MATH 105 101
Assignment 2 Solutions
Due date: October 2, 2014
MATH 105 101 Assignment 2 Solutions
All work must be shown for full marks.
1. Find both
first-order partial derivatives of the function g(x, y) = y sin (x
√
point ( 2π , 49). Simplify your answers. (4 marks)
2− π )
12
at the
Solution: Compute the first-order partial derivatives:
π
ln(y) cos(x2 − )2x
12
π sin(x2 − π )−1
2
12
gy (x, y) = sin(x − )y
12
gx (x, y) = y sin(x
2− π )
12
√
Evaluate at the point (
π
, 49):
2
√ √
π
π
π
π
π
sin( π4 − 12
)
gx (
, 49) = 49
ln(49) cos( − )2
2
4 12
2
√
π
π
= 49sin( 6 ) ln(49) cos( ) π
6
√ !
1
3 √
= 49 2 ln(49)
π
2
√
√
7 3π
ln(49) = 7 3π ln(7)
=
2
√
π
π
π
π
π
gy (
, 49) = sin( − )49sin( 4 − 12 )−1
2
4 12
π sin( π )−1
= sin( )49 6
6
1 −1
1
= 49 2 = .
2
14
2. Let f (x, y) be a two-variable function defined on R2 such that all first-, second- and
third- order partial derivatives are continuous on R2 . Suppose that at the point (3, 2),
the third order partial derivatives of f satisfy the equation:
2(fxyx (3, 2))2 − 5fyxx (3, 2) = 7
Clearly state any theorem that is used.(4 marks)
Solution: Since the second order partial derivatives of f are continuous, by Clairaut’s
Theorem, fxy = fyx . Therefore, fxyx = (fxy )x = (fyx )x = fyxx . On the other
Page 1 of 4
MATH 105 101
Assignment 2 Solutions
Due date: October 2, 2014
hand, consider the function fx , which is also a two-variable function defined on
R2 . The second-order partial derivatives of fx are the third-order partial derivatives of f , which are continuous. So, it follows from Clairaut’s Theorem that
(fx )yx = (fx )xy , which is equivalent to fxyx = fxxy . Thus, fxyx = fxxy = fyxx . Let
A = fxyx (3, 2) = fxxy (3, 2) = fyxx (3, 2). Then,
2A2 − 5A − 7 = 0
(2A − 7)(A + 1) = 0
A = 3.5, −1
So, the possible values for fxxy (3, 2) are 3.5 and −1.
3. Find and classify all critical points of the function f (x, y) = x3 − 12x + y 3 + 3y 2 − 9y.
(6 marks)
Solution: We have the domain of f is Df = {(x, y) ∈ R2 }, and the first order
partial derivatives are:
fx (x, y) = 3x2 − 12 = 3(x − 2)(x + 2)
fy (x, y) = 3y 2 + 6y − 9 = 3(y + 3)(y − 1)
Since both first-order derivatives are defined at every point in R2 , if (a, b) is a
critical point of f , then fx (a, b) = fy (a, b) = 0. We have that fx (x, y) = 0 if
x = ±2, and fy (x, y) = 0 if y = −3, 1. So, we get the following four critical points:
(2, −3), (2, 1), (−2, −3) and (−2, 1).
Compute the second-order partial derivatives and discriminant:
fxx (x, y) = 6x
fyy (x, y) = 6y + 6
fxy (x, y) = fyx (x, y) = 0
D(x, y) = (6x)(6y + 6) = 36x(y + 1)
Evaluate the discriminant and fxx at each of those points:
D(2, −3) = −144
D(2, 1) = 144
D(−2, −3) = 144
D(−2, 1) = −144
fxx (2, −3) = 12
fxx (2, 1) = 12
fxx (−2, −3) = −12
fxx (−2, 1) = −12
By the Second Derivative Test, (2, −3) and (−2, 1) are saddle points, (2, 1) is a
local minimum, and (−2, −3) is a local maximum.
Page 2 of 4
MATH 105 101
Assignment 2 Solutions
Due date: October 2, 2014
1 3
2
4. Find the absolute maximum and minimum values of the function f (x, y) = e− 3 x +x−y
in the square S = {(x, y) ∈ R2 : −2 ≤ x ≤ 2, −2 ≤ y ≤ 2}. Indicate those points at
which f attains those values. (6 marks)
Solution: We first find the critical points of f inside the square. We have that
the domain of f is Df = R2 , and the first order partial derivatives are:
1 3
+x−y 2
fx (x, y) = e− 3 x
(−x2 + 1)
1 3
+x−y 2
fy (x, y) = −e− 3 x
2y
Since both partial derivatives are defined on all of R2 , the critical points of f are
such that fx = fy = 0 at that point. We have fx = 0 for x = ±1, an fy = 0 for
y = 0. So, there are two critical points (1, 0) and (−1, 0), both of which lie in the
square.
We now find the maximum and minimum of f on the boundary. We will consider
each side of the square separately:
1 3
• The top side is y = 2 for −2 ≤ x ≤ 2, then g(x) = f (x, 2) = e− 3 x +x−4 . So,
1 3
g 0 (x) = e− 3 x +x−4 (−x2 + 1), which equals 0 when x = ±1. Thus, we get two
points (1, 2) and (−1, 2), and two end points (−2, 2) and (2, 2).
• The bottom side is y = −2 for −2 ≤ x ≤ 2, then g(x) = f (x, −2) =
1 3
e− 3 x +x−4 , which has two critical points x = ±1 as we solved above. Thus,
we also get two points (1, −2) and (−1, −2), and two end points (−2, −2)
and (2, −2).
2
2
• The left side is x = −2 for −2 ≤ y ≤ 2, then h1 (y) = f (−2, y) = e 3 −y .
2
2
So, h01 (y) = −e 3 −y (2y) = 0 for y = 0. We get (−2, 0) and two end points
(−2, −2) and (−2, 2).
2
2
• The right side is x = 2 for −2 ≤ y ≤ 2, then h2 (y) = f (2, y) = e− 3 −y . So,
2
2
h02 (y) = −e− 3 −y (2y) = 0 for y = 0. Again, we get (2, 0) and two end points
(2, −2) and (2, 2).
Evaluate f at all points obtained above:
2
2
f (−1, 0) = e− 3
f (1, 0) = e 3
2
2
f (2, 0) = e− 3
f (−2, 0) = e 3
10
f (−1, 2) = e− 3
10
f (2, 2) = e− 3
10
f (−1, −2) = e− 3
14
f (−2, −2) = e− 3
14
f (1, 2) = e− 3
14
f (−2, 2) = e− 3
14
f (1, −2) = e− 3
10
f (2, −2) = e− 3
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MATH 105 101
Assignment 2 Solutions
Due date: October 2, 2014
2
Thus, f attains the maximum value of e 3 at (1, 0) and (−2, 0) and the minimum
14
value of e− 3 at (−1, 2), (2, 2), (−1, −2) and (2, −2).
5. Find the largest area of a rectangle that can be inscribed in the ellipse x2 + 2y 2 = 1,
and that its sides are parallel to the axes. (5 marks)
Solution: The area of such a rectangle is given by the function f (x, y) = 4xy,
where (x, y) is a point on the ellipse x2 + 2y 2 = 1 and x ≥ 0, y ≥ 0. Using Lagrange
Multiplier method to maximize f with respect to g(x, y) = x2 + 2y 2 − 1 = 0, we
want to solve the following system:
∇f (x, y) = λ∇g(x, y)
g(x, y) = 0
So, we obtain the following system of equations:
4y = 2λx
4x = 4λy
2
x + 2y 2 = 1
Substituting the second equation x = λy into the first equation, we √
get 2y = λ2 y,
2
2
which can be factored into y(2 − λ ) = 0. So, either y = 0 or λ = ± 2. Consider
three cases:
• If y = 0, then by the second equation, we get x = 0, but x = y = 0 does not
satisfy the last equation. So, there is no solution in this case.
√
√
• If λ = 2, then the second equation gives x = 2y. Substituting that into
the third equation, we get 4y 2 = 1, which gives y = ± 12 . Thus, x = ± √12 . So,
−1 −1
we get two points ( √12 , 12 ) and ( √
, ).
2 2
√
√
• If λ = − 2, then the second equation gives x = − 2y. Substituting that
into the third equation, we get 4y 2 = 1, which gives y = ± 12 . Thus, x = ∓ 2√1 2 .
−1 1
So, we get two points ( √
, ) and ( √12 , −1
).
2
2 2
Compare the values of f at the four points we found, we get that f attains its
−1 −1
, ).
maximum of √22 at the points ( √12 , 21 ) and ( √
2 2
Total: 25 marks.
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