ha lvh262 Homework 14 8 karakurt 56295 This print out should have 13 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Use the method of Lagrange multipliers to minimize f x y 2x2 6y 2 subject to the constraint x y 2 1 min value 6 correct 2 min value 27 4 6 min value 13 2 4x i 12y j i j After comparing coefficients this reduces to the pair of equations 4x 12y 1 i e y x But we still have the constraint 3 equation g x y x y 2 0 Substituting y 1 x gives 3 4 1 x x 2 x 2 0 3 3 f g g x y 0 occurs at 3 1 x y 2 2 4 no min value exists 25 4 the equation f g thus becomes Consequently the only solution of 23 3 min value 4 5 min value 1 and at this point 3 1 6 f 2 2 Explanation Set g x y x y 2 But is this a maximum or a minimum value There are several ways of deciding this algebraically and graphically Which to use depends on f and g Then by the method of Lagrange multipliers the extreme values of f under the constraint g 0 occur at solutions of Notice first that if we solve for y in the constraint equation then y 2 x and so on g 0 the function f becomes f g g x y 0 But when f x 2 x 2x2 6 2 x 2 f 4x i 12x j Its graph is a parabola opening upwards and such a parabola has a minimum value Of course we could have solved the problem by finding this minimum value with single variable calculus techniques without using Lagrange multipliers but the problem did ask us to use Lagrange multipliers g i j Graphically we could draw the contour map for f superimpose the graph of g 0 f x y 2x2 6y 2 we see that Since ha lvh262 Homework 14 8 karakurt 56295 and then look for points where the the contour lines for f are tangential to the graph of g 0 Alternatively we can note that the graph of 2 1 min value 2 2 min value 2 3 z f x y 2x2 6y 2 3 no min value exists is a paraboloid 4 min value z 3 correct 5 min value 1 Explanation Set g x y x y 2 y x Then by the method of Lagrange multipliers the extreme values of f under the constraint g 0 occur at the solutions of f g while the graph of But when g x y x y 2 0 in 3 space is a vertical plane Minimizing f on g 0 corresponds to finding the height of the lowest point on the intersection of this vertical plane with the paraboloid Since the intersection will be a parabola opening upwards again we see that f has a minimum on g 0 and this min value 6 keywords Lagrange multipliers optimization gradient quadratic function constraint constrained optimization 002 g x y 0 10 0 points f x y we see that f p Since p 3x2 y 2 y i p j 3x2 y 2 3x2 y 2 3x g i j the equation f g thus becomes 1 p 3x i y j i j 3x2 y 2 After comparing coefficients this reduces to the pair of equations p 3x 3x2 y 2 y p 3x2 y 2 Use the method of Lagrange multipliers to minimize p 3x2 y 2 f x y i e y 3x But we still have the constraint equation x y 2 x 3x 2 4x 2 0 subject to the constraint g x y x y 2 0 Substituting y 3x gives ha lvh262 Homework 14 8 karakurt 56295 Consequently the only solution of f g g x y 0 keywords Lagrange multipliers optimization gradient quadratic function constraint constrained optimization occurs at 1 3 x y 2 2 and at this point 1 3 3 f 2 2 3 003 10 0 points Finding the minimum value of f x y 2x y 1 subject to the constraint But is this a maximum or a minimum value We can decide algebraically or graphically the best choice depending on f and g Let s do it graphically because the graphs in 3 space of f and g 0 are easy to describe Indeed the graph of p z f x y 3x2 y 2 g x y 4x2 3y 2 3 0 is equivalent to finding the height of the lowest point on the curve of intersection of the graphs of f and g shown in z is a cone z x y Use Lagrange multipliers to determine this minimum value x y while the graph of g x y x y 2 0 in 3 space is a vertical plane Minimizing f on g 0 corresponds to finding the height of the lowest point on the intersection of this vertical plane with the cone Since the intersection will be half of a hyperbola opening upwards we see that f has a minimum on g 0 and this min value 3 1 min value 5 2 min value 4 3 min value 3 correct 4 min value 2 5 min value 6 Explanation The extreme values occur at solutions of f x y g x y Now f x y h 2 1 i ha lvh262 Homework 14 8 karakurt 56295 4 while 5 max value 10 g x y h 8x 6y i Explanation The extreme values of f subject to the constraint g 0 occur at solutions of Thus 2 8 x 1 6 y f x y g x y and so g x y 0 Now 1 1 4x 6y i e x 3 y 2 But 1 3 y y 12y 2 3 0 i e y g 2 2 Consequently the extreme points are 3 1 3 1 4 2 4 2 Since 3 1 1 f 4 2 while g x y h 2x 2y i Thus 4 2 x 3 1 f 3 4 2 3 2 y and so 2 3 x 2y 3 i e y x 4 But 9 3 g x x x2 x2 1 0 4 16 we thus see that min value 3 keywords 004 f x y h 4 3 i 10 0 points Determine the maximum value of f x y 4x 3y 3 i e x 4 5 Consequently the extreme points are 4 3 4 3 5 5 5 5 Since 4 3 f 8 5 5 we thus see that subject to the constraint 2 4 3 f 2 5 5 max value 8 2 g x y x y 1 0 1 max value 8 correct 2 max value 12 keywords 005 10 0 points 3 max value 9 If the method of Lagrange multipliers is used to maximize 4 max value 11 f x …