use method of lagrange multipliers to determine which point on the surface 1/x +
ID: 3082166 • Letter: U
Question
use method of lagrange multipliers to determine which point on the surface 1/x +1/y +1/z =1 with x,y,z >0 is closest to the origin please solve it clearlyBYUSING LAGRANGE MULTIPLIERSASAP THANKSExplanation / Answer
(If the coords can be complex-valued), then z=i, x=y=0 actually gives a shorter distance than z=0, x=y=±1 This minimizes d at d=1 (instead of v2 for husoski) Obviously z is complex-valued. If z must be real then it's going to be x=y=±1, z=0 You didn't say which. [Both the following methods prove there is no global minimum, thus the minimum d occurs at the endpoint. Justification: it looks like symmetry requires the solution to have x=y. Then d² = x² + y² + z² = x² + y² + xy-1 becomes d² = 3x²-1 d = v(3x²-1) ?d/?x = 6x / 2v(3x²-1) = 3x / v(3x²-1) ?d/?x=0 => x=0 z² = xy-1 then means x=y=0 * Method of Lagrange multipliers, which locates the the global minimum as z=i, x=y=0, thus again there is no real-valued global minimum. to prove this is Note also we can eliminate variable z entirely as follows Call the distance d(x,y,z) We must have that d² = x² + y² + z² and with the constraint z² = xy-1 => d² = x² + y² + (xy-1) d² = x² + y² + 2xy -xy -1 d² = (x+y)² -xy -1 d = v(x² + y² + (xy-1)) ?d = (?d/?x, ?d/?y) = (2x+y, 2y+x) / 2v(x² + y² + xy-1) Use Lagrange multipliers: d = v(x² + y² + xy-1) - ?(xy-1 -z²) ?d = (?d/?x, ?d/?y) = (2x+y, 2y+x) / 2v(x² + y² + xy-1) - (?y, ?x) ?d/?? always just returns the constraint, so we ignore it. Solve ?d = 0 => solve ?d/?x =0 and note the local minimum (if it exists) must be symmetric in x and y: (2x+y, 2y+x) / 2v(x² + y² + xy-1) = (?y, ?x) => (2x+y) / 2v(x² + y² + xy-1) = ?y (2x+y) = 2?y v(x² + y² + xy-1) (2x+y) = 2?y v(x² + y² + xy-1) The only symmetric solution (?d/?x, ?d/?y) = (0,0) must have x=y=0, (?= don't-care) This then gives z=1 (from the constraint)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.