎦ t

(b) Hence if P =

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎢⎢

⎢⎣

√1 6

√1 3

√1 2

√1 6

√1 3

√−1 2

√−2 6

√1

3 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎥⎥

⎥⎦

and D = ⎡⎢

⎢⎢⎢⎢

⎣

0 0 0

0 3 0

0 0 −4

⎤⎥⎥⎥

⎥⎥⎦

, then P is orthogonal and P^TAP =

D.

(c) A+5I3 =⎡⎢

⎢⎢⎢⎢

⎣

4 3 1 3 4 1 1 1 6

⎤⎥⎥⎥

⎥⎥⎦

is Positive Definite. The eigenvalues of A+5I3 are 1, 5, 8, all positive.

They can also convert to v^T(A + 5I3)v = 4x²+ 4y² + 6z²+ 6xy + 2xz + 2yz and complete the squares:

4x²+ 4y²+ 6z²+ 6xy + 2xz + 2yz = 3(x + y)²+ (x + z)²+ (y + z)²+ 4z²≥ 0

Or use Sylvester’s criterion for the matrix A+ 5I3=⎡⎢

⎢⎢⎢⎢

⎣

4 3 1 3 4 1 1 1 6

⎤⎥⎥⎥

⎥⎥⎦ .

detA₁= 4 > 0, detA2 = 16 − 9 = 7 > 0, detA3 = 96 + 3 + 3 − 4 − 4 − 54 = 40 > 0.

(d) A− 5I3 =⎡⎢

⎢⎢⎢⎢

⎣

−6 3 1

3 −6 1

1 1 −4

⎤⎥⎥⎥

⎥⎥⎦

is Negative Definite. The eigenvalues of A− 5I3 are −9, −5, −2, all negative.

They can also convert to v^T(A − 5I3)v = −6x²− 6y²− 4z²+ 6xy + 2xz + 2yz and complete the squares:

−6x²− 6y²− 4z²+ 6xy + 2xz + 2yz = −3(x + y)²− (x + z)²− (y + z)²− 2x²− 2y²− 2z² ≤ 0

Or use Sylvester’s criterion for the matrix A− 5I3=⎡⎢

⎢⎢⎢⎢

⎣

−6 3 1

3 −6 1

1 1 −4

⎤⎥⎥⎥

⎥⎥⎦ .

detA₁= −6 < 0, detA2= 36 − 9 = 27 > 0, detA3= −144 + 3 + 3 + 6 + 6 + 36 = −90 > 0.

Grading:

(a) 2 points for eigenvalues. 2 points for each corresponding eigenvector.

(b) 3 points for P and 1 point for D.

(c) 1 point for the correct conclusion. 3 points for justifying their result.

(d) 1 point for the correct conclusion. 3 points for justifying their result.

In general, 1 point off for each minor mistake (algebra, miscopy). 2 points off for each major mistake (inventing new math).

If they get a zero vector as eigenvector, then they get no points for the eigenvector AND lose points for (b).

3. Maximize f(x, y, z) = yz subject to x + z = 1, x²+ y² ≤ 6, z ≥ 0.

(a) (4 pts) Check whether the NDCQ is satisfied.

(b) (8 pts) Write out the Lagrangian function and the first order conditions.

(c) (8 pts) Solve the constrained optimization problem given that the constraints form a closed and bounded region.

Solution:

(a) The equality constraint is h(x, y, z) = x + z = 1. The inequality constraints are

g₁(x, y, z) = x² + y² ≤ 6 and g2(x, y, z) = −z ≤ 0. ∇h = (1, 0, 1), ∇g1(x, y, z) = (2x, 2y, 0), and

∇g2(x, y, z) = (0, 0, −1) (2 pt)

If both g₁ and g₂ are binding, then z = 0, x = 1, and y = ±√

5. In this case, ∇h = (1, 0, 1),

∇g1= (2, ±2√

5, 0) and ∇g2= (0, 0, −1) are linearly independent. (0.5 pt)

If only g1 is binding, then x² + y² = 6 and ∇g1(x, y, z) = (2x, 2y, 0) ≠ (0, 0, 0). In this case,

∇h = (1, 0, 1) and ∇g1 = (2x, 2y, 0) are linearly independent. (0.5 pt)

If only g₂ is binding, then z = 0, x = 1. In this case, ∇h = (1, 0, 1) and ∇g2 = (0, 0, −1) are linearly independent. (0.5 pt)

If neither g1 nor g2 are binding, then we just need to check∇h = (1, 0, 1) which is not (0, 0, 0).

(0.5 pt)

The above discussion shows that on the constraint set, ∇h and gradient(s) of binding inequality constraint(s) are linearly independent. Hence the NDCQ is satisfied.

(b) Lagrangian function L(x, y, z, µ, λ1, λ2) = yz − µ(x + z − 1) − λ1(x²+ y²− 6) + λ2z. (1 pt) First order conditions:

L_x= −µ − 2xλ1= 0 (1pt) L_y= z − 2yλ1= 0 (1pt) L_z= y − µ + λ2= 0 (1pt) λ₁L_λ1 = λ1(6 − x²− y²) = 0 (1pt)

λ₂z= 0 (1pt)

1− x − z = 0, x²+ y²≤ 6, z ≥ 0 (1pt) λ1 ≥ 0, λ2≥ 0 (1pt)

(c) Since f(x, y, z) = yz is zero when z = 0, we can assume z > 0 and λ2 = 0.

(1 pt for ruling out the case λ₂> 0.)

Now we have µ+ 2xλ1= 0, 2yλ1 = z, y = µ, x + z = 1, λ1(6 − x²− y²) = 0.

If λ₁= 0, then x = 1, y = 0, z = 0, µ = 0 and f(1, 0, 0) = 0. (1 pt for the case λ1= 0.)

Page 6 of 13

If λ₁ ≠ 0, then x² + y² = 6. Replace µ and z to get 2xλ1 + y = 0, 2yλ1 = 1 − x. Hence 2xyλ₁ = −y²= x − x². So we get 2x²− x − 6 = (2x + 3)(x − 2) = 0.

If x= 2, then z = −1(not valid). If x = −3

2 , then z= 5

2. Then we get y = µ =

√15

2 , λ₁ = 5 2√

15. (If y is negative then λ₁< 0.) The solution satisfies all our conditions.

(5 pts for the case λ₁> 0 and the complete solution (x^∗, y^∗, z^∗) and µ^∗, λ₁. ) The maximum of f(x, y, z) = yz is 5√

15

4 at (−3 2 ,

√15 2 ,5

2). (1 pt)

4. Suppose that you keep t hours a day as leisure time and 16− t hours to tutor with wage 400 dollars per hour. Your daily budget is 200+400(16−t) and you spend money on food and clothes with prices 250 and 350, respectively, per unit. If you consume x units of food and y units of clothes, then your utility function U(x, y, t) depends on x, y and hours of leisure time t, where ∂U

∂x > 0, ∂U

∂y > 0, and

∂U

∂t > 0. Now you want to maximize U(x, y, t) under the constraints 250x+350y ≤ 200+400(16−t), t≤ 16, t ≥ 0, x ≥ 0, y ≥ 0.

(a) (8 pts) Write down the Kuhn-Tucker Lagrangian function, ˜L(x, y, t, λ1, λ₂), and the first order conditions in the Kuhn-Tucker formulation.

(b) (4 pts) Show that if(x^∗, y^∗, t^∗) is a maximizer, then the constraint 250x+350y ≤ 200+400(16−t) is binding at (x^∗, y^∗, t^∗).

(c) (6 pts) Show that if(x^∗, y^∗, t^∗) is a maximizer satisfying x^∗> 0, y^∗ > 0, and 0 < t^∗ < 16, then

∂U

∂t(x^∗, y^∗, t^∗) 1

400 =∂U

∂x(x^∗, y^∗, t^∗) 1

250 = ∂U

∂y(x^∗, y^∗, t^∗) 1 350.

Solution:

(a) ˜L(x, y, t, λ1, λ₂) = U(x, y, t) − λ1(250x + 350y − 200 − 400(16 − t)) − λ2(t − 16) (1 pt) At the maximizer (x^∗, y^∗, t^∗), there are λ^∗1 and λ^∗₂ s.t.

1. ∂ ˜L

∂x = ∂U

∂x(x^∗, y^∗, t^∗) − 250λ^∗1 ≤ 0

∂ ˜L

∂y = ∂U

∂y(x^∗, y^∗, t^∗) − 350λ^∗1 ≤ 0

∂ ˜L

∂t = ∂U

∂t(x^∗, y^∗, t^∗) − 400λ^∗1− λ^∗2 ≤ 0 (2 pts) 2. x^∗∂ ˜L

∂x = x^∗(∂U

∂x(x^∗1, x^∗₂, t^∗) − 250λ^∗1) = 0 y^∗∂ ˜L

∂y = y^∗(∂U

∂y(x^∗, y^∗, t^∗) − 350λ^∗1) = 0 t^∗∂ ˜L

∂t = t^∗(∂U

∂t(x^∗, y^∗, t^∗) − 400λ^∗1 − λ^∗2) = 0 (2 pts) 3. ∂ ˜L

∂λ1 = −250x^∗− 350y^∗+ 200 + 400(16 − t^∗) ≥ 0, ∂ ˜L

∂λ2 = −t^∗+ 16 ≥ 0 (1 pt) 4. λ^∗₁ ∂ ˜L

∂λ₁ = λ^∗1(−250x^∗− 350y^∗+ 200 + 400(16 − t^∗)) = 0, λ^∗2

∂ ˜L

∂λ₂ = λ^∗2(−t^∗+ 16) = 0 (1 pt) 5. λ^∗₁ ≥ 0 and λ^∗2 ≥ 0 (1 pt)

(Students get full credits about the FOC with (x, y, t), λ1, λ₂ instead of (x^∗, y^∗, t^∗), λ^∗1, λ^∗₂.) (b) If 250x^∗+ 350y^∗< 200 + 400(16 − t^∗), then λ^∗1 = 0 (1 pt)

However, this implies that ∂U

∂x(x^∗, y^∗, t^∗) ≤ 0 and ∂U

∂y(x^∗, y^∗, t^∗) ≤ 0 . (2 pts) This contradicts the assumptions that ∂U

∂x, ∂U

∂x > 0. (1 pt) Hence we must have 250x^∗+ 350y^∗ = 200 + 400(16 − t^∗).

(c) Since t^∗< 16, we have λ^∗2 = 0 (1 pt).

Because x^∗ > 0, y^∗> 0, we derive ∂U

∂x(x^∗, y^∗, t^∗) − 250λ^∗1 = 0 and ∂U

∂y(x^∗, y^∗, t^∗) − 350λ^∗1 = 0.

Page 8 of 13

(2 pts)

Also, 0< t^∗ implies that ∂U

∂t(x^∗, y^∗, t^∗) − 400λ^∗1− λ^∗2 = ∂U

∂t(x^∗, y^∗, t^∗) − 400λ^∗1 = 0 (1 pt) Thus at the maximizer (x^∗, y^∗, t^∗), ∂U

∂x 1

250 = ∂U

∂y 1

350 = ∂U

∂t 1

400 = λ^∗1 (2 pts)

5. Consider the problem of maximizing f(x, y, z) = xyz subject to 2x + y + z = 18, and x + 2y + z = 18.

(a) (1 pts) Write down the Lagrangian function for this problem, L(x, y, z, µ1, µ₂), where µ1 and µ₂ are the Lagrange multipliers.

Solution:

Grading scheme: 1 pt for the correct answer no partial credit for this part.

Solution: The Lagrangian function is

L(x, y, z, µ1, µ2) = xyz − µ1(2x + y + z − 18) − µ2(x + 2y + z − 18).

(b) (2 pts) Check whether the NDCQ is satisfied.

Solution:

Grading scheme: 0.75 pt for computing ∇h1, 0.75 pt for computing∇h2, 0.5 pt for stating the rank is 2.

Solution: We have two constraint equalities h₁(x, y, z) = 2x + y + z = 18 and h2(x, y, z) = x+ 2y + z = 18. [∇h₁

∇h2] = [2 1 1

1 2 1]. The rank of [2 1 1

1 2 1] iis 2. So NDCQ is satisfied.

(c) (4 pts) Write down the first order conditions for this problem.

Solution:

Grading scheme: 0.8 pt for each part Solution: We have to compute

∂L

∂x = yz − 2µ1− µ2= 0

∂L

∂y = xz − µ1− 2µ2= 0

∂L

∂z = xy − µ1− µ2= 0

∂L

∂µ₁ = −(2x + y + z − 18) = 0

∂L

∂µ₂ = −(x + 2y + z − 18) = 0

(d) (7 pts) Show that the solution of the first order conditions are(x, y, z, µ1, µ2) = (4, 4, 6, 8, 8) or (x, y, z, µ1, µ₂) = (0, 0, 18, 0, 0). (You have to show your steps to get complete credits).

Solution:

Grading scheme: There are two solution. Each solution 3.5 point. Given partial credit accordingly.

Solution: We have xyz− (2µ1+ µ2)x = xyz − (µ1+ 2µ2)y = xyz − (µ1+ µ2)z = 0.

Using 2x+ y + z = 18 and x + 2y + z = 18, we add previous three equations to get 3xyz= µ1(2x + y + z) + µ2(x + 2y + z) = 18(µ1+ µ2).

From xyz− (µ1+ µ2)z = 0 and xyz = 6(µ1+ µ2) we have z = 6 if µ1+ µ2≠ 0. Now 2x + y = 12 and x+ 2y = 12. Then x = 4 and y = 4. Then 2µ1+ µ2 = 24 and µ1+ µ2 = 16. Then µ1 = 8 and µ2= 8. So (x, y, z, µ1, µ2) = (4, 4, 6, 8, 8).

Page 10 of 13

If µ₁+µ2 = 0 then xyz = 0. Hence x = 0, y = 0 or z = 0. If x = 0 then µ1+2µ2 = 0, µ1+µ2= 0.

This gives µ₁ = µ2 = 0. If x = 0 then y + z = 18 and 2y + z = 18. Thus y = 0 z = 18 Thus (x, y, z, µ1, µ₂) = (0, 0, 18, 0, 0).

If y = 0 then µ1 = µ2 = 0, x = 0, z = 18. Thus (x, y, z, µ1, µ₂) = (0, 0, 18, 0, 0) If z = 0 then µ₁= µ2= 0, y = 6, x = 6. But this doesn’t satisfy xy = 0.

(e) (7 pts) Check the second order conditions at(x, y, z, µ1, µ₂) = (4, 4, 6, 8, 8) and (x, y, z, µ1, µ₂) = (0, 0, 18, 0, 0). Show that (4, 4, 6) a local maximizer and (0, 0, 18) is a local minimizer.

Solution:

Grading scheme: There are two solution. Each solution 3.5 point. Given partial credit accordingly.

Solution:

The constraint equations are h1(x, y.z) = 2x + y + z = 18 and h2(x, y.z) = x + 2y + z = 18

∇h1= (2, 1, 1) and ∇h2 = (1, 2, 1). rank [∇h₁

∇h2] = rank [2 1 1

1 2 1] =2. So NDCQ holds.

Recall

∂L

∂x = yz − 2µ1− µ2 = 0

∂L

∂y = xz − µ1− 2µ2= 0

∂L

∂z = xy − µ1− µ2 = 0 We have

∂²L

∂x² = 0, ∂²L

∂x∂y = z, ∂²L

∂x∂z = y

∂²L

∂y² = 0, ∂²L

∂y∂z = x, ∂²L

∂z² = 0.

So the bordered hessian matrix of Lagrangian is H =

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣

0 0 2 1 1 0 0 1 2 1 2 1 0 z y 1 2 z 0 x 1 1 y x 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦ At (x, y, z, µ1, µ₂) = (4, 4, 6, 8, 8), we have H =

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣

0 0 2 1 1 0 0 1 2 1 2 1 0 6 4 1 2 6 0 4 1 1 4 4 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦ Now n= 3 k = 2. 2k + 1 = 5 n + k = 5.

⎡⎢⎢⎢

⎢⎢

0 0 2 1 1 0 0 1 2 1⎤⎥

⎥⎥⎥⎥

⎡⎢⎢⎢

⎢⎢

0 0 2 1 1

0 0 1 2 1⎤⎥

⎥⎥⎥⎥

[−3 −1

6 14] = −36< 0

(−1)^k= 1 (−1)ⁿ= −1. So it is negative definite and it is a local maximizer.

At (x, y, z, µ1, µ₂) = (0, 0, 18, 0, 0), we have H =

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣

0 0 2 1 1

0 0 1 2 1

2 1 0 18 0

1 2 18 0 0

1 1 0 0 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦ detH = det

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣

0 0 2 1 1

0 0 1 2 1

2 1 0 18 0

1 2 18 0 0

1 1 0 0 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦

= det

⎡⎢⎢⎢

⎢⎢⎢⎢

⎢⎢⎣

0 0 2 1 1

0 0 1 2 1

0 −1 0 18 0

0 1 18 0 0

1 1 0 0 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎥⎥⎦

= det

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

0 2 1 1

0 1 2 1

−1 0 18 0

1 18 0 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

= det

⎡⎢⎢⎢

⎢⎢⎢⎢

⎣

0 2 1 1

0 1 2 1

0 18 18 0 1 18 0 0

⎤⎥⎥⎥

⎥⎥⎥⎥

⎦

= (−1)det⎡⎢

⎢⎢⎢⎢

⎣

2 1 1

1 2 1

18 18 0

⎤⎥⎥⎥

⎥⎥⎦

= (−18)det⎡⎢

⎢⎢⎢⎢

⎣

2 1 1 1 2 1 1 1 0

⎤⎥⎥⎥

⎥⎥⎦

= (−18)det⎡⎢

⎢⎢⎢⎢

⎣

0 −1 1

0 1 1

1 1 0

⎤⎥⎥⎥

⎥⎥⎦

= (−18)det [−1 1 1 1] =36.

Now k= 2. (−1)^k= 1 > 0. So it is positive definite and it is a local minimizer.

(f) (1 pt) Does f(x, y, z) = xyz have a global maximum or global minimum subject to 2x+y+z = 18, and x+ 2y + z = 18?

Solution:

Grading scheme: 0.3 point for finding the parametric equation of the line. 0.3 pt for finding f in one variable. 0.4 point to explain f goes to ∞ as the point goes to ∞.

Solution: The constraint set 2x+ y + z = 18, and x + 2y + z = 18 is a line. First, we have 2x+ y + z − 2(x + 2y + z) = 18 − 36, i.e. −3y − z = −18. Hence z = −3y + 18, x = 18 − 2y − z = 18− 2y + 3y − 18 = y. Thus the constraint set can be expresses ad x = y, z = −3y + 18.

f(x, y, z) = y ⋅y ⋅(−3y +18) = −3y³+18y² on the constraint set. lim_y→∞f = −∞ and lim_y→−∞f= ∞.

So f doesn’t have a global maximum and a global minimum.

(g) (4 pts) Estimate the value of the local maximum of the following function f(x, y, z) = xyz subject to 2x+ y + z = 18.1, and x + 2y + z = 18.2.

Solution:

Grading scheme: One point for setting up the right problem.

One point for pointing ∂f(x^∗(a1, a₂), y^∗(a1, a₂), z^∗(a1, a₂))

∂a₁ = µ^∗1(a1, a₂) and one point for pointing ∂f(x^∗(a1, a2), y^∗(a1, a2), z^∗(a1, a2))

∂a₂ = µ^∗2(a1, a₂). One point for getting the right answer.

Solution: (x^∗(a1, a₂), y^∗(a1, a₂), z^∗(a1, a₂), µ^∗1(a1, a₂), µ^∗2(a1, a₂)) be the local maximizer and the multipliers of the following function.

Let f(x, y, z, a) = xyz subject to h1(x, y, z) = 2x + y + z = 18 + a1 and x+ 2y + z = 18 + a2. We have

(x^∗(0, 0), y^∗(0, 0), z^∗(0, 0), µ^∗1(0, 0), µ^∗2(0, 0)) = (4, 4, 6, 8, 8)

Page 12 of 13

We have

∂f(x^∗(a1, a₂), y^∗(a1, a₂), z^∗(a1, a₂))

∂a₁ = µ^∗1(a1, a₂) and

∂f(x^∗(a1, a2), y^∗(a1, a2), z^∗(a1, a2))

∂a₂ = µ^∗2(a1, a₂).

Thus

∂f(x^∗(a1, a₂), y^∗(a1, a₂), z^∗(a1, a₂))

∂a₁ ∣

(0,0)= µ^∗1(0, 0) = 8 and

∂f(x^∗(a1, a₂), y^∗(a1, a₂), z^∗(a1, a₂))

∂a₂ = µ^∗2(0, 0) = 8.

Thus

f(x^∗(0.1, 0.2), y^∗(0.1, 0.2), z^∗(0.1, 0.2))

≈ f(x^∗(0, 0), y^∗(0, 0), z^∗(0, 0) + µ^∗1(0, 0) ⋅ 0.1 + µ^∗2(0, 0) ⋅ 0.2

= 96 + 0.8 + 1.6 = 98.4

(h) (4 pts) Estimate the value of the local maximum of the following function f(x, y, z) = xyz+0.1x subject to 2x+ y + 1.1z = 18, and x + 2y + z = 18.1.

Solution:

Grading scheme: One point for setting up the right problem. One point for getting the Lagrangian L(x, y, z, µ1, µ₂, a) = xyz + ax − µ1(2x + y + (1 + a)z − 18) − µ2(x + 2y + z − 18 − a) One points for computing the right ∂L

∂a. One point for getting the right answer.

Solution: Let f(x, y, z, a) = xyz+ax, h1(x, y, z, a) = 2x+y+(1+a)z = 18 and x+2y+z = 18+a.

Let (x^∗(a), y^∗(a), z^∗(a).µ^∗1(a), µ^∗2) be the local maximizer and the multipliers of the fol- lowing function Let f(x, y, z, a) = xyz + ax subject to h1(x, y, z, a) = 2x + y + (1 + a)z = 18 and x+ 2y + z = 18 + a.

We know that (x^∗(0), y^∗(0), z^∗(0), µ^∗1(0), µ^∗2(0)) = (4, 4, 6, 8, 8).

We also have df((x^∗(a), y^∗(a), z^∗(a), a)

da = ∂L

∂a(x^∗(a), y^∗(a), z^∗(a), µ^∗1(a), µ^∗2(a), a) Note that

L(x, y, z, µ1, µ₂, a) = xyz + ax − µ1(2x + y + (1 + a)z − 18) − µ2(x + 2y + z − 18 − a).

Then ∂L

∂a = x − µ1z+ µ2. So df((x^∗(a), y^∗(a), z^∗(a), a)

da ∣a=0= x^∗(0) − µ^∗1(0)z^∗(0) + µ^∗2(0) = 4 − 8 ⋅ 6 + 8 = −36.

Then

f(x^∗(0.1), y^∗(0.1), z^∗(0.1), 0.1)