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1. Partial derivatives of f(x, y) = xy− x2y− xy2 are fx = y− 2xy− y2 andfy = x− x2 − 2xy. Critical points are found from the system{

y − 2xy − y2 = 0x− x2 − 2xy = 0

It is equivalent to {y(1− 2x− y) = 0x(1− x− 2y) = 0

It follows from the first equation that either y = 0, or 1−2x−y = 0. In the firstcase 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 thesecond case we get the system{

2x+ y = 1x+ 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).Let us find fxxfyy−f2xy. We have fxx = −2y, fyy = −2x, fxy = 1−2x−2y.

Therefore, fxxfyy − f2xy = 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 onthe boundary x2 + 4y2 = 1 of the region. They are found from the system{

x2 + 4y2 = 1〈−ye−xy,−xe−xy〉 = λ〈2x, 8y〉

or x2 + 4y2 = 1−ye−xy = 2λx−xe−xy = 8λy

It follows from the second equation that either x = 0, or

λ = − y

2xe−xy.

In the case x = 0 we get 4y2 = 1 from the first equation, but this contradictswith −ye−xy = 2λx = 0, so this is impossible.

1

Substituting the expression for λ into the third equation of the system, weget

−xe−xy = −4y

xy,

or (dividing by −e−xy and multiplying by x)

x2 = 4y2.

Using x2+4y2 = 1, we get x2 = 4y2 = 1/2, so that x = ±1/√

2 and y = ±1/2√

2,where all four combinations of sign are possible.

We have obtained five special points to test: (0, 0), ±(1/√

2, 1/2√

2), and±(1/

√2,−1/2

√2). The corresponding values of the function are 1, e1/4, and

e−1/4. It follows that (1/√

2, 1/2√

2) and (−1/√

2,−1/2√

2) are points of max-imum, and (1/

√2,−1/2

√2) and (−1/

√2, 1/2

√2) are points of minimum.

3. We have∫∫D

1/x dA =∫ e1

∫ y4y2

1/x dx dy =∫ e1

lnx|y4

x=y2 dy =∫ e1

(4 ln y −2 ln y) dy =

∫ e1

2 ln y dy = 2y ln y − 2y|e1 = (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

0

∫ x/3

0

ex2

dy dx =

∫ 3

0

xex2

/3 dx =

∫ 3

0

ex2

/6 dx2 = ex2

/6∣∣∣30

=e9 − 1

6

5. It is equal to the integral∫∫D

√4×2 + 4y2 + 1 dA, where D is the region

bounded by circle x2 + y2 = 4 (equal to intersection of the paraboloid with thexy-plane). Let us pass to polar coordinates to evaluate the integral. It is equalto ∫ 2π

0

∫ 2

0

√4r2 + 1 · r dr dθ = 2π

∫ 2

0

√4r2 + 1

8d(4r2 + 1) =

1

4

2

3(4r2 + 1)3/2

∣∣∣∣20

=1

6(173/2 − 1).

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

∫ 2π

0

∫ π/2

0

∫ 1

0

ρ2 sin2 φ · ρ2 sinφ dρ dφ dθ = 2π

∫ π/2

0

sin3 φ dφ

∫ 1

0

ρ4 dρ =

− 2π

5

∫ π/2

0

(1− cos2 φ) d cosφ = −2π

5

(cosφ− cos3 φ

3

)∣∣∣∣π/20

=2π

5· 2

3=

4π

15.

2