1. Partial derivatives of f (x, y) = xy − x2 y − xy 2 are fx = y − 2xy − y 2 and
fy = x − x2 − 2xy. Critical points are found from the system
y − 2xy − y 2 = 0
x − x2 − 2xy = 0
It is equivalent to
y(1 − 2x − y) = 0
x(1 − x − 2y) = 0
It follows from the first equation that either y = 0, or 1 − 2x − y = 0. In the first
case the second equation becomes x(1 − x) = 0, which gives us two solutions
(x, y) = (0, 0) and (x, y) = (1, 0).
Consider now the second case 1 − 2x − y = 0, i.e., 2x + y = 1.
It follows from the second equation x(1 − x − 2y) = 0 that either x = 0, or
1 − x − 2y = 0. In the first case we get y = 1, so that (x, y) = (0, 1). In the
second case we get the system
2x + y = 1
x + 2y = 1
Its solution is (x, y) = (1/3, 1/3).
We get six critical points (0, 0), (1, 0), (0, 1), (1/3, 1/3).
2
. We have fxx = −2y, fyy = −2x, fxy = 1 − 2x − 2y.
Let us find fxx fyy − fxy
2
Therefore, fxx fyy − fxy = 4xy − (1 − 2x − 2y)2 .
Its values at (0, 0), (1, 0), (0, 1), (1/3, 1/3) are −1, −1, −1, 4/9−(1−4/3)2 =
4/9 − 1/9 = 1/3. It follows that the points (0, 0), (1, 0) and (0, 1) are saddles,
and the point (1/3, 1/3) is a point of local maximum, since fxx (1/3, 1/3) =
−2/3 < 0.
2. Partial derivatives of f (x, y) = e−xy are fx = −ye−xy , fy = −xe−xy .
They are both equal to zero only at (x, y) = (0, 0), which is inside the region.
Let us use the method of Lagrange multipliers to find extremal points on
the boundary x2 + 4y 2 = 1 of the region. They are found from the system
x2 + 4y 2 = 1
−xy
h−ye
, −xe−xy i = λh2x, 8yi
or
 2
 x + 4y 2
−ye−xy

−xe−xy
= 1
= 2λx
= 8λy
It follows from the second equation that either x = 0, or
λ=−
y −xy
e
.
2x
In the case x = 0 we get 4y 2 = 1 from the first equation, but this contradicts
with −ye−xy = 2λx = 0, so this is impossible.
1
Substituting the expression for λ into the third equation of the system, we
get
4y
y,
x
and multiplying by x)
−xe−xy = −
or (dividing by −e−xy
x2 = 4y 2 .
√
√
Using x2 +4y 2 = 1, we get x2 = 4y 2 = 1/2, so that x = ±1/ 2 and y = ±1/2 2,
where all four combinations of sign are possible.
√
√
We
five special points to test: (0, 0), ±(1/ 2, 1/2 2), and
√ have obtained
√
±(1/ 2, −1/2 2). The √
corresponding
values √
of the function
are 1, e1/4 , and
√
√
−1/4
e
. It follows
√ that (1/√ 2, 1/2 2) and
√ (−1/√ 2, −1/2 2) are points of maximum, and (1/ 2, −1/2 2) and (−1/ 2, 1/2 2) are points of minimum.
RR
R e R y4
Re
Re
y4
3. We have D 1/x dA = 1 y2 1/x dx dy = 1 ln x|x=y2 dy = 1 (4 ln y −
Re
e
2 ln y) dy = 1 2 ln y dy = 2y ln y − 2y|1 = (2e − 2e) − (0 − 2) = 2.
4. The region of integration is the triangle with vertices (0, 0), (3, 0), (3, 1).
Therefore, we can write the integral as
3
Z
x/3
Z
0
2
ex dy dx =
0
Z
3
2
xex /3 dx =
0
Z
0
3
3
2
2
e9 − 1
ex /6 dx2 = ex /6 =
6
0
RR p
5. It is equal to the integral D 4x2 + 4y 2 + 1 dA, where D is the region
bounded by circle x2 + y 2 = 4 (equal to intersection of the paraboloid with the
xy-plane). Let us pass to polar coordinates to evaluate the integral. It is equal
to
Z 2π Z 2 p
Z 2√ 2
4r + 1
2
4r + 1 · r dr dθ = 2π
d(4r2 + 1) =
8
0
0
0
2
1
1 2 2
(4r + 1)3/2 = (173/2 − 1).
4 3
6
0
6. The region is described in spherical coordinates as 0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π,
0 ≤ φ ≤ π/2. The integral is then equal to
Z
2π
π/2
Z
Z
1
2
2
Z
2
ρ sin φ · ρ sin φ dρ dφ dθ = 2π
0
−
2π
5
0
Z
0
0
π/2
(1 − cos2 φ) d cos φ = −
cos φ −
2
3
Z
sin φ dφ
0
2π
5
π/2
1
ρ4 dρ =
0
π/2
cos3 φ 2π 2
4π
=
· =
.
3
5
3
15
0