# UT M 408C - HW 14.8-solutions-1 (8 pages)

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## HW 14.8-solutions-1

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## HW 14.8-solutions-1

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Pages:
8
School:
University of Texas at Austin
Course:
M 408c - Differential and Integral Calculus
##### Differential and Integral Calculus Documents

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lewis scl876 HW 14 8 radin 55315 This print out should have 11 questions Multiple choice questions may continue on the next column or page find all choices before answering 1 while g x y h 8x 6y i Thus 001 10 0 points 2 8 x Finding the maximum value of and so f x y 2x y 1 subject to the constraint g x y 4x2 3y 2 3 0 is equivalent to finding the height of the highest point on the curve of intersection of the graphs of f and g shown in 1 1 4x 6y 3 y 2 Consequently the extreme points are 3 1 4 2 Since 3 1 f 3 4 2 y i e x But 1 3 g y y 12y 2 3 0 i e y 2 2 z x 1 6 y 3 1 4 2 3 1 f 1 4 2 we thus see that max value 3 Use Lagrange multipliers to determine this maximum value 1 max value 6 2 max value 5 keywords 002 10 0 points Use Lagrange Multipliers to determine the maximum value of 3 max value 7 f x y 6xy 4 max value 3 correct subject to the constraint 5 max value 4 Explanation The extreme values occur at solutions of f x y g x y Now g x y x2 y 2 1 0 9 16 1 maximum 36 correct 2 maximum 34 f x y h 2 1 i 3 maximum 37 lewis scl876 HW 14 8 radin 55315 2 we thus see that f has 4 maximum 33 max value 36 5 maximum 35 Explanation By the Method of Lagrange multipliers the extreme values of f occur at the common solutions of f x y g x y subject to the constraint g x y 0 keywords 003 g x y 0 Now f x y h 6y 6x i The temperature T at a point x y z on the surface x2 y 2 z 2 48 while g x y D2 1 E x y 9 8 is given by T x y z x y z But then by the condition on f and g 6y 2 x 9 6x 1 y 8 which after simplification gives 48x 27y x y 4 i e y x 3 Thus by the constraint equation x2 in degrees centigrade Find the maximum temperature on this surface Correct answer 12 Explanation We have to maximize the function T subject to the constraint x2 y 2 z 2 48 x2 4 g x x 1 0 3 9 9 3 2 i e x Consequently the extreme 2 points are 3 2 3 2 2 2 2 2 2 2 The corresponding Lagrange function is F x y z x y z x2 y 2 z 2 48 At the critical point of F therefore Fx 1 2 x 0 Fy 1 2 y 0 and 3 2 2 2

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