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M 408C
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Gary, Berg
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Mathematics

M 408C

Gary, Berg

Fall

Description

taboada (lat2278) – HW12 – berg – (56260) 1
This print-out should have 18 questions.
Multiple-choice questions may continue on 1. maximum = 7
the next column or page – ﬁnd all choices
before answering. 2. maximum = 10
001 10.0 points 3. maximum = 8
4. maximum = 9
Finding the maximum value of
f(x, y) = x + 2y + 1 5. maximum = 11
subject to the constraint
003 10.0 points
g(x, y) = 2x + y − 2 = 0
Use Lagrange Multipliers to determine the
maximum value of
is equivalent to ﬁnding the height of the high-
est point on the curve of intersection of the
graphs of f and g shown in f(x, y) = 2xy
z
subject to the constraint
x2 y2
g(x, y) = + − 1 = 0.
1 4
1. maximum = 6
x
y 2. maximum = 5
3. maximum = 2
Use Lagrange multipliers to determine this
4. maximum = 4
maximum value.
1. max value = 3 5. maximum = 3
004 10.0 points
2. max value = 0
3. max value = 1 The temperature T at a point (x,y,z) on
the surface
4. max value = 2
x + y + z = 75
5. max value = 4
is given by
002 10.0 points
Determine the maximum value of T(x,y,z) = x + y + z
f(x, y) = 4x 1/y 3/4
in degrees centigrade.Find the maximum
subject to the constraint temperature on this surface.
g(x, y) = x + 3y − 8 = 0. 005 10.0 points taboada (lat2278) – HW12 – berg – (56260) 2
A rectangular box with edges parallel to the
3
axes is inscribed in the ellipsoid 3. volume =
8
6x + 2y + 9z 2 = 1
4. volume = 1
24
similar to the one shown in
1
5. volume =
16
007 10.0 points
The graph of g(x, y) = 0 is shown as a
dashed line in
-1 -2 -3 -3 -2 -10
y 1
2
3
Use Lagrange multipliers to determine the
x
maximum volume of this box. P
Note: all 8 vertices of the box will lie on the
ellipsoid when the volume is maximized.
3
4
1. volume = cu. units 2
9 1
2
2. volume = 27 cu. units
1 while the level curves f(x, y) = k for a func-
3. volume = 9 cu. units tion z = f(x, y) are shown as continuous
4. volume = 2 cu. units curves with values of k listed at the edges.
9 If g(P) = 0, which one of the following
properties does f have?
5. volume = 4 cu. units
27
1. neither local max nor local min at P
006 10.0 points
Use Lagrange multipliers to ﬁnd the volume 2. a local min at P
of the largest rectangular box in the ﬁrst oc-
tant with three faces in the coordinate planes 3. a local max at P
and one vertex in the plane
008 10.0 points
1x + 4y + 2z = 3.
Find the value of the integral
Z 1
1 I = f (x, y) dx
1. volume = 4
0
1 when
2. volume = 2
8 f(x, y) = 4x − x y . taboada (lat2278) – HW12 – berg – (56260) 3
1
1. I = 2 − y 5. I = 2
3
2. I = 4y − y 2 011 10.0 points
Evaluate the iterated integral
1 2
3. I = 4y − 2y Z Z
2 4 1
I = 2dxdy .
4. I = 2 − y 1 0 (x + y)

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