Midterm IV November 26, 2009
No Calculators, books or notes. Only your number 2 pencil.
I. Use Lagrange multipliers to find the maximum and minimum values of the function f (x, y, z) =
xyz on the surface 3x2 + y 2 + 2z 2 = 6.
Let g(x, y, z) = 3x2 + y 2 + 2z 2 . Then ∇f = hyz, xz, xyi, ∇g = h6x, 2y, 4zi and ∇f = λ∇g gives the
Lagrange multiplier equations:
yz = λ6x
xz = λ2y
xy = λ4z
The equations give:
xyz = λ6x2
xyz = λ2y 2
xyz = λ4z 2
There are two cases to consider. The first case
is λ 6= 0.√Then
3x2 = y 2 = 2z 2 and the equation of
√
√
the surface gives y 2 + y 2 + y 2 = 6. So y = ± 2,
2/ 3 and z = ±1 giving eight points on
√x =
√± √
the surface where extreme values can occur, (± 2/ 3, ± 2, ±1).
The second case is λ = 0. The solutions of ∇f = 0∇g are the critical points of f , the points where
yz = 0, xz = 0 and xy = 0 which are the points on the three coordinate axes.
√ This gives 6√more
2
2
2
points on
√ the surface 3x + y + 2z = 6 where extreme values may occur, (± 2, 0, 0), (0, ± 6, 0),
(0, 0, ± 3).
√
√
The √
values of f (x, y, z) = xyz on the the
√ 14 points are 2/ 3, −2/ 3 and 0. The maximum value
is 2/ 3 and the minimum value is −2/ 3.
II. Find the absolute maximum and minimum of the function f (x, y) = x2 − y 2 on the disk (x −
1)2 + y 2 ≤ 4.
An extreme values of f (x, y) must occur at a critical point of f or at a point on the edge of
the region. The gradient of f is ∇f (x, y) = h2x, −2yi and the only critical point is (0, 0). Let
g(x, y) = (x − 1)2 + y 2 . Extreme values on the edge of the region occur at points that satisfy the
Lagrange multiplier equations ∇f = λ∇g. Now ∇g = h2(x − 1), 2yi and the Lagrange multiplier
equations are
2x = λ2(x − 1)
−2y = λ2y
2
(x − 1) + y 2 = 4
The second equation yields two cases: The first case is y = 0. Then (x − 1)2 = 4 and so x = 3 or
x = −1. This gives two points (3, 0) and (−1, 0) where the extreme values can occur. The √
second
2
case is y 6= 0 giving λ = −1 and 2x = −2(x − 1). Then
x
=
1/2,
1/4
+
y
=
4
and
y
=
±
15/2.
√
The extreme values could occur at the points (1/2, ± 15/2).
Next evaluate
f (x, y) = x2 − y 2 on the five test points. f (0, 0) = 0, f (3, 0) = 9, f (−1, 0) = 1 and
√
f (1/2, ± 15/2) = 1/4 − 15/4 = −14/4 = −7/2. The maximum value is 9 and the minimum value
is −7/2.
III. Evaluate the double integral
Z Z
x cos(y) dA
D
where D is the region bounded by y = 0, x = 1 and y = x2 .
The region is both type I and type II. As type I, the region is 0 ≤ x ≤ 1, 0 ≤ y ≤ x2 and the double
integral is
Z 1
Z 1
Z 1
Z 1 Z x2
2
x2
x(sin(x ) − sin(0)) =
x sin(x2 )
x cos(y) dy dx =
x sin(y) |0 =
0
0
0
0
0
2
Then using the substitution u = x ,
1
Z
1 1
1
1
x sin(x ) =
sin u = − cos(x2 ) |10 = (1 − cos(1)
2
2
2
0
√ 0
As Type II, the region is 0 ≤ y ≤ 1, y ≤ x ≤ 1 and the double integral is
Z 1Z 1
Z 1
Z 1
1
2
1
(1 − y) cos(y))
x cos(y) dx dy =
(x /2) cos(y) |√y =
√
0
y
0
0 2
Z
2
which is harder to evaluate but gives the same answer.
IV. Do NOT evaluate the integral. (A diagram is not required but is highly recommended) Give a
single iterated integral that computes the area of the region bounded by y = x/2, y = 1/x − 1/2
and the x-axis.
Begin with one of the basic properties of double integrals: the area of a closed bounded region D is
the double integral over D of the constant function 1,
Z Z
dA.
D
In this case, D is the Type II region given by
0 ≤ y ≤ 1/2
2
2y ≤ x ≤ 2y+1
So the area is
Z
1/2
Z
2/(2y+1)
1 dA
0
2y
(There was a typo in the original version of this problem.)
V. Interchange the order of integration, expressing the answer as an iterated integral.
Z 9 Z √9−y
f (x, y) dx dy
0
In the other order
Z
0
3
Z
9−x2
f (x, y) dy dx
0
0