# 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 2 F x2 y 2 z 2 48 0 Thus 3 2 3 2 2 2 f 2 2 36 2 2 while f Fz 1 2 z 0 3 2 2 2 2 Since f 10 0 points 3 2 3 2 2 2 f 2 2 36 2 2 x y z p 48 3 4 Substituting in the original definition of T we see that Tmax 12 C lewis scl876 HW 14 8 radin 55315 004 10 0 points A rectangular box with edges parallel to the axes is inscribed in the ellipsoid 6x2 2y 2 4z 2 1 similar to the one shown in 3 so we have to maximize V subject to this last restriction on x y and z To use Lagrange multipliers set F x y z 8xyz 6x2 2y 2 4z 2 1 Then F will have a critical point when 8yz 12 x 8zx 4 y 8xy 8 z and 6x2 2y 2 4z 2 1 From the first set of equations we see that 8xyz 12 x2 8yzx 4 y 2 8zxy 8 z 2 in which case 12x2 4y 2 8z 2 From the fact that P lies on the ellipsoid therefore it thus follows that F has a critical point at Use Lagrange multipliers to determine the maximum volume of this box Note all 8 vertices of the box will lie on the ellipsoid when the volume is maximized 1 volume 2 volume 3 volume 1 6 2 9 2 3 1 3 1 9 x2 1 18 y2 1 6 1 12 At this critical point V will be maximized Consequently the box has cu units maximum volume cu units correct cu units z2 005 2 cu units 9 10 0 points cu units If the method of Lagrange multipliers is used to maximize cu units f x y 3xy Explanation The rectangular box will be centered at the origin so if the corner lying in the first octant is P x y z then the box will have sidelengths 2x 2y and 2z Thus the box will have volume subject to the constraint x y 8 which of the following is true V 8xyz III The maximum value of f x y occurs when y 4 4 volume 5 volume But P lies on the ellipsoid 6x2 2y 2 4z 2 1 I The maximum value of f x y is less than or equal to 49 II The maximum value of f x y occurs when x 8 1 III only lewis scl876 HW 14 8 radin 55315 2 II and III only 4 5 maximum 4 correct Explanation By the Method of Lagrange multipliers the extreme values occur at the common solutions of 3 I and III only correct 4 I only f x y g x y g x y 0 5 I II and III Now Explanation The method of Lagrange multipliers looks for the critical points of the Lagrange function F x y 3xy x y 8 Now these critical points will be the common solutions of the equations 1 f x 3 4 y 3 4 x 2 Thus f x y D1 2 x 3 4 y 3 4 3 1 4 1 4 E x y 2 On the other hand Fx 3y 0 Fy 3x 0 and f 1 x 3 4 y 3 4 y 2 g x y h 1 3 i But then by the condition on f and g F x y 8 0 Thus there is only one set of solutions and these are x y 4 3x 12 Since f 4 4 48 we see that I and III are the correct statements 006 10 0 points 1 3 4 3 4 x y 2 which after simplification gives 1 3 4 3 4 1 x y x1 4 y 1 4 2 2 i e y x Thus by the constraint equation g x x x 3x 8 0 i e x 2 Consequently Determine the maximum value of f x y 2x 3 1 4 1 4 x y 3 2 2 2 1 4 3 4 y subject …

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