How could I have prepared for this test?

As written in the Term Test 2 instructions, we suggested you complete all lecture worksheets, solve textbook exercises, review the problem sets, review the readings, and memorize definitions. Please read these comments carefully and the detailed remarks in correspond solutions. Use them to reflect on the feedback from your test.

Q1) No part marks were awarded for any part.

(1a) was similar to worksheet C6 Q1.

(1b) was similar to worksheet C8 Q4 and C9 Q1.

(1c) was a basic definition and was similar to worksheet C8 Q1 and C9 Q1.

(1d) was similar to worksheet C9 Q4, and Problem Set 4 Q4. Practice with optimization problems in C9, D6, and D7 would also help.

Q2) No part marks were awarded for any part.

(2a) was similar to worksheet D6 Q4.

(2b) was similar to worksheet D6 Q5 and D7 Q2. Practice with optimization problems in D6 and D7 would also help.

(2c) was similar to worksheet D1 Q3 and D2 Q2.

Q3) Part marks were awarded if only 1 error was made.

(3a) was similar to D4 Q3 and D4 Q9.2.

(3b) was similar to D3 Q4.

(3c) was similar to D4 Q8 and D5 Q2.

Q4) This was a routine optimization problem. It is similar to many such problems, notably C9 Q5 and C9 Q6. Q5) This was similar to E4 Q6 and Q10.1.

Q6) This was similar to E1 Q6. No part marks were awarded except for (6e) and (6f), where you received partial

credit if your answers were incorrect but consistent with Clairaut’s theorem.

Q7) This was similar to E3 Q2 and Q3.

Q8) This was based on two of the three equivalent definitions of Taylor polynomials which can be studied in E3

and E5. Related problems include E3 Q7 and E5 Q4. As always, writing quality matters.

We informed you that the term test will be roughly 40-60% based on Module E. Questions Q5, Q6, Q7, and Q8 are all based on Module E; their combined points constitute 52% of the total score.

1.  (4 points) The parts of this question are unrelated. No justification is necessary for any part.

Fill in EXACTLY ONE circle. (unfilled filled )

(1a) Let F : Rn - Rm be a C 1 map. Fix a ∈ Rn .

Which statement is FALSE?

There exists a linear map L : Rn - Rm such that h(l)im-0 = 0.

The limit lim F (a + h) - F (a) exists.

The Jacobian of F at a exists and is an m x n matrix.

For v ∈ Rn , Dv F (a) = dFa (v).

None of the above statements are false.

(1b) Let A C Rn be a set and let f : A - R be differentiable on the interior of A. Fix a ∈ A.

Which statement is TRUE?

If Vf (a) = 0, then f has a local extremum at a.

If Vf (a) = 0 and a is an interior point of A, then f has a local extremum at a. If f has a local extremum at a, then Vf (a) = 0.

If f has a global extremum at a, then Vf (a) = 0.

None of the above statements are true.

(1c) Let A C Rn be a set and let f : A - R be a C & real-valued function. Fix a ∈ A.

Which statement is EQUIVALENT to “f has a local maximum at a”?

￥x ∈ A, f (x) < f (a).

刁e > 0 s.t. ￥x ∈ Rn , |x - a| < e =→ f (x) < f (a).

刁e > 0 s.t. ￥x ∈ A, |x - a| < e =→ f (x) < f (a).

Vf (a) = 0.

Vf (a) = 0 and Hf (a) has only negative eigenvalues.

None of the above statements are equivalent.

(1d) The SOCK-O company sells two styles of socks: Artemis Anklets and Luna Low-Cuts. They are trying to maximize their profit. They have determined that if they spend A hundreds of thousands of dollars on Artemis Anklets, spend L hundreds of thousands of dollars on Luna Low-Cuts, and sell all of the socks they produce, then they will achieve a profit of P hundreds of thousands of dollars.  They know the following facts about P :

• P is differentiable on {(A, L) ∈ R2  : A > 0, L > 0} and continuous on {(A, L) ∈ R2  : A > 0, L > 0}.

• The only critical points of P are (A, L) = (40, 15) and (A, L) = (12, 98). Note P > 0 at both points.

•  If A+ L > 100, then P < 0.

What is the strongest possible conclusion that SOCK-O can make about the maximum profit?

They maximize profit by spending $4.0 million on anklets and $1.5 million on low-cuts. They maximize profit by spending $1.2 million on anklets and $9.8 million on low-cuts.

They maximize profit by spending either $4.0 million on anklets and $1.5 million on low-cuts, or $1.2 million on anklets and $9.8 million on low-cuts.

They can maximize profit, but there is not enough information to decide how.

They cannot maximize profit.

Nothing can be concluded since a maximum profit may or may not exist.

2.  (5 points) The parts of this question are unrelated. No justification is necessary for any part.

(2a) Let f : R2 - R and g : R2 - R be C 1 real-valued functions. The graph below shows the gradient vector field Vf (x , y) and the constraint curve g(x , y) = 237.

Label approximately where are the possible local extrema of f on the curve.

(2b)  You are optimizing a function f : R3  - R on the closed ball B = {(x , y, z) ∈ R3  : x 2 + y 2 + z2  < 81}.

You have done many calculations and determined all of the following.

• f is C 1 on R3 .

•  Vf (x , y, z) = (0, 0, 0) if and only if (x , y, z) = (2, 3, 5), and (6, 6, 6).

• f (2, 3, 5) = 2022, and f (6, 6, 6) = -X.

Vf (x , y, z) = (2入x, 2入y, 2入z) x2 + y2 + z2 = 81

are (x , y, z) = (9, 0, 0) with 入 = -2 and (0, 9, 0) with 入 = 3.

• f (9, 0, 0) = 137 and f (0, 9, 0) = 224

What is the maximum of f on B?

2022 224 137 -X Not enough information to decide

What is the minimum of f on B?

2022 224 137 -X Not enough information to decide

(2c) Let F : R3 - R3 be a C 1 map. Fix a, b, c ∈ R3 . You have computed that

At which points is F a local diffeomorphism?

Fill in ALL boxes that apply. If none apply, leave it blank.                                       (unfilled 口 filled ■)

■ a 口 b 口 c

Which statement is TRUE? Fill in EXACTLY ONE circle. (unfilled filled )

F is a diffeomorphism.

F is not a diffeomorphism.

There is not enough information to decide whether F is a diffeomorphism.

3.  (3 points) The parts of this question are unrelated. No justification is necessary. Fill in ALL boxes that apply. If none apply, leave it blank.

(unfilled 口 filled ■)

Based on this information, which of these four statements must be TRUE?

口 G locally defines (x1 , x2 ) as a C 1 function of (x3 , x4 ) near p.

■ G locally defines (x1 , x3 ) as a C 1 function of (x2 , x4 ) near p.

■ G locally defines (x2 , x3 ) as a C 1 function of (x1 , x4 ) near p.

■ G locally defines (x3 , x4 ) as a C 1 function of (x1 , x2 ) near p.

(3b) Let g : R2 - R. The curve below is defined by the equation g(x , y) = 0.

Four points A, B, C , D ∈ R2 are labelled on the curve.

At which of these points can x be described locally as a C 1 function of y?

■ A 口 B 口 C 口 D

(3c)  Eseosa wishes to prove that curve x3 + y3 = 1 is a regular curve. She gives the following argument:

Select all valid critiques of this argument. If none apply, do not select any.

口 Line 1 is flawed since the set S is empty.

■ Line 2 is flawed since f is not C 1 .

口 Line 3 is flawed since there exists at least one point (x , y) ∈ S which does not satisfy f (x) = y . 口 Line 4 is flawed since a regular curve does not need to be a graph.

4.  (6 points) Find the global extrema of the function f (x , y) = x2 + 3xy + y2 - 6x + 6y on the right halfdisk A = {(x , y) ∈ R2  : x2 + y2 < 16, x > 0}. As always, remember to justify your argument.

•  Since A is a compact set and f is a polynomial (and hence continuous), f attains a global maximum and a global minimum on A by the EVT. These global extrema occur either on aA or Ao .

Optimizing on the interior:

•  Since f is differentiable on Ao , the local EVT tells us that if f attains a global extremum on the interior of A then it must occur at a critical point. As such, we compute:

Vf (x , y) = 0 4→ (2x + 3y - 6, 3x + 2y + 6) = (0, 0) 4→ (x , y) = (-6, 6)

• Therefore (-6, 6) is the only critical point of f on R2 .  However, it is not contained in the interior of A. Therefore, f does not attain its global extrema on the interior of A.

Optimizing on the boundary:

•  Decompose it into two subsets:

L = {(0,y ) ∈ R2  : -4 < y < 4} and C = {(x y, ) ∈ R2  : x2 + y2 = 16, x > 0}.

As L and C are compact, f attains global extrema on each L and C by the global EVT.

• Parametrize L by o : [-4, 4] - R2 given by o(t) = (0, t). Then the global extrema of f on L occur at the global extrema of f o o on [-4, 4].

•  Since f o o is differentiable by the chain rule, the extrema of f on L occurs at either an endpoint of [-4, 4] or at a critical point of f o o . Notice (f o o)(t) = t2 + 6t so (f o o)′ (t) = 2t + 6.

• The only critical point of f o o is at t = -3. Therefore, the global extrema of f on L occur at either t = -4, t = 4, or t = -3. These give us the points (0, -4), (0, 4), and (0, -3).

• For C , define the open set U = (0, &) x R and define g : U - R by g(x , y) = x2 + y2 - 16.  Then C = {(x , y) ∈ U : g(x , y) = 0} U {(0, 士4)}.

• The global extrema of f on C occur at either the ends of the arc ((0, -4) and (0, 4)) or are solutions to the Lagrange multiplier system: (x , y) ∈ U and 入 ∈ R such that

2x + 3y - 6 = 2入x

3x + 2y + 6 = 2入y

x2 + y2 = 16

•  By WolframAlpha, the solutions to this system occur at (x , y) = (57 - 1, 1 +57) and (2 52, -2 52 ).

• Therefore, the global extrema of f on aA occur among the points (0, -4), (0, 4), (0, -3), (57 - 1, 1 +57 ), and (2 52, -2 52). Checking the value of f at these points gives:

f (0, -4) = -8 f (0, 4) = 40 f (0, -3) = -9

f (57 - 1, 1 +57) = 46 f (2 52, -2 52) = -8 - 24 52

• Therefore, the maximum of f on aA is 46 and the minimum of f on aA is -8 - 24 52.

•  Since the global extrema of f on A must occur on the boundary, the global extrema of f on aA found above are also the global extrema of f on A.

5.  (4 points) You are classifying the local extrema of a C3 real-valued function f on R3 . You find it has exactly three critical points a, b, c ∈ R3 . You compute the Hessian matrices

Hf (a) ='(．)0(5)   -X(0)   0(0)'(┐) , Hf (b) ='(．)0(2)   1(0)   0(0)'(┐) .

You compute the second Taylor polynomial of f at c to be

P (x , y, z) = 237 - 2x2 - z2 .

For each part below, no justification necessary. Fill in EXACTLY ONE circle. (unfilled filled ) (5a)  Classify the critical point a.

local max

local min

saddle point

not enough information to decide

none of these

(5b)  Classify the critical point b.

local max

local min

saddle point

not enough information to decide

none of these

(5c)  Classify the critical point c .

local max

local min

saddle point

not enough information to decide

none of these

(5d) You verify that f (a) = 2022 and f (b) = 244. At which point does f achieve a global minimum?

a

b

c

not enough information to decide