2x+4=lambda(2x) 2y-4=lambda(2y) (x^2)+(y^2)=1 Solve for x and y

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

2x+4=lambda(2x) 2y-4=lambda(2y) (x^2)+(y^2)=1 Solve for x and y

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Yay for La Grange multipliers! I just took this test this morning.... So if we isolate x and y in terms of lambda, we can solve for lambda in the third equation and then find x and y using the first two equations again... So the first equation becomes: x( 1- lambda) + 2 = 0 -> x= -2/(1- lambda) The second equation becomes: y ( 1- lambda) -2 = 0 -> y= 2/(1 - lambda) Then if we use these x and y values in the third equation we get: 4/(1- lambda)^2 + 4/(1- lambda)^2 = 1 Solve for lambda and we get : lambda= 1- sqrt(8) Since we now know lambda, we can sub that value back into the equations where we isolated x and y in terms of lambda, and we get: x= -(sqrt2)/2 y= (sqrt 2)/2

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Not the answer you are looking for?

Search for more explanations.

Ask your own question