A community for students.
Here's the question you clicked on:
 0 viewing
richyw
 4 years ago
Find the points on \(x^2+xy+y^2=3\) which are closest and farthest from \((0,0)\)
richyw
 4 years ago
Find the points on \(x^2+xy+y^2=3\) which are closest and farthest from \((0,0)\)

This Question is Closed

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.2Hmm... What sort of class are you doing this problem for?

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0refresher course in calculus.

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0this is a multivariable optimization problem.

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.2Ah. I had a feeling it was calc. ;P I was thinking of maybe using rotation of axes and trying to use conic section properties. I'm not as familiar with the multivar calc methods I saw before.

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0Would you be able to do it without calc? Any method that get's the answer is fine by me haha.

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.2It's been a while since I did rotation of axes, but what it basically does is removes the obnoxious 'xy' term by using more obnoxious trig. With any luck though, we could possibly get a nice conic section like an ellipse, where we can identify the major and minor axes for those maximum/minimum distances

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.2This PDF has some pretty sweet pictures and explanation, I thought it was really helpful: http://www.stewartcalculus.com/data/CALCULUS%20Early%20Vectors/upfiles/RotationofAxes.pdf

cruffo
 4 years ago
Best ResponseYou've already chosen the best response.0I have a suggestion that may not require much more than calc 1...

cruffo
 4 years ago
Best ResponseYou've already chosen the best response.0We want to maximize and minimize the distance between the curve and the point. \[\large \text{Distance}^2 = (x0)^2 + (y0)^2 = x^2 + y^2\] problem is that the curve is not easily solved for y. But If you use quadratic formula and solve for y thus, \[\large y^2 +xy + (x^2 3) = 0\] you may have some luck..

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0hmm. that could work. but will keep bumping for a calc answer! Thanks for all your help though.

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.2Alright, I got my method to work out, although it is quite long and tedious. lol  w = r cos t z = r sin t x = r cos(t + p) y = r sin(t + p) x = r cos t cos p  r sin t sin p y = r sin t cos p + r cos t sin p x = w cos p  z sin p y = z cos p + w sin p (w cos p  z sin p)^2 + (w cos p  z sin p)(z cos p + w sin p) + (z cos p + w sin p)^2 = 3 w^2 cos^2 p  2 wz sin p cos p + z^2 sin^2 p + wz cos^2 p + w^2 sin p cos p  z^2 sin p cos p  wz sin^2 p + z^2 cos^2 p + 2 wz cos p sin p + w^2 sin^2 p = 3 w^2 (cos^2 p + sin p cos p + sin^2 p) + wz(2sin p cos p + cos^2 p  sin^2 p + 2sin p cos p) + z^2 (sin^2 p  sin p cos p + cos^2 p) = 3 w^2 (1 + sin p cos p) + wz(cos^2 p  sin^2 p) + z^2(1  sin p cos p) = 3 w^2 (1 + sin p cos p) + wz cos(2p) + z^2 (1  sin p cos p) = 3 w^2 (1 + 1/2 sin(2p)) + wz cos(2p) + z^2 (1  1/2 sin(2p)) = 3 cos(2p) = 0 2p = pi/2 p = pi/4 w^2 (1 + 1/2 sin 2(pi/4)) + wz cos(2 pi/4) + z^2 (1  1/2 sin(2 pi/4)) = 3 w^2 (1 + 1/2) + 0wz + z^2(1  1/2) = 3 3/2 w^2 + 1/2 z^2 = 3 1/2 w^2 + 1/6 z^2 = 1 Our rotation of axes doesn't affect (0, 0)'s position, so we can ignore messing with it. sqrt(2) is 1/2 the length of our minor axis, and it is along the xaxis. (+ sqrt(2), 0) are our minor vertices. sqrt(6) is 1/2 the length of our major axis, and it is along the yaxis. (0, + sqrt(6) are our major vertices. So, the last thing to do now is convert back to the original system. (w, z) = (sqrt(2), 0) x = sqrt(2) cos(pi/4)  0 sin(pi/4) y = 0 cos(pi/4) + sqrt(2) sin (pi/4) x = sqrt(2)* sqrt(2)/2 = 1 y = sqrt(2) sqrt(2)/2 = 1 (1, 1) is the minimum point, along with (1, 1). (w, z) = (0, sqrt(6)) x = 0 cos(pi/4)  sqrt(6) sin(pi/4) y = sqrt(6) cos(pi/4) + 0 sin(pi/4) x = sqrt(6) * sqrt(2)/2 = sqrt(3) y = sqrt(6) * sqrt(2)/2 = sqrt(3) (sqrt(3), sqrt(3)), and (sqrt(3), sqrt(3)) Conclusion: Points closest to (0, 0): (1, 1) and (1, 1). Points farthest from (0, 0): (sqrt(3), sqrt(3)) and (sqrt(3), sqrt(3)). 

cruffo
 4 years ago
Best ResponseYou've already chosen the best response.0Got the same thing using the same method :)

richyw
 4 years ago
Best ResponseYou've already chosen the best response.0holy! thanks for putting in that effort. I really appreciate it.

AccessDenied
 4 years ago
Best ResponseYou've already chosen the best response.2I looked into a method of solving this using Lagrange Multipliers. I'm no expert with this multivariable calc, but here's my attempt: Let \(D = d^2 = x^2 + y^2\) be the equation we are trying to optimize, and our constraint is \(x^2 + xy + y^2 = 3\). In general, this Lagrange Multiplier stuff says that: \(\nabla f(x,y) = \lambda \nabla g(x,y) \) We can then break this up into individual partial derivatives in each variable. \( f_x(x,y) = \lambda g_x(x,y) \), and \( f_y(x,y) = \lambda g_y(x,y) \). Pairing this up with our constraint, we have a threevariable system of equations: \( f_x(x,y) = \lambda g_x(x,y) \), \( f_y(x,y) = \lambda g_y(x,y) \), \( g(x,y) = k \) Now, we'll go back to our problem and we'll apply this information. \(f(x,y) = D\), \(g(x,y) = x^2 + xy + y^2\), and \(k = 3\). \(D_x = \lambda g_x\), \(D_y = \lambda g_y\) \(g = k\) \(2x = \lambda \left(2x + y\right)\) \(2y = \lambda \left(2y + x\right) \) \( x^2 + xy + y^2 = 3 \) From what I can see, solving this system gives you the answers. I suppose there are some checks that should be carried out to verify that your answers are correct from there. http://www.wolframalpha.com/input/?i=solve+system+%7B2x%3Dz%282x%2By%29%2C+2y%3Dz%282y%2Bx%29%2C+x%5E2%2Bxy%2By%5E2%3D3%7D
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.