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程序代写案例-C4
时间：2022-01-12
1. Let f : !3→ ! be a C4 function and suppose its 3rd Taylor polynomial at (0,0,0) is given by
P3(x , y, z) = 1+ 2x + 2y + 3z + x2 − 5x y + 2y2 − 7yz + z3.
Let P0, P1, P2, and P4 be the 0th, 1st, 2nd and 4th Taylor polynomials of f at (0,0,0).
No justification is necessary.
(1a) (1 point) Estimate the value f (0,0,0.2) with a quadratic approximation of f at (0,0,0).
f (0,0,0.2)≈
(1b) (4 points) If possible, determine each of the following quantities. If it is not possible, write “N/A".
f (0,0,0) =
f (0,0,1) =
∂ f
∂ x
(0,0,0) =
∂ 3 f
∂ x∂ y∂ z
(0,0,0) =
H f (0,0,0) =
lim
(x ,y,z)→(0,0,0)
f (x , y, z)− P3(x , y, z)
||(x , y, z)||2 =
lim
(x ,y,z)→(0,0,0)
f (x , y, z)− P3(x , y, z)
||(x , y, z)||3 =
MAT237 Term Test 2 - Page 2 of 11 January 23, 2020
2. (5 points) For each part below, select the BEST answer. Fill in EXACTLY ONE circle. (unfilled filled )
No justification is necessary.
(2a) For a C3 function f : !2→ !, you plot the quadratic form at each of its critical points a, b, c, d ∈ !2.
quadratic form of f at a quadratic form of f at b quadratic form of f at c quadratic form of f at d
If you apply the second derivative test to each critical point, at most one of them gives an inconclusive
answer. Identify which one (if any) gives the inconclusive answer.
© Point a © Point b © Point c © Point d © None of them will be inconclusive.
(2b) Let U ⊆ !n be an open non-empty set and let a, b ∈ U . Let L be the line segment between a and b.
Which of the following is TRUE?
© If f : U → !237 is differentiable and L lies in U then ∃c ∈ L s.t. f (b)− f (a) =∇ f (c) · (b− a).
© If f : U → !237 is differentiable and U is convex then ∃c ∈ L s.t. f (b)− f (a) = f ′(c)(b− a).
© If f : U → !237 is C1 and L lies in U then ∃c ∈ L s.t. f (b)− f (a) =∇ f (c) · (b− a).
© If f : U → !237 is C1 and U is convex then ∃c ∈ L s.t. f (b)− f (a) = f ′(c)(b− a)
© None of the above are true.
(2c) Let f : !3→ ! be C∞. What is the difference between ∂ (3,4) f and ∂ (3,0,4) f ?
© The order of the derivatives is different.
© There is no difference between the two derivatives.
© Both derivatives exist but they are not necessarily equal.
© One of the derivatives does not make sense.
© There is not enough information to decide.
(2d) Let f : !3→ ! be C3. Which of the following is NOT equal to ∂1∂2∂3 f ?
© ∂1∂3∂2 f
© ∂3∂1∂2 f
© ∂2∂3∂1 f
© ∂3∂2∂1 f
© None of the above. They are all equal to ∂1∂2∂3 f .
(2e) Let f : !n→ ! be CN+1 at a ∈ !n. Let PN be the N th Taylor polynomial of f at a. Which statement best
describes Taylor’s theorem from an analytic viewpoint?
© f (x) is well approximated by PN (x) for N large.
© f (x) is well approximated by PN (x) for x near a.
© f (x) is equal to PN (x) with error that decays faster than ||x − a||N as N grows.
© f (x) is equal to PN (x) with error that decays faster than ||x − a||N as x approaches a.
© PN defines a polynomial surface of degree ≤ N which best fits the graph of f near a.
MAT237 Term Test 2 - Page 3 of 11 January 23, 2020
3. (3 points) Below is a contour plot of the C2 function f : !2→ ! and the origin a = (0,0) ∈ !2.
x
y
−2
−1
0
1
2
a
Determine whether each quantity is positive, negative, or zero. Select the most plausible answer.
No justification is necessary. Fill in EXACTLY ONE circle. (unfilled filled )
(3a)
∂ f
∂ x
(a) © positive © negative © zero
(3b)
∂ f
∂ y
(a) © positive © negative © zero
(3c)
∂ 2 f
∂ x2
(a) © positive © negative © zero
(3d)
∂ 2 f
∂ y2
(a) © positive © negative © zero
(3e)
∂ 2 f
∂ y∂ x
(a) © positive © negative © zero
(3f)
∂ 2 f
∂ x∂ y
(a) © positive © negative © zero
MAT237 Term Test 2 - Page 4 of 11 January 23, 2020
4. The parts of this question are unrelated.
(4a) (3 points) Let f (x , y) = (x2 − y2, 2x y). Identify the set S of points where f has a local C1 inverse.
(4b) (2 points) Let g : !2→ !. The curve below is defined by the equation g(x , y) = 237.
Five points A,B,C ,D, E ∈ !2 are labelled on the curve. No justification is necessary.
Fill in ALL boxes that apply. (unfilled " and filled #)
At which of these points can y be described locally as a function of x?
" A " B " C " D " E
At which of these points can y be described locally as a C1 function of x?
" A " B " C " D " E
MAT237 Term Test 2 - Page 5 of 11 January 23, 2020
5. (5 points) For each part below, select the BEST answer. Fill in EXACTLY ONE circle. (unfilled filled )
No justification is necessary.
(5a) Let D ⊆ !n be a set. Let f : D→ ! and let a ∈ D.
Which of the following statements are equivalent to " f has a local minimum at a"?
© ∀x ∈ D, f (x)≥ f (a)
© ∃" > 0 s.t. ∀x ∈ B"(a), f (x)≥ f (a)
© ∃" > 0 s.t. ∀x ∈ B"(a)∩ D, f (x)≥ f (a)
© None of the above are equivalent.
(5b) Let q : !n→ ! be a quadratic form. Which of the following is FALSE?
© There exists a symmetric matrix A such that q(v) = vTAv for all v ∈ !n.
© q(v) = 12 vTHq(0)v for all v ∈ !n.
© q is exactly one of definite, semidefinite, or indefinite.
© q is at least one of positive definite, negative definite, semidefinite, or indefinite.
© None of the above are false.
(5c) The range of a quadratic form q is all of !. Which of the following is TRUE?
© The determinant of q is not positive.
© The matrix associated to q has at least one positive eigenvalue and one negative eigenvalue.
© q is both positive definite and negative definite.
© q is semidefinite but not indefinite.
© None of the above are true.
(5d) Let D ⊆ !n be open and f : D→ ! be C3. Fix a ∈ D. Consider the four statements:
I. If ∇ f (a) = 0 then a is a local extremum of f .
II. If a is a local extremum of f then ∇ f (a) = 0.
III. If ∇ f (a) = 0 and H f (a) is definite then a is a local extremum of f .
IV. If a is a local extremum of f then ∇ f (a) = 0 and H f (a) is definite.
Which of these four statements are TRUE?
© I, II, III, and IV
© II, III, and IV
© I and III
© II and III
© III and IV
© Only II
© Only IV
© None of them
(5e) You have set aside 20 hours to work on two class projects. You want to maximize your grade (measured
in points), which depends on how you divide your time between the two projects. Suppose that you
use the method of Lagrange multipliers to solve this optimization problem and obtain the multiplier λ.
What is the practical meaning of the statement λ = 3?
© If you set aside about 3 more hours, you could gain about one more point on your two projects
combined.
© If you set aside about 1 more hour, you could gain about 3 more points on your two projects
combined.
© If you shift 3 hours from one project to the other, you could gain about 1 more point.
© If you shift 1 hour from one project to the other, you could gain about 3 more points.
© None of the above.
MAT237 Term Test 2 - Page 6 of 11 January 23, 2020
6. The parts of this question are unrelated.
(6a) (3 points) Let f : !2→ !2 and g : !2→ ! be differentiable functions. You have the following data:
(x , y) f (x , y) g(x , y) f 1x (x , y) f
1
y (x , y) f
2
x (x , y) f
2
y (x , y) gx(x , y) gy(x , y)
(0,0) (1,2) 1 1 2 3 −1 2 1
(1,2) (4,−2) 2 4 −12 3 0 −2 1
Let h= g ◦ f . Use the chain rule to estimate the value of h(0.1,−0.2).
(6b) (1 point) The graph below shows a gradient vector field ∇F(x) and a constraint curve G(x) = 0.
−2 −1 0 1 2
−2
−1
0
1
2
By labelling on the graph, approximately identify where the local extrema of F on the curve are located.
No justification is necessary.
MAT237 Term Test 2 - Page 7 of 11 January 23, 2020
7. For (x1, x2, y1, y2) ∈ !4, define
F(x1, x2, y1, y2) =
!
x21 − x22 − y31 + y22 + 4, 2x1x2 + x22 − 2y21 + 3y42 + 8
"
.
The point a = (2,−1,2,1) satisfies F(a) = (0,0). You may use WolframAlpha to do any basic matrix compu-
tations without justification.
(7a) (3 points) Show the equation F(x1, x2, y1, y2) = 0 defines (y1, y2) locally as a C1 function φ of (x1, x2)
near a.
(7b) (2 points) Compute the Jacobian of φ at a.
MAT237 Term Test 2 - Page 8 of 11 January 23, 2020
8. You are classifying the critical points of a C3 real-valued function f on the closed ball S = B237(0) ⊆ !3. You
find it has exactly two critical points a, b ∈ S and compute the Hessian matrices at each point to be
H f (a) =
−5 0 00 −9 0
0 0 −16
, H f (b) =
0 0 00 2 0
0 0 −3
.
Fill in the circle to select your answer. (unfilled filled )
(8a) (1 point) Classify the critical point a. No justification necessary.
© local max © local min © saddle point © not enough information © none of these
(8b) (1 point) Classify the critical point b. No justification necessary.
© local max © local min © saddle point © not enough information © none of these
(8c) (2 points) If possible, identify the global extrema of f on S.
© Both a and b are global extrema.
© One of a or b is a global extremum.
© There are no global extrema but there are local extrema.
© There is not enough information to determine any global extrema.
© None of the above statements are true.
Briefly justify your answer to (8c) in the box below.
MAT237 Term Test 2 - Page 9 of 11 January 23, 2020
9. (6 points) A power station company, Team Rocket, has two generators named Pikachu and Zapdos. Pikachu
outputs x megawatts of power and Zapdos outputs y megawatts of power. The monthly operating costs are
CP(x) = 140,000+
x2
100
, CZ(y) = 90,000+
y3
1200
,
measured in dollars for Pikachu and Zapdos respectively. If Team Rocket is contracted to output power at a
rate of 880 megawatts, how should they balance the generating loads to minimize their costs? Explain your
setup and summarize your conclusion in a full sentence. Use the method of Lagrange multipliers.
MAT237 Term Test 2 - Page 10 of 11 January 23, 2020
10. (6 points) Let m< n and let F : !n→ !m be differentiable. Let S = {x ∈ !n : F(x) = 0} and p ∈ S.
Prove that TpS ⊆ ker dFp.
MAT237 Term Test 2 - Page 11 of 11 January 23, 2020
1. Let f : !3→ ! be a C4 function and suppose its 3rd Taylor polynomial at (0,0,0) is given by
P3(x , y, z) = 1+ 2x + 2y + 3z + x2 − 5x y + 2y2 − 7yz + z3.
Let P0, P1, P2, and P4 be the 0th, 1st, 2nd and 4th Taylor polynomials of f at (0,0,0).
No justification is necessary.
(1a) (1 point) Estimate the value f (0,0,0.2) with a quadratic approximation of f at (0,0,0).
f (0,0,0.2)≈ 1.6
(1b) (4 points) If possible, determine each of the following quantities. If it is not possible, write “N/A".
f (0,0,0) = 1
f (0,0,1) = N/A
∂ f
∂ x
(0,0,0) =2
∂ 3 f
∂ x∂ y∂ z
(0,0,0) = 0
H f (0,0,0) =
2 −5 0−5 4 −7
0 −7 0
lim
(x ,y,z)→(0,0,0)
f (x , y, z)− P3(x , y, z)
||(x , y, z)||2 = 0
lim
(x ,y,z)→(0,0,0)
f (x , y, z)− P3(x , y, z)
||(x , y, z)||3 =0
MAT237 Term Test 2 solutions - Page 2 of 11 January 23, 2020
2. (5 points) For each part below, select the BEST answer. Fill in EXACTLY ONE circle. (unfilled filled )
No justification is necessary.
(2a) For a C3 function f : !2→ !, you plot the quadratic form at each of its critical points a, b, c, d ∈ !2.
quadratic form of f at a quadratic form of f at b quadratic form of f at c quadratic form of f at d
If you apply the second derivative test to each critical point, at most one of them gives an inconclusive
answer. Identify which one (if any) gives the inconclusive answer.
© Point a © Point b © Point c Point d © None of them will be inconclusive.
(2b) Let U ⊆ !n be an open non-empty set and let a, b ∈ U . Let L be the line segment between a and b.
Which of the following is TRUE?
© If f : U → !237 is differentiable and L lies in U then ∃c ∈ L s.t. f (b)− f (a) =∇ f (c) · (b− a).
© If f : U → !237 is differentiable and U is convex then ∃c ∈ L s.t. f (b)− f (a) = f ′(c)(b− a).
© If f : U → !237 is C1 and L lies in U then ∃c ∈ L s.t. f (b)− f (a) =∇ f (c) · (b− a).
© If f : U → !237 is C1 and U is convex then ∃c ∈ L s.t. f (b)− f (a) = f ′(c)(b− a)
None of the above are true.
(2c) Let f : !3→ ! be C∞. What is the difference between ∂ (3,4) f and ∂ (3,0,4) f ?
© The order of the derivatives is different.
© There is no difference between the two derivatives.
© Both derivatives exist but they are not necessarily equal.
One of the derivatives does not make sense.
© There is not enough information to decide.
(2d) Let f : !3→ ! be C3. Which of the following is NOT equal to ∂1∂2∂3 f ?
© ∂1∂3∂2 f
© ∂3∂1∂2 f
© ∂2∂3∂1 f
© ∂3∂2∂1 f
None of the above. They are all equal to ∂1∂2∂3 f .
(2e) Let f : !n→ ! be CN+1 at a ∈ !n. Let PN be the N th Taylor polynomial of f at a. Which statement best
describes Taylor’s theorem from an analytic viewpoint?
© f (x) is well approximated by PN (x) for N large.
© f (x) is well approximated by PN (x) for x near a.
© f (x) is equal to PN (x) with error that decays faster than ||x − a||N as N grows.
f (x) is equal to PN (x) with error that decays faster than ||x − a||N as x approaches a.
© PN defines a polynomial surface of degree ≤ N which best fits the graph of f near a.
MAT237 Term Test 2 solutions - Page 3 of 11 January 23, 2020
3. (3 points) Below is a contour plot of the C2 function f : !2→ ! and the origin a = (0,0) ∈ !2.
x
y
−2
−1
0
1
2
a
Determine whether each quantity is positive, negative, or zero. Select the most plausible answer.
No justification is necessary. Fill in EXACTLY ONE circle. (unfilled filled )
(3a)
∂ f
∂ x
(a) © positive © negative zero
(3b)
∂ f
∂ y
(a) © positive negative © zero
(3c)
∂ 2 f
∂ x2
(a) positive © negative © zero
(3d)
∂ 2 f
∂ y2
(a) © positive negative © zero
(3e)
∂ 2 f
∂ y∂ x
(a) © positive © negative zero
(3f)
∂ 2 f
∂ x∂ y
(a) © positive © negative zero
MAT237 Term Test 2 solutions - Page 4 of 11 January 23, 2020
4. The parts of this question are unrelated.
(4a) (3 points) Let f (x , y) = (x2 − y2, 2x y). Identify the set S of points where f has a local C1 inverse.
Solution:
• By the inverse function theorem f has a local C1 inverse precisely when its Jacobian is invertible.
• The Jacobian of f at (x , y) is
%
2x −2y
2y 2x
&
which has determinant 4x2 + 4y2.
• The determinant 4x2 + 4y2 vanishes if and only if (x , y) = (0,0).
• Therefore, by the inverse function theorem, f has a local C1 inverse near (x , y) precisely when
(x , y) ,= (0,0) so that S = !2 \ {(0,0)}.
Remark: Strictly speaking, the IFT cannot conclude anything at the point (0,0) so the argument above still
needs to show that f does not have a local C1 inverse at (0,0). No marks were deducted if you did not provide
this explanation. Instead, you can use the true statement from F3 worksheet Q6.1 to prove (0,0) ,∈ S. This
lemma is a consequence of the chain rule.
(4b) (2 points) Let g : !2→ !. The curve below is defined by the equation g(x , y) = 237.
Five points A,B,C ,D, E ∈ !2 are labelled on the curve. No justification is necessary.
Fill in ALL boxes that apply. (unfilled " and filled #)
At which of these points can y be described locally as a function of x?
" A # B " C # D # E
At which of these points can y be described locally as a C1 function of x?
" A # B " C " D # E
MAT237 Term Test 2 solutions - Page 5 of 11 January 23, 2020
5. (5 points) For each part below, select the BEST answer. Fill in EXACTLY ONE circle. (unfilled filled )
No justification is necessary.
(5a) Let D ⊆ !n be a set. Let f : D→ ! and let a ∈ D.
Which of the following statements are equivalent to " f has a local minimum at a"?
© ∀x ∈ D, f (x)≥ f (a)
© ∃" > 0 s.t. ∀x ∈ B"(a), f (x)≥ f (a)
∃" > 0 s.t. ∀x ∈ B"(a)∩ D, f (x)≥ f (a)
© None of the above are equivalent.
(5b) Let q : !n→ ! be a quadratic form. Which of the following is FALSE?
© There exists a symmetric matrix A such that q(v) = vTAv for all v ∈ !n.
© q(v) = 12 vTHq(0)v for all v ∈ !n.
q is exactly one of definite, semidefinite, or indefinite.
© q is at least one of positive definite, negative definite, semidefinite, or indefinite.
© None of the above are false.
(5c) The range of a quadratic form q is all of !. Which of the following is TRUE?
© The determinant of q is not positive.
The matrix associated to q has at least one positive eigenvalue and one negative eigen-
value.
© q is both positive definite and negative definite.
© q is semidefinite but not indefinite.
© None of the above are true.
(5d) Let D ⊆ !n be open and f : D→ ! be C3. Fix a ∈ D. Consider the four statements:
I. If ∇ f (a) = 0 then a is a local extremum of f .
II. If a is a local extremum of f then ∇ f (a) = 0.
III. If ∇ f (a) = 0 and H f (a) is definite then a is a local extremum of f .
IV. If a is a local extremum of f then ∇ f (a) = 0 and H f (a) is definite.
Which of these four statements are TRUE?
© I, II, III, and IV
© II, III, and IV
© I and III
II and III
© III and IV
© Only II
© Only IV
© None of them
(5e) You have set aside 20 hours to work on two class projects. You want to maximize your grade (measured
in points), which depends on how you divide your time between the two projects. Suppose that you
use the method of Lagrange multipliers to solve this optimization problem and obtain the multiplier λ.
What is the practical meaning of the statement λ = 3?
© If you set aside about 3 more hours, you could gain about one more point on your two projects
combined.
If you set aside about 1 more hour, you could gain about 3 more points on your two
projects combined.
© If you shift 3 hours from one project to the other, you could gain about 1 more point.
© If you shift 1 hour from one project to the other, you could gain about 3 more points.
© None of the above.
MAT237 Term Test 2 solutions - Page 6 of 11 January 23, 2020
6. The parts of this question are unrelated.
(6a) (3 points) Let f : !2→ !2 and g : !2→ ! be differentiable functions. You have the following data:
(x , y) f (x , y) g(x , y) f 1x (x , y) f
1
y (x , y) f
2
x (x , y) f
2
y (x , y) gx(x , y) gy(x , y)
(0,0) (1,2) 1 1 2 3 −1 2 1
(1,2) (4,−2) 2 4 −12 3 0 −2 1
Let h= g ◦ f . Use the chain rule to estimate the value of h(0.1,−0.2).
Solution:
• We will approximate
h(0.1,−0.2)≈ h(0,0) + dh(0,0)(0.1,−0.2).
• In matrix form the chain rule together with the given data yields
dh(0,0)(x , y) =
'
gx( f (0,0)) gy( f (0,0))
() f 1x (0,0) f 1y (0,0)
f 2x (0,0) f
2
y (0,0)
*%
x
y
&
=
'−2 1(%1 23 −1&%xy&
= x − 5y.
• Putting this together,
h(0.1,−0.2)≈ h(0,0) + dh(0,0)(0.1,−0.2)
= g( f (0,0)) + 0.1− 5 · (−0.2)
= g(1,2) + 1.1
= 2+ 1.1
= 3.1.
(6b) (1 point) The graph below shows a gradient vector field ∇F(x) and a constraint curve G(x) = 0.
−2 −1 0 1 2
−2
−1
0
1
2
By labelling on the graph, approximately identify where the local extrema of F on the curve are located.
No justification is necessary.
MAT237 Term Test 2 solutions - Page 7 of 11 January 23, 2020
7. For (x1, x2, y1, y2) ∈ !4, define
F(x1, x2, y1, y2) =
+
x21 − x22 − y31 + y22 + 4, 2x1x2 + x22 − 2y21 + 3y42 + 8
,
.
The point a = (2,−1,2,1) satisfies F(a) = (0,0). You may use WolframAlpha to do any basic matrix compu-
tations without justification.
(7a) (3 points) Show the equation F(x1, x2, y1, y2) = 0 defines (y1, y2) locally as a C1 function φ of (x1, x2)
near a.
Solution:
• The Jacobian of F is
%
2x1 −2x2 −3y21 2y2
2x2 2x1 + 2x2 −4y1 12y32
&
.
• At (x1, x2, y1, y2) = (2,−1,2,1) = a, the submatrix
∂ F
∂ y
=
%−3y21 2y2−4y1 12y32
&
=
%−12 2
−8 12
&
has non-zero determinant −144+ 16= −128.
• Therefore by the implicit function theorem, we can solve for (y1, y2) as a C1 function of (x1, x2) near
a.
(7b) (2 points) Compute the Jacobian of φ at a.
Solution:
• By the implicit function theorem, the Jacobian of φ satisfies
Dφ(2,−1) = −
-
∂ F
∂ y
(2,−1,2,1)
.−1 ∂ F
∂ x
(2,−1,2,1).
• Substituting the expressions from part (a) and computing with WolframAlpha yields
Dφ(2,−1) = −
%−12 2
−8 12
&−1 % 4 2
−2 2
&
=
/ 13
32
5
32
7
16 − 116
0
.
MAT237 Term Test 2 solutions - Page 8 of 11 January 23, 2020
8. You are classifying the critical points of a C3 real-valued function f on the closed ball S = B237(0) ⊆ !3. You
find it has exactly two critical points a, b ∈ S and compute the Hessian matrices at each point to be
H f (a) =
−5 0 00 −9 0
0 0 −16
, H f (b) =
0 0 00 2 0
0 0 −3
.
Fill in the circle to select your answer. (unfilled filled )
(8a) (1 point) Classify the critical point a. No justification necessary.
local max © local min © saddle point © not enough information © none of these
(8b) (1 point) Classify the critical point b. No justification necessary.
© local max © local min saddle point © not enough information © none of these
(8c) (2 points) If possible, identify the global extrema of f on S.
© Both a and b are global extrema.
© One of a or b is a global extremum.
© There are no global extrema but there are local extrema.
There is not enough information to determine any global extrema.
© None of the above statements are true.
Briefly justify your answer to (8c) in the box below.
We don’t know anything about the values of f on the boundary of S = B237(0) so without more infor-
mation the global maximum and/or the global minimum could be on the boundary.
Remark: The phrase “determine any global extrema" means that you must be able to identify those points
exactly. It is not enough to know that some exist but you cannot locate them.
MAT237 Term Test 2 solutions - Page 9 of 11 January 23, 2020
9. (6 points) A power station company, Team Rocket, has two generators named Pikachu and Zapdos. Pikachu
outputs x megawatts of power and Zapdos outputs y megawatts of power. The monthly operating costs are
CP(x) = 140,000+
x2
100
, CZ(y) = 90,000+
y3
1200
,
measured in dollars for Pikachu and Zapdos respectively. If Team Rocket is contracted to output power at a
rate of 880 megawatts, how should they balance the generating loads to minimize their costs? Explain your
setup and summarize your conclusion in a full sentence. Use the method of Lagrange multipliers.
Solution:
• If Pikachu outputs x megawatts of power and Zapdos outputs y megawatts of power, the total cost
to operate the generators is
C(x , y) = Cp(x) + CZ(y) = 230,000+
x2
100
+
y3
1200
.
• Team Rocket is contracted to output power at a rate of 880 megawatts so (x , y) is subject to the
contstraint f (x , y) = x + y = 880. Additionally x and y must both be non-negative.
• We wish to use Lagrange multipliers to minimize C(x , y) subject to the constraint f (x , y) = 880 on
the open set U = {(x , y) ∈ !2 : x , y > 0}.
• The gradients of C and f are
∇C(x , y) =
1
x
50
,
y2
400
2
and ∇ f (x , y) = (1,1).
• Suppose λ ∈ ! satisfies 1
x
50
,
y2
400
2
=∇C(x , y) = λ∇ f (x , y) = (λ,λ)
where f (x , y) = 880. Writing x50 = λ =
y2
400 and plugging into x + y = f (x , y) = 880 then yields
y2
8 + y − 880= 0.
• Solving the above quadratic equation yields y = 80 or y = −88 but y cannot be negative. This yields
(x , y) = (800,80) as a possible solution inside U with corresponding value C(800,80)≈ 236826.67.
• We need to check the boundary points (x , y) = (880,0) and (x , y) = (0,880). The corresponding
operating costs are C(880,0) = 237744 and C(0,880)≈ 797893.33 respectively.
• Comparing these values, we see that the minimummonthly operation costs are C(x , y) = 236826.67
dollars achieved when Pikachu outputs 800 megawatts of power and Zapdos outputs 80 megawatts
of power.
Remark: The boundary points (880,0) and (0,880) must be checked separately since (880,0), (0,880) ,∈ U .
Lagrange multipliers can only be used for functions and constraints defined on an open set U which does not
include these points in ∂ U on the constraint curve {(x , y) ∈ !2 : x + y = 880, x ≥ 0, y ≥ 0}.
MAT237 Term Test 2 solutions - Page 10 of 11 January 23, 2020
10. (6 points) Let m< n and let F : !n→ !m be differentiable. Let S = {x ∈ !n : F(x) = 0} and p ∈ S.
Prove that TpS ⊆ ker dFp.
Solution:
• Let v ∈ TpS.
• By definition of TpS this means that there exists a differentiable path γ : !→ !n with image lying in
S such that γ(0) = p and γ′(0) = v.
• Since S is the 0-level set of F and the image of γ lies in S, for any t ∈ !, (F ◦ γ)(t) = F(γ(t)) = 0.
• This means that the composition F ◦ γ is constant and hence has derivative always equal to zero.
• In particular, since F and γ are differentiable everywhere, differentiating at t = 0 and using the chain
rule yields
dFp(γ′(0)) = F ′(p)γ′(0) = F ′(γ(0))γ′(0) = (F ◦ γ)′(0) = 0
which says that v = γ′(0) ∈ ker dFp.
• Therefore TpS ⊆ ker dFp as required.
Remark: You cannot use D6 Theorem 1 or D7 Theorem 2 at all because those assumptions are not satisfied.
Notice you cannot assume F is C1 nor can you assume dFp is full rank. Incorrectly applying these theorems
significantly affects the rest of your proof so very little credit was awarded in this case. This question is almost
identical to Q7 on the Lecture D8 worksheet.
MAT237 Term Test 2 solutions - Page 11 of 11 January 23, 2020
学霸联盟
P3(x , y, z) = 1+ 2x + 2y + 3z + x2 − 5x y + 2y2 − 7yz + z3.
Let P0, P1, P2, and P4 be the 0th, 1st, 2nd and 4th Taylor polynomials of f at (0,0,0).
No justification is necessary.
(1a) (1 point) Estimate the value f (0,0,0.2) with a quadratic approximation of f at (0,0,0).
f (0,0,0.2)≈
(1b) (4 points) If possible, determine each of the following quantities. If it is not possible, write “N/A".
f (0,0,0) =
f (0,0,1) =
∂ f
∂ x
(0,0,0) =
∂ 3 f
∂ x∂ y∂ z
(0,0,0) =
H f (0,0,0) =
lim
(x ,y,z)→(0,0,0)
f (x , y, z)− P3(x , y, z)
||(x , y, z)||2 =
lim
(x ,y,z)→(0,0,0)
f (x , y, z)− P3(x , y, z)
||(x , y, z)||3 =
MAT237 Term Test 2 - Page 2 of 11 January 23, 2020
2. (5 points) For each part below, select the BEST answer. Fill in EXACTLY ONE circle. (unfilled filled )
No justification is necessary.
(2a) For a C3 function f : !2→ !, you plot the quadratic form at each of its critical points a, b, c, d ∈ !2.
quadratic form of f at a quadratic form of f at b quadratic form of f at c quadratic form of f at d
If you apply the second derivative test to each critical point, at most one of them gives an inconclusive
answer. Identify which one (if any) gives the inconclusive answer.
© Point a © Point b © Point c © Point d © None of them will be inconclusive.
(2b) Let U ⊆ !n be an open non-empty set and let a, b ∈ U . Let L be the line segment between a and b.
Which of the following is TRUE?
© If f : U → !237 is differentiable and L lies in U then ∃c ∈ L s.t. f (b)− f (a) =∇ f (c) · (b− a).
© If f : U → !237 is differentiable and U is convex then ∃c ∈ L s.t. f (b)− f (a) = f ′(c)(b− a).
© If f : U → !237 is C1 and L lies in U then ∃c ∈ L s.t. f (b)− f (a) =∇ f (c) · (b− a).
© If f : U → !237 is C1 and U is convex then ∃c ∈ L s.t. f (b)− f (a) = f ′(c)(b− a)
© None of the above are true.
(2c) Let f : !3→ ! be C∞. What is the difference between ∂ (3,4) f and ∂ (3,0,4) f ?
© The order of the derivatives is different.
© There is no difference between the two derivatives.
© Both derivatives exist but they are not necessarily equal.
© One of the derivatives does not make sense.
© There is not enough information to decide.
(2d) Let f : !3→ ! be C3. Which of the following is NOT equal to ∂1∂2∂3 f ?
© ∂1∂3∂2 f
© ∂3∂1∂2 f
© ∂2∂3∂1 f
© ∂3∂2∂1 f
© None of the above. They are all equal to ∂1∂2∂3 f .
(2e) Let f : !n→ ! be CN+1 at a ∈ !n. Let PN be the N th Taylor polynomial of f at a. Which statement best
describes Taylor’s theorem from an analytic viewpoint?
© f (x) is well approximated by PN (x) for N large.
© f (x) is well approximated by PN (x) for x near a.
© f (x) is equal to PN (x) with error that decays faster than ||x − a||N as N grows.
© f (x) is equal to PN (x) with error that decays faster than ||x − a||N as x approaches a.
© PN defines a polynomial surface of degree ≤ N which best fits the graph of f near a.
MAT237 Term Test 2 - Page 3 of 11 January 23, 2020
3. (3 points) Below is a contour plot of the C2 function f : !2→ ! and the origin a = (0,0) ∈ !2.
x
y
−2
−1
0
1
2
a
Determine whether each quantity is positive, negative, or zero. Select the most plausible answer.
No justification is necessary. Fill in EXACTLY ONE circle. (unfilled filled )
(3a)
∂ f
∂ x
(a) © positive © negative © zero
(3b)
∂ f
∂ y
(a) © positive © negative © zero
(3c)
∂ 2 f
∂ x2
(a) © positive © negative © zero
(3d)
∂ 2 f
∂ y2
(a) © positive © negative © zero
(3e)
∂ 2 f
∂ y∂ x
(a) © positive © negative © zero
(3f)
∂ 2 f
∂ x∂ y
(a) © positive © negative © zero
MAT237 Term Test 2 - Page 4 of 11 January 23, 2020
4. The parts of this question are unrelated.
(4a) (3 points) Let f (x , y) = (x2 − y2, 2x y). Identify the set S of points where f has a local C1 inverse.
(4b) (2 points) Let g : !2→ !. The curve below is defined by the equation g(x , y) = 237.
Five points A,B,C ,D, E ∈ !2 are labelled on the curve. No justification is necessary.
Fill in ALL boxes that apply. (unfilled " and filled #)
At which of these points can y be described locally as a function of x?
" A " B " C " D " E
At which of these points can y be described locally as a C1 function of x?
" A " B " C " D " E
MAT237 Term Test 2 - Page 5 of 11 January 23, 2020
5. (5 points) For each part below, select the BEST answer. Fill in EXACTLY ONE circle. (unfilled filled )
No justification is necessary.
(5a) Let D ⊆ !n be a set. Let f : D→ ! and let a ∈ D.
Which of the following statements are equivalent to " f has a local minimum at a"?
© ∀x ∈ D, f (x)≥ f (a)
© ∃" > 0 s.t. ∀x ∈ B"(a), f (x)≥ f (a)
© ∃" > 0 s.t. ∀x ∈ B"(a)∩ D, f (x)≥ f (a)
© None of the above are equivalent.
(5b) Let q : !n→ ! be a quadratic form. Which of the following is FALSE?
© There exists a symmetric matrix A such that q(v) = vTAv for all v ∈ !n.
© q(v) = 12 vTHq(0)v for all v ∈ !n.
© q is exactly one of definite, semidefinite, or indefinite.
© q is at least one of positive definite, negative definite, semidefinite, or indefinite.
© None of the above are false.
(5c) The range of a quadratic form q is all of !. Which of the following is TRUE?
© The determinant of q is not positive.
© The matrix associated to q has at least one positive eigenvalue and one negative eigenvalue.
© q is both positive definite and negative definite.
© q is semidefinite but not indefinite.
© None of the above are true.
(5d) Let D ⊆ !n be open and f : D→ ! be C3. Fix a ∈ D. Consider the four statements:
I. If ∇ f (a) = 0 then a is a local extremum of f .
II. If a is a local extremum of f then ∇ f (a) = 0.
III. If ∇ f (a) = 0 and H f (a) is definite then a is a local extremum of f .
IV. If a is a local extremum of f then ∇ f (a) = 0 and H f (a) is definite.
Which of these four statements are TRUE?
© I, II, III, and IV
© II, III, and IV
© I and III
© II and III
© III and IV
© Only II
© Only IV
© None of them
(5e) You have set aside 20 hours to work on two class projects. You want to maximize your grade (measured
in points), which depends on how you divide your time between the two projects. Suppose that you
use the method of Lagrange multipliers to solve this optimization problem and obtain the multiplier λ.
What is the practical meaning of the statement λ = 3?
© If you set aside about 3 more hours, you could gain about one more point on your two projects
combined.
© If you set aside about 1 more hour, you could gain about 3 more points on your two projects
combined.
© If you shift 3 hours from one project to the other, you could gain about 1 more point.
© If you shift 1 hour from one project to the other, you could gain about 3 more points.
© None of the above.
MAT237 Term Test 2 - Page 6 of 11 January 23, 2020
6. The parts of this question are unrelated.
(6a) (3 points) Let f : !2→ !2 and g : !2→ ! be differentiable functions. You have the following data:
(x , y) f (x , y) g(x , y) f 1x (x , y) f
1
y (x , y) f
2
x (x , y) f
2
y (x , y) gx(x , y) gy(x , y)
(0,0) (1,2) 1 1 2 3 −1 2 1
(1,2) (4,−2) 2 4 −12 3 0 −2 1
Let h= g ◦ f . Use the chain rule to estimate the value of h(0.1,−0.2).
(6b) (1 point) The graph below shows a gradient vector field ∇F(x) and a constraint curve G(x) = 0.
−2 −1 0 1 2
−2
−1
0
1
2
By labelling on the graph, approximately identify where the local extrema of F on the curve are located.
No justification is necessary.
MAT237 Term Test 2 - Page 7 of 11 January 23, 2020
7. For (x1, x2, y1, y2) ∈ !4, define
F(x1, x2, y1, y2) =
!
x21 − x22 − y31 + y22 + 4, 2x1x2 + x22 − 2y21 + 3y42 + 8
"
.
The point a = (2,−1,2,1) satisfies F(a) = (0,0). You may use WolframAlpha to do any basic matrix compu-
tations without justification.
(7a) (3 points) Show the equation F(x1, x2, y1, y2) = 0 defines (y1, y2) locally as a C1 function φ of (x1, x2)
near a.
(7b) (2 points) Compute the Jacobian of φ at a.
MAT237 Term Test 2 - Page 8 of 11 January 23, 2020
8. You are classifying the critical points of a C3 real-valued function f on the closed ball S = B237(0) ⊆ !3. You
find it has exactly two critical points a, b ∈ S and compute the Hessian matrices at each point to be
H f (a) =
−5 0 00 −9 0
0 0 −16
, H f (b) =
0 0 00 2 0
0 0 −3
.
Fill in the circle to select your answer. (unfilled filled )
(8a) (1 point) Classify the critical point a. No justification necessary.
© local max © local min © saddle point © not enough information © none of these
(8b) (1 point) Classify the critical point b. No justification necessary.
© local max © local min © saddle point © not enough information © none of these
(8c) (2 points) If possible, identify the global extrema of f on S.
© Both a and b are global extrema.
© One of a or b is a global extremum.
© There are no global extrema but there are local extrema.
© There is not enough information to determine any global extrema.
© None of the above statements are true.
Briefly justify your answer to (8c) in the box below.
MAT237 Term Test 2 - Page 9 of 11 January 23, 2020
9. (6 points) A power station company, Team Rocket, has two generators named Pikachu and Zapdos. Pikachu
outputs x megawatts of power and Zapdos outputs y megawatts of power. The monthly operating costs are
CP(x) = 140,000+
x2
100
, CZ(y) = 90,000+
y3
1200
,
measured in dollars for Pikachu and Zapdos respectively. If Team Rocket is contracted to output power at a
rate of 880 megawatts, how should they balance the generating loads to minimize their costs? Explain your
setup and summarize your conclusion in a full sentence. Use the method of Lagrange multipliers.
MAT237 Term Test 2 - Page 10 of 11 January 23, 2020
10. (6 points) Let m< n and let F : !n→ !m be differentiable. Let S = {x ∈ !n : F(x) = 0} and p ∈ S.
Prove that TpS ⊆ ker dFp.
MAT237 Term Test 2 - Page 11 of 11 January 23, 2020
1. Let f : !3→ ! be a C4 function and suppose its 3rd Taylor polynomial at (0,0,0) is given by
P3(x , y, z) = 1+ 2x + 2y + 3z + x2 − 5x y + 2y2 − 7yz + z3.
Let P0, P1, P2, and P4 be the 0th, 1st, 2nd and 4th Taylor polynomials of f at (0,0,0).
No justification is necessary.
(1a) (1 point) Estimate the value f (0,0,0.2) with a quadratic approximation of f at (0,0,0).
f (0,0,0.2)≈ 1.6
(1b) (4 points) If possible, determine each of the following quantities. If it is not possible, write “N/A".
f (0,0,0) = 1
f (0,0,1) = N/A
∂ f
∂ x
(0,0,0) =2
∂ 3 f
∂ x∂ y∂ z
(0,0,0) = 0
H f (0,0,0) =
2 −5 0−5 4 −7
0 −7 0
lim
(x ,y,z)→(0,0,0)
f (x , y, z)− P3(x , y, z)
||(x , y, z)||2 = 0
lim
(x ,y,z)→(0,0,0)
f (x , y, z)− P3(x , y, z)
||(x , y, z)||3 =0
MAT237 Term Test 2 solutions - Page 2 of 11 January 23, 2020
2. (5 points) For each part below, select the BEST answer. Fill in EXACTLY ONE circle. (unfilled filled )
No justification is necessary.
(2a) For a C3 function f : !2→ !, you plot the quadratic form at each of its critical points a, b, c, d ∈ !2.
quadratic form of f at a quadratic form of f at b quadratic form of f at c quadratic form of f at d
If you apply the second derivative test to each critical point, at most one of them gives an inconclusive
answer. Identify which one (if any) gives the inconclusive answer.
© Point a © Point b © Point c Point d © None of them will be inconclusive.
(2b) Let U ⊆ !n be an open non-empty set and let a, b ∈ U . Let L be the line segment between a and b.
Which of the following is TRUE?
© If f : U → !237 is differentiable and L lies in U then ∃c ∈ L s.t. f (b)− f (a) =∇ f (c) · (b− a).
© If f : U → !237 is differentiable and U is convex then ∃c ∈ L s.t. f (b)− f (a) = f ′(c)(b− a).
© If f : U → !237 is C1 and L lies in U then ∃c ∈ L s.t. f (b)− f (a) =∇ f (c) · (b− a).
© If f : U → !237 is C1 and U is convex then ∃c ∈ L s.t. f (b)− f (a) = f ′(c)(b− a)
None of the above are true.
(2c) Let f : !3→ ! be C∞. What is the difference between ∂ (3,4) f and ∂ (3,0,4) f ?
© The order of the derivatives is different.
© There is no difference between the two derivatives.
© Both derivatives exist but they are not necessarily equal.
One of the derivatives does not make sense.
© There is not enough information to decide.
(2d) Let f : !3→ ! be C3. Which of the following is NOT equal to ∂1∂2∂3 f ?
© ∂1∂3∂2 f
© ∂3∂1∂2 f
© ∂2∂3∂1 f
© ∂3∂2∂1 f
None of the above. They are all equal to ∂1∂2∂3 f .
(2e) Let f : !n→ ! be CN+1 at a ∈ !n. Let PN be the N th Taylor polynomial of f at a. Which statement best
describes Taylor’s theorem from an analytic viewpoint?
© f (x) is well approximated by PN (x) for N large.
© f (x) is well approximated by PN (x) for x near a.
© f (x) is equal to PN (x) with error that decays faster than ||x − a||N as N grows.
f (x) is equal to PN (x) with error that decays faster than ||x − a||N as x approaches a.
© PN defines a polynomial surface of degree ≤ N which best fits the graph of f near a.
MAT237 Term Test 2 solutions - Page 3 of 11 January 23, 2020
3. (3 points) Below is a contour plot of the C2 function f : !2→ ! and the origin a = (0,0) ∈ !2.
x
y
−2
−1
0
1
2
a
Determine whether each quantity is positive, negative, or zero. Select the most plausible answer.
No justification is necessary. Fill in EXACTLY ONE circle. (unfilled filled )
(3a)
∂ f
∂ x
(a) © positive © negative zero
(3b)
∂ f
∂ y
(a) © positive negative © zero
(3c)
∂ 2 f
∂ x2
(a) positive © negative © zero
(3d)
∂ 2 f
∂ y2
(a) © positive negative © zero
(3e)
∂ 2 f
∂ y∂ x
(a) © positive © negative zero
(3f)
∂ 2 f
∂ x∂ y
(a) © positive © negative zero
MAT237 Term Test 2 solutions - Page 4 of 11 January 23, 2020
4. The parts of this question are unrelated.
(4a) (3 points) Let f (x , y) = (x2 − y2, 2x y). Identify the set S of points where f has a local C1 inverse.
Solution:
• By the inverse function theorem f has a local C1 inverse precisely when its Jacobian is invertible.
• The Jacobian of f at (x , y) is
%
2x −2y
2y 2x
&
which has determinant 4x2 + 4y2.
• The determinant 4x2 + 4y2 vanishes if and only if (x , y) = (0,0).
• Therefore, by the inverse function theorem, f has a local C1 inverse near (x , y) precisely when
(x , y) ,= (0,0) so that S = !2 \ {(0,0)}.
Remark: Strictly speaking, the IFT cannot conclude anything at the point (0,0) so the argument above still
needs to show that f does not have a local C1 inverse at (0,0). No marks were deducted if you did not provide
this explanation. Instead, you can use the true statement from F3 worksheet Q6.1 to prove (0,0) ,∈ S. This
lemma is a consequence of the chain rule.
(4b) (2 points) Let g : !2→ !. The curve below is defined by the equation g(x , y) = 237.
Five points A,B,C ,D, E ∈ !2 are labelled on the curve. No justification is necessary.
Fill in ALL boxes that apply. (unfilled " and filled #)
At which of these points can y be described locally as a function of x?
" A # B " C # D # E
At which of these points can y be described locally as a C1 function of x?
" A # B " C " D # E
MAT237 Term Test 2 solutions - Page 5 of 11 January 23, 2020
5. (5 points) For each part below, select the BEST answer. Fill in EXACTLY ONE circle. (unfilled filled )
No justification is necessary.
(5a) Let D ⊆ !n be a set. Let f : D→ ! and let a ∈ D.
Which of the following statements are equivalent to " f has a local minimum at a"?
© ∀x ∈ D, f (x)≥ f (a)
© ∃" > 0 s.t. ∀x ∈ B"(a), f (x)≥ f (a)
∃" > 0 s.t. ∀x ∈ B"(a)∩ D, f (x)≥ f (a)
© None of the above are equivalent.
(5b) Let q : !n→ ! be a quadratic form. Which of the following is FALSE?
© There exists a symmetric matrix A such that q(v) = vTAv for all v ∈ !n.
© q(v) = 12 vTHq(0)v for all v ∈ !n.
q is exactly one of definite, semidefinite, or indefinite.
© q is at least one of positive definite, negative definite, semidefinite, or indefinite.
© None of the above are false.
(5c) The range of a quadratic form q is all of !. Which of the following is TRUE?
© The determinant of q is not positive.
The matrix associated to q has at least one positive eigenvalue and one negative eigen-
value.
© q is both positive definite and negative definite.
© q is semidefinite but not indefinite.
© None of the above are true.
(5d) Let D ⊆ !n be open and f : D→ ! be C3. Fix a ∈ D. Consider the four statements:
I. If ∇ f (a) = 0 then a is a local extremum of f .
II. If a is a local extremum of f then ∇ f (a) = 0.
III. If ∇ f (a) = 0 and H f (a) is definite then a is a local extremum of f .
IV. If a is a local extremum of f then ∇ f (a) = 0 and H f (a) is definite.
Which of these four statements are TRUE?
© I, II, III, and IV
© II, III, and IV
© I and III
II and III
© III and IV
© Only II
© Only IV
© None of them
(5e) You have set aside 20 hours to work on two class projects. You want to maximize your grade (measured
in points), which depends on how you divide your time between the two projects. Suppose that you
use the method of Lagrange multipliers to solve this optimization problem and obtain the multiplier λ.
What is the practical meaning of the statement λ = 3?
© If you set aside about 3 more hours, you could gain about one more point on your two projects
combined.
If you set aside about 1 more hour, you could gain about 3 more points on your two
projects combined.
© If you shift 3 hours from one project to the other, you could gain about 1 more point.
© If you shift 1 hour from one project to the other, you could gain about 3 more points.
© None of the above.
MAT237 Term Test 2 solutions - Page 6 of 11 January 23, 2020
6. The parts of this question are unrelated.
(6a) (3 points) Let f : !2→ !2 and g : !2→ ! be differentiable functions. You have the following data:
(x , y) f (x , y) g(x , y) f 1x (x , y) f
1
y (x , y) f
2
x (x , y) f
2
y (x , y) gx(x , y) gy(x , y)
(0,0) (1,2) 1 1 2 3 −1 2 1
(1,2) (4,−2) 2 4 −12 3 0 −2 1
Let h= g ◦ f . Use the chain rule to estimate the value of h(0.1,−0.2).
Solution:
• We will approximate
h(0.1,−0.2)≈ h(0,0) + dh(0,0)(0.1,−0.2).
• In matrix form the chain rule together with the given data yields
dh(0,0)(x , y) =
'
gx( f (0,0)) gy( f (0,0))
() f 1x (0,0) f 1y (0,0)
f 2x (0,0) f
2
y (0,0)
*%
x
y
&
=
'−2 1(%1 23 −1&%xy&
= x − 5y.
• Putting this together,
h(0.1,−0.2)≈ h(0,0) + dh(0,0)(0.1,−0.2)
= g( f (0,0)) + 0.1− 5 · (−0.2)
= g(1,2) + 1.1
= 2+ 1.1
= 3.1.
(6b) (1 point) The graph below shows a gradient vector field ∇F(x) and a constraint curve G(x) = 0.
−2 −1 0 1 2
−2
−1
0
1
2
By labelling on the graph, approximately identify where the local extrema of F on the curve are located.
No justification is necessary.
MAT237 Term Test 2 solutions - Page 7 of 11 January 23, 2020
7. For (x1, x2, y1, y2) ∈ !4, define
F(x1, x2, y1, y2) =
+
x21 − x22 − y31 + y22 + 4, 2x1x2 + x22 − 2y21 + 3y42 + 8
,
.
The point a = (2,−1,2,1) satisfies F(a) = (0,0). You may use WolframAlpha to do any basic matrix compu-
tations without justification.
(7a) (3 points) Show the equation F(x1, x2, y1, y2) = 0 defines (y1, y2) locally as a C1 function φ of (x1, x2)
near a.
Solution:
• The Jacobian of F is
%
2x1 −2x2 −3y21 2y2
2x2 2x1 + 2x2 −4y1 12y32
&
.
• At (x1, x2, y1, y2) = (2,−1,2,1) = a, the submatrix
∂ F
∂ y
=
%−3y21 2y2−4y1 12y32
&
=
%−12 2
−8 12
&
has non-zero determinant −144+ 16= −128.
• Therefore by the implicit function theorem, we can solve for (y1, y2) as a C1 function of (x1, x2) near
a.
(7b) (2 points) Compute the Jacobian of φ at a.
Solution:
• By the implicit function theorem, the Jacobian of φ satisfies
Dφ(2,−1) = −
-
∂ F
∂ y
(2,−1,2,1)
.−1 ∂ F
∂ x
(2,−1,2,1).
• Substituting the expressions from part (a) and computing with WolframAlpha yields
Dφ(2,−1) = −
%−12 2
−8 12
&−1 % 4 2
−2 2
&
=
/ 13
32
5
32
7
16 − 116
0
.
MAT237 Term Test 2 solutions - Page 8 of 11 January 23, 2020
8. You are classifying the critical points of a C3 real-valued function f on the closed ball S = B237(0) ⊆ !3. You
find it has exactly two critical points a, b ∈ S and compute the Hessian matrices at each point to be
H f (a) =
−5 0 00 −9 0
0 0 −16
, H f (b) =
0 0 00 2 0
0 0 −3
.
Fill in the circle to select your answer. (unfilled filled )
(8a) (1 point) Classify the critical point a. No justification necessary.
local max © local min © saddle point © not enough information © none of these
(8b) (1 point) Classify the critical point b. No justification necessary.
© local max © local min saddle point © not enough information © none of these
(8c) (2 points) If possible, identify the global extrema of f on S.
© Both a and b are global extrema.
© One of a or b is a global extremum.
© There are no global extrema but there are local extrema.
There is not enough information to determine any global extrema.
© None of the above statements are true.
Briefly justify your answer to (8c) in the box below.
We don’t know anything about the values of f on the boundary of S = B237(0) so without more infor-
mation the global maximum and/or the global minimum could be on the boundary.
Remark: The phrase “determine any global extrema" means that you must be able to identify those points
exactly. It is not enough to know that some exist but you cannot locate them.
MAT237 Term Test 2 solutions - Page 9 of 11 January 23, 2020
9. (6 points) A power station company, Team Rocket, has two generators named Pikachu and Zapdos. Pikachu
outputs x megawatts of power and Zapdos outputs y megawatts of power. The monthly operating costs are
CP(x) = 140,000+
x2
100
, CZ(y) = 90,000+
y3
1200
,
measured in dollars for Pikachu and Zapdos respectively. If Team Rocket is contracted to output power at a
rate of 880 megawatts, how should they balance the generating loads to minimize their costs? Explain your
setup and summarize your conclusion in a full sentence. Use the method of Lagrange multipliers.
Solution:
• If Pikachu outputs x megawatts of power and Zapdos outputs y megawatts of power, the total cost
to operate the generators is
C(x , y) = Cp(x) + CZ(y) = 230,000+
x2
100
+
y3
1200
.
• Team Rocket is contracted to output power at a rate of 880 megawatts so (x , y) is subject to the
contstraint f (x , y) = x + y = 880. Additionally x and y must both be non-negative.
• We wish to use Lagrange multipliers to minimize C(x , y) subject to the constraint f (x , y) = 880 on
the open set U = {(x , y) ∈ !2 : x , y > 0}.
• The gradients of C and f are
∇C(x , y) =
1
x
50
,
y2
400
2
and ∇ f (x , y) = (1,1).
• Suppose λ ∈ ! satisfies 1
x
50
,
y2
400
2
=∇C(x , y) = λ∇ f (x , y) = (λ,λ)
where f (x , y) = 880. Writing x50 = λ =
y2
400 and plugging into x + y = f (x , y) = 880 then yields
y2
8 + y − 880= 0.
• Solving the above quadratic equation yields y = 80 or y = −88 but y cannot be negative. This yields
(x , y) = (800,80) as a possible solution inside U with corresponding value C(800,80)≈ 236826.67.
• We need to check the boundary points (x , y) = (880,0) and (x , y) = (0,880). The corresponding
operating costs are C(880,0) = 237744 and C(0,880)≈ 797893.33 respectively.
• Comparing these values, we see that the minimummonthly operation costs are C(x , y) = 236826.67
dollars achieved when Pikachu outputs 800 megawatts of power and Zapdos outputs 80 megawatts
of power.
Remark: The boundary points (880,0) and (0,880) must be checked separately since (880,0), (0,880) ,∈ U .
Lagrange multipliers can only be used for functions and constraints defined on an open set U which does not
include these points in ∂ U on the constraint curve {(x , y) ∈ !2 : x + y = 880, x ≥ 0, y ≥ 0}.
MAT237 Term Test 2 solutions - Page 10 of 11 January 23, 2020
10. (6 points) Let m< n and let F : !n→ !m be differentiable. Let S = {x ∈ !n : F(x) = 0} and p ∈ S.
Prove that TpS ⊆ ker dFp.
Solution:
• Let v ∈ TpS.
• By definition of TpS this means that there exists a differentiable path γ : !→ !n with image lying in
S such that γ(0) = p and γ′(0) = v.
• Since S is the 0-level set of F and the image of γ lies in S, for any t ∈ !, (F ◦ γ)(t) = F(γ(t)) = 0.
• This means that the composition F ◦ γ is constant and hence has derivative always equal to zero.
• In particular, since F and γ are differentiable everywhere, differentiating at t = 0 and using the chain
rule yields
dFp(γ′(0)) = F ′(p)γ′(0) = F ′(γ(0))γ′(0) = (F ◦ γ)′(0) = 0
which says that v = γ′(0) ∈ ker dFp.
• Therefore TpS ⊆ ker dFp as required.
Remark: You cannot use D6 Theorem 1 or D7 Theorem 2 at all because those assumptions are not satisfied.
Notice you cannot assume F is C1 nor can you assume dFp is full rank. Incorrectly applying these theorems
significantly affects the rest of your proof so very little credit was awarded in this case. This question is almost
identical to Q7 on the Lecture D8 worksheet.
MAT237 Term Test 2 solutions - Page 11 of 11 January 23, 2020
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