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Lecture

Calc Hw 12.pdf

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Department
Mathematics
Course
M 408C
Professor
Gary, Berg
Semester
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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