微乙小考三 (2016/4/14)
1. (5分) 求 f (x, y) = x2 y 在曲線 C : x2 + 2y 2 = 6 上的最大值與最小值。
sol: Let g(x, y) = x2 + 2y 2 − 6. Using the method of Lagrange multipliers, we have




2xy = 2λx


 ∇f = λ∇g
⇒
x2 = 4λy

 x2 + 2y 2 = 6


 x2 + 2y 2 = 6
√
Solving it we obtain (x, y) = (0, ± 3), (±2, 1), (±2, −1), which leads to
√
f (0, ± 3) = 0 , f (±2, 1) = 4 (max),f (±2, −1) = −4 (min).
2. (7分) 求 f (x, y) = x2 + 2y 2 在圓盤 D = {x2 + y 2 ≤ 1} 上的最大值與最小值。
sol:
1. We first find the extreme values of f (x, y) = x2 + 2y 2 on R2 .
(a) ∇f (x, y) = 0 gives (x, y) = (0, 0)
(b) D(0, 0) = fxx (0, 0)fyy (0, 0) − (fxy (0, 0))2 = 8 > 0 and fxx (0, 0) = 2 > 0
By (a) and (b), we see that (0, 0) is the minimum with f (0, 0) = 0 on R2 .
2. We find the extreme values of f (x, y) = x2 + 2y 2 on the circle g(x, y) = x2 + y 2 − 1 = 0:

 ∇f = λ∇g
⇒ (x, y) = (0, ±1), (±1, 0)
 x2 + y 2 = 1
which gives us f (0, ±1) = 2, f (±1, 0) = 1
By (1) and (2), we conclude that the extreme values of f (x, y) on the disk D : x2 + y 2 ≤ 1
is 2 and 0.
3. (4分) 計算
RR
T
xydA, T 是一個三角形，三頂點座標為 (0, 0) , (1, 0) , (1.1).
sol:
Z Z
1
Z
1
xydydx
y=x
1 2 xy dx
2
y=0
Z
1
Z
xydA =
T
0
Z
0
=
0
1
=
2
1
=
8
1
x
0
x3 dx
4. (4分) 計算
RR
x2 +y 2 ≤1
ln (x2 + y 2 ) dA.
(提示: 利用極座標轉換)
sol: Let x = r cos θ, y = r sin θ
Z Z
2
2π
Z
2
ln(x + y )dA =
x2 +y 2 ≤1
0
Z
Z
1
ln(r2 ) · rdrdθ
0
2π Z 1
1
ln(r2 )dr2 dθ
2 0
0
Z 2π
1
1
2
2
2 =
r ln r − r dθ
2 0
0
=
= −π
2